4-Chapter Complex Number

8/9/2019 4-Chapter Complex Number

1/35

omplex Numbers

INTRODUCTION

Ø Consider a simple quadratic equation x2 + 4 = 0. Clearlythere is no solution of this equation in the set of realnumbers. To permit the solution of such equations the setof complex numbers is introduced.

Ø The solutions of the above equation are given by x2 = –4

.21 – 24x ±=±=-±=Þ

Ø Swiss Mathematician Euler introduced the symbol i (iota)

for positive square root of – 1, i.e., i = 1- .

The square root of a negative number is called an imaginarynumber.

Ø Now for any two real numbers x and y, we can form a newnumber x + iy. This number x + iy is called a complexnumber. The set C of complex numbers is therefore defined by C = {x + iy | x Î R, y Î R}The extension of concept of numbers from real numbersto complex numbers enabled us to solve any polynomialequation.

Ø A complex number is deonoted by a single letter such as z,w etc. Given a complex number z = x + iy, x is called itsReal part and y its Imaginary Part and we denote

x = Re(z) and y = Im (z)If y = 0, then z = x is a purely real number.If x = 0, then z = iy is a purely imaginary number.

The complex number 0 = 0 + i0 is both purely real as wellas purely imaginary.

Ø Two complex numbers z1 = x1 + iy1 and z2 = x2 + iy2 aresaid to be equal if and only if x1 = x2 and y1 = y2.

AXIOMATIC APPROACH TOWARDS THE COMPLEX

NUMBER SYSTEM

Ø A complex number is defined as an ordered pair (x, y) of real numbers x and y. Thus,

C = {(x, y) | x Î R, y Î R}

If z = (x, y) then x = Re(z) and y = Im (z), z = (x, y) isequivalent to z = x + iy.

Thus, i is equivalent to (0, 1) and z = (x, –y).

i (greek letter iota) represents positive square root of

–1, so, 1-=i . It is called imaginary unit. We have

1. .etc........,,1,,1

5432

iiiiii ==-=-=Thus for any integer k,i4k = 1, i4k+1 = i, i4k+2 = –1, i4k+3 = – i.

That is if power of i is m, Nm Î , then divide m by 4and find the remainder.

If the remainder is zero, then im = 1If the remainder is one, then im = iIf the remainder is two, then im = –1If the remainder is three, then im = – i

2. The sum of four consecutive powers of i is zero, for example, i12 + i13 + i14 + i15 = 0

3. For any two real numbers a and b, ab ba =´ is

true only if at least one number is non negative or zero.

If both a and b are negative then ab ba ¹´In fact if a > 0 and b > 0 then

1 1a b a b i a i b ab- ´ - = - ´ - = ´ = -

OPERATIONS ON COMPLEX NUMBERS

Suppose that z1 = (x1, y1) and z2 = (x2, y2) be two complexnumbers, that is, z1 = x1 + iy1 and z2 = x2 + iy2(i) Equality : z1 = z2, if x1 = x2 and y1 = y2(ii) Addition : z

1

+ z2

= (x1

+ x2

, y1

+ y2

) or equivalently z1

+z2 = (x1 + x2) + i (y1 + y2)


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2 Phys ics

(iii) Subtraction : z1 – z2 = (x1 – x2, y1 – y2) or equivalently z1 – z2 = (x1 – x2) + i (y1 – y2)

(iv) Multiplication : z1z2 = (x1x2 – y1y2, x1y2 + x2y1) or equivalently z1z2 = (x1x2 – y1y2) + i (x1y2 + x2y1)

(v) Division : If z2 ¹ 0 then

÷÷

ø

ö

çç

è

æ

+

-

+

+=

22

22

122122

22

2121

2

1

yx

xyxy,

yx

yyxx

z

z

or equivalently ÷÷

ø

ö

çç

è

æ

+

-+

÷÷

ø

ö

çç

è

æ

+

+=

22

22

122122

22

2121

2

1

yx

xyxyi

yx

yyxx

z

z

(vi) Multiplication by a real number : If z = (x, y) and m ÎR then mz = (mx, my) or equivalently if z = x + iy then mz = mx + imy

ALGEBRA OF OPERATIONS

If z1, z2 and z3 belong to set C of complex numbers, thenfollowing properties hold.

(1) Closure Property : z1 ± z2; z1z2 and2

1

z

z, z2 ¹ 0 all also

belong to C.

(2) Commutative Property : z1 + z2 = z2 + z1 and z1z2 = z2z1(3) Associative Property : z1 + (z2 + z3) = (z1 + z2) + z3 and

z1 (z2z3) = (z1z2)z3(4) Cancellation Property : z1 + z3 = z2 + z3 Þ z1 = z2 and

z1z3 = z2z3 Þ z3 = 0 or z1 = z2(5) Distributive Property : z

1

(z2

+ z3

) = z1

z2

+ z1

z3(6) Existence of Identity : 0 = (0, 0) is additive identity, i.e. 0

+ z = z + 0 = z " z Î C

1 = (1, 0) is multiplicative identity, i.e. 1(z) = (z)1 = z " z Î C(7) Existence of Inverse : For every complex number z = (x,

y), we may get a unique number–z = (–x, –y) such that z + (–z) = (–z) + z = 0. (–z) is Additive Inverse.

For every complex number z = (x, y), z 0¹ we may get a

unique number z –1 orz

1 = ÷

÷ ø

öççè

æ

++ 2222 yx

y – ,

yx

x such that

1zz

1

z

1z =÷ ø öç

è æ =÷

ø öç

è æ . ÷

ø öç

è æ

z

1 is multiplicative inverse.

[NOTE : A set with two operations on it satisfying all above

properties is called a Field.]

(8) The order relations 'greater than' and 'less than' arenot defined for non real complex numbers. Theinequalities like –2i < 0; 1 + 2i > 1; i – 1< i are meaningless.

The Modulus and the conjugate of a Complex Number

Let z = x + iy be a complex number. Then the modulus(absolute value) of z, denoted by | z | is defined as follows :

| z | = 2 2 2 2{Re( )} {Im( )}a b z z+ = +

Clearly, modulus of a complex number is a real number.Again let z = x + iy be a complex number. Then the complexnumber x – iy is called the conjugate of z and is denoted by

z or z*.

Thus, we have Re ( z ) = Re ( z) and Im ( z ) = – Im ( z).Note :

1. The additive inverse of ‘ z’ is ‘– x – iy’ and conjugateof ‘ z’ is ‘ x – iy’.

2. The multiplicative inverse of a non-zero complexnumber ‘ z’ can be given by

12 2 2

1

| |

x iy z z

x iy x y z

- -= = =+ +

or 2| | z z z=

(i) z)z( = (ii) 2121 zzzz ±=±

(iii) nn2121 )z()z(zzzz =Þ= , n NÎ

(iv) z)z( -=i (v) 0z,zz

z

z2

2

1

2

1 ¹=÷÷ ø

öççè

æ

(vi) )zRe(2zz =+ , which is a purely real number

(vii) )zIm(2zz =- , which is a purely imaginary number..

(viii) zz = if and only if z is purely real.(ix) zz -= if and only if z is purely imaginary..(x) If f(z) is a polynomial in a complex variable z, then

)z(f )z(f = [where f means the complex coefficients

are replaced by their conjugate is]

(xi) )zzRe(2)zzRe(2zzzz 21212121 ==+

(xii) If

321

321

321

ccc

b b b

aaa

z = then

321

321

321

ccc

b b b

aaa

z =

where ai

, bi

, ci

(i = 1, 2, 3) are complex numbers.

ILLUSTR TIVE EX MPLES

1. The value of the sum å+

=

7n4

1k

k i is

(a) 0 (b) 1

(c) –1 (d) i or – i, depending on n is even or odd

Sol. We have, 1017n4

4k

k 327n4

1k

k -=+--=+++= åå +

=

+

=

iiiiiii

[Note that, k = 4 to 4n + 7 contains 4n + 4 terms, a multipleof 4] Answer (c)


3/35

Laws of Motion 3

2. If R y,x,3

y)32(

3

2x)1(Î=

-+-

++

-+i

i

ii

i

ii, then

(a) x = 3, y = –1 (b) x = –3, y = 1(c) x = 3, y = 1 (d) x = –3, y = –1

Sol. We have ii

ii

i

ii =-

+-++

-+3

y)32(3

2x)1(

Multiplying both sides by (3 + i) (3 – i), we get[(1 + i)x – 2i] (3 – i) + [(2 – 3i)y + i] (3 + i) = (3 + i) (3 – i)i

Þ (4x + 9y – 3) + (2x – 7y – 3)i = 10iÞ 4x + 9y – 3 = 0 and 2x – 7y – 13 = 0Solving these equations, we get x = 3, y = –1 Answer (a)

3. If(cos x sin x)(cos y sin y)

A B(cot u )(1 tan v)

+ += +

+ +

i ii

i i, then

(a) A = sinu cosv cos (x + y – u – v)(b) B = sinu cosv sin(x + y – u – v)

(c) A = cosu sinv cos(x + y – u – v)(d) B = cosu sinv sin(x + y – u – v)

Sol . We have,(cos x sin x)(cos y sin y)

(cot u )(1 tan v)

+ +

+ +

i i

i i

÷ ø

öçè

æ +÷

ø

öçè

æ +

++=

vcos

vsin1

usin

ucos

)ysiny)(cosxsinx(cos

ii

ii

)vsinv)(cosusinu(cos

)ysiny)(cosxsinx(cosvcosusin

ii

ii

++

++=

)]vusin()vu[cos(

)]yxsin()yx[cos(vcosusin

+++

+++=

i

i

)]vusin()vu[cos(

)]vusin()vu[cos(

+-+

+-+´

i

i

)vu(sin)vu(cos

)]vuyxsin()vuyx[cos(vcosusin22 +++

--++--+=

i

)vuyxcos(vcosusin --+= )vuyxsin(vcosusin --++i

)vuyxcos(vcosusinA --+=\

and )vuyxsin(vcosusinB --+= Answers (a, b)4. If z1z2z3 are there complex numbers then the value of

z1 Im )zzIm(z)zzIm(z)zz( 2313232 1++ is

(a) Re (z1z2z3) (b) Im (z1z2z3)(c) Re (z1 + z2 + z3) (d) 0

Sol. 1 1 1 2 2 2 3 3 3, , Let z x iy z x iy z x iy= + = + = +

11 2 1 1 2 2 3 3Im( ) ( ) Im ( – )( )Then z z z x iy x iy x iy= + +é ùë û

1 1 2 3 2 3 2 3 3 2( ) Im ( ) ( – ) x iy x x y y i x y x y= + + +é ùë û

1 1 2 3 3 2( )( – ) x iy x y x y= +

1 2 3 3 2 1 2 3 3 2( – ) ( – ) x x y x y iy x y x y= +Similarly, )yx – yx(iy)yx – yx(x)zzIm(z 3113231132232 +=

and )yx – yx(iy)yx – yx(x)zzIm(z 122131221323 1 +=

Clearly 0)zzIm(z)zzIm(z)zzIm(z 231231 132 =++

Answer (d)5. If ( x + iy)1/3 = a + ib, where x, y, a, b Î R, show that

2 22( ) x y

a ba b

- = - +

Sol . ( x + iy)1/3 = a + ibÞ x + iy = (a + ib)3

i.e., x + iy = a3

+ i3

b3

+ 3iab (a + ib)= a3 – ib3 + i3a2b – 3ab2 = a3 – 3ab2 + i (3a2b – b3)Þ x = a3 – 3ab2 and y = 3a2b – b3

Thus x

a = a2 – 3b2 and

y

b = 3a2 – b2

So, x y

a b- = a2 – 3b2 – 3a2 + b2

= – 2a2 – 2b2 = – 2(a2 + b2).

6. Solve the equation z2 = z , where z = x + iy

Sol. z2 = z Þ x2 – y2 + i2 xy = x – iyTherefore, x2 – y2 = x ...... (1)and 2 xy = – y ...... (2)

From (2), we have y = 0 or x = –1

2When y = 0, from (1), we get

x2 – x = 0, i.e., x = 0 or x = 1.

When x = –1

2, from (1), we get

y2 =1 1

4 2+ or y2 =

3

4, i.e., y =

3

2± .

Hence, the solutions of the given equation are

0 + i0, 1 + i0, –1

2 + i

3

2, –

1

2 – i

3

2

4.1

Solve following problems with the help of above text and

examples.

1. The value of 1iiiii

iiiii574576578580582

584586588590592

-++++

++++ is

(a) 2 (b) –2 (c) 1 (d) –12. 1 + i2 + i4 ... + i2n is

(a) Positive (b) Negative(c) 0 (d) Can’t be determined

3. The value of å=

++13

1n

1nn )ii( , where i = 1- equals

(a) i (b) i – 1 (c) – i (d) 0

4. The least positive integer n such that

n

i1

i2÷ ø

öçè

æ +

is a positive

integer is(a) 2 (b) 4 (c) 8 (d) 16


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4 Phys ics

5. The multiplicative inverse of 2)i56( + is

(a) i61

60

61

11- (b) i

61

60

61

11+

(c) i61

60

61

9- (d) None of these

6. The multiplicative inverse ofi54

i43

-+

is

(a) i25

31

25

8+- (b) i

25

31

25

8-

(c) i25

31

25

8-- (d) None of these

7. The conjugate complex number of 2)i21(

i2

-

- is

(a) i25

11

25

2

÷ ø

ö

çè

æ

+÷ ø

ö

çè

æ (b) i25

11

25

2

÷ ø

ö

çè

æ

-÷ ø

ö

çè

æ

(c) i25

11

25

2÷

ø

öçè

æ +÷

ø

öçè

æ - (d) i

25

11

25

2÷

ø

öçè

æ -÷

ø

öçè

æ -

8. The value of ÷ ø

öçè

æ -

+÷ ø

öçè

æ +

+- i42

i43

i1

3

i21

1 is

(a) i

4

9

4

1- (b) i

4

9

4

1+

(c) i4

9

4

1+- (d) i

4

9

4

1--

9. The value ofq+q- sini2cos1

1 is

(a) q+

q

-q+ cos35

2tan2

icos35

1(b)

q+

q

+q+ cos35

2tan2

icos35

1

(c)q+

q

-q+ cos35

2cot2

icos35

1(d) None of these

10. If ,iyxiba3 -=- then =+3 iba

(a) iyx + (b) iyx - (c) ixy + (d) ixy -

GRAPHICAL REPRESENTATION OF COMPLEX NUMBERS

N(0, y)

P (x,y)

O M (x,o)

y

Q (x,–y)

x

Ø Every complex number x+iy can be represented geometrically as a unique point P (x,y) in the xoy planewith x-coordinate representing its real part and y-coordinaterepresenting its imaginary part.

Ø The Point (x, 0) on the x-axis represents the purely realnumber x. As such x-axis is called the real axis. Similarly,the point (0, y) on the y-axis represents purely imaginarynumber iy. Therefore, y-axis is called the imaginary axis.

Ø The plane having a complex number assigned to each of its points, is called the complex plane or Argand planeor Guassian plane. This representation of complexnumbers as points in the plane is known as Arganddiagram.

Ø The distance from the origin to the point P(x, y) is defined as the MODULUS (or absolute value) of the complex

number z = x + iy, denoted by | z |, thus | z | = 22 yx +

Ø The conjugate z of complex number z is represented bythe point Q, which is the mirror image of P on the x–axis.

1. 0|z| ³ , 0|z| = 0z =Û .

2. 2|z|zz = ( majority of the complex equations are solved using this property)

3. |z||z||z||z| -=-==

4. nn2121 |z||z||z||z||zz| =Þ=

5. |z|

|z|

z

z

2

1

2

1 = , 0z2 ¹

6.|z|

z1

|z|

zÞ= , is a unimodular complex number

( 0z ¹ ).

7. )zzRe(2|z||z||zz| 212

22

12

21 ++=+

8. )zzRe(2|z||z||zz| 212

22

12

21 -+=-

9. )|z||z(|2|zz||zz| 222

12

212

21 +=-++

10.

2 2 2 2 2 21 2 1 2 1 2| az bz | | bz az | (a b )(| z | | z | ),+ + - = + +

where a, b Î R

11. If 0z,z 21 ¹ then2

22

12

21 |z||z||zz| +=+ 2

1

z

zÛ is

purely imaginary.

ANSWER KEY

1. (b) 2. (d) 3. (b) 4. (c) 5. (a) 6. (c) 7. (d) 8. (b) 9. (c) 10. (a)


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Laws of Motion 5

POLAR FORM (OR TRIGONOMETRICAL FORM) OF

COMPLEX NUMBERS

P (z)

O M

y

XX

q

r = | z |

y

Let P represents the nonzerocomplex number z = x+iy. Let the

directed line regment OP be of length r and makes an angle q withthe positive direction of the x–axis(q in radians)

The point P is uniquely determined by the ordered pair of realnumbers (r, q) called the polar coordinates of the point P. Clearly,

x = r cos q, y = r sin q,

r= 22 yx + , tanq =x

y

Thus z=r(cosq + isin q) is the poar from of z. r is the modulus of

the number z and q is called the ARGUMENT (or AMPLITUDE) of the number z, denoted by arg (z) or amp (z)Hence.

22 yx|z|r +== andx

ytan)z(arg 1-==q

z = r(cosq + i sinq) is also written as r cis(q)Ø Note that q is not defined uniquely, In fact q is the solutionof simultaneous equations

22 yx

xcos

+=q and

22 yx

ysin

+=q

If z = 1 + i,211yxr

2222 =+=+= ;

41

x

ytan

p=qÞ==q

Polar from of ÷ ø

öçè

æ p+

p=

4sini

4cos2z

Clearly the possible arguments of the number z=1+i are thefollowing angles :

IÎp+p

p+pp

k ,k 24

..,,.........24

,4

Any two arguments of a complex number differ by a number

which is a multiple of 2 .

The unique value of q, such that p£q 0, y > 0, then the point P lies in the first quadrant

and then a==q zarg

For example, if i1z += , then arg(z) =4

p

Case II : If x < 0, y > 0, then the point P lis in the second

quadrant and then -p==q zarg

For example, if i31z +-= , then 31

3tan =

-=a

so3

2

3)zarg(

p=

p-p=

y¢

q

P

xO

x¢

y

a

Case III : If x < 0, y < 0 then the point P lies in the third quadrant and then

p-a=a-p-==q )(zarg

y¢

q

P

xOx¢

y

aFor example, if i1z --=

then 111tan =

--=a

so4

3)

4()zarg(

p-=

p-p-=

Case IV : If x > 0, y < 0 then the point P lies in the fourth

quadrant and then a-==q zarg

For example, if ,i3z -= then

3

1

3

1tan =

-=a

y¢

q

P

xOx¢

y

a

so, 6)zarg( p-=

Case V : If y = 0, then z is purely real and P lies on real axis, and z = x,so arg (z) = 0 if x > 0; arg (z) = p if x < 0

For example arg(3) = 0 and p=÷ ø

öçè

æ -

2

1arg

Case VI : If x = 0, then z = iy is a purely imaginary number and P lies on imaginary axis

So, arg (z) =2

p if y > 0 and arg (z) =

2

p- if y < 0

For example,2

)i2arg( p= and2

)i100arg( p-=-


6/35

6 Phys ics

1. )zarg()zarg( -=

2. p++= k 2)zarg()zarg()zzarg( 2121

3. p+-=÷÷ ø ö

ççè æ k 2)zarg()zarg(

z

zarg 212

1

4.z

arg 2arg(z) 2k z

æ ö= + pç ÷

è ø

5. p+= k 2)zarg(n)zarg( n

Where k = 0, –1 or 1 would be taken so that thevalue comes out into the principal value region

6. If q=÷÷ ø

öççè

æ

1

2

z

zarg the q-p=÷÷

ø

öççè

æ k 2

z

zarg

2

1 , Ik Î

7. p±=-- )zarg()zarg(

8. )zarg(2

)izarg( +p

=

9. 2/)zarg()zarg(|zz||zz| 212121 p=-Û-=+

10. )zarg()zarg(|z||z||zz| 212121 =Û+=+

EULER'S NOTATION

It can be shown that q-q=q+q= q-q sinicose,sinicose ii

\ )ysiniy(cosee.eee xiyxiyxz +=== +

Also q=q+q ire)sini(cosr

Again,i2

eesinand

2

eecos

iiii q-qq-q -=q

+=q

LOGARITHM OF A COMPLEX NUMBER

Let q=q+q=+= ire)sini(cosr iyxz

Then q+== q ir log)re(log)z(log ei

ee

So, )zarg(i|z|log)z(log ee +=

As such the argument of a complex number is not unique, thelog of a complex number cannot be unique. In general,

)],zarg(k 2[i|z|log)z(log ee +p+= Ik Î

For example, Ik ,2

k 2ielog)ilog( 2/i Î÷ ø

öçè

æ p+p== p So,

2i p

is

one of the values of log (i)

Also, ÷ ø

öçè

æ p+

p=÷

ø

öçè

æ p+=

p=

2log

2i

2logilog)

2ilog()ilog(log

[Taking principal value only]

7. If z1 and z2 are any two complex numbers then

|zzz||zzz| 22211

22

211 --+-+ is equal to

(a) |z| 1 (b) |z| 2

(c) |zz|21

+ (d) |zz||zz|2121

-++Sol. (Trick ) The nature of the problem suggests at once that

we shold use the formula

)|z||z(|2|zz||zz| 222

12

212

21 +=-++

We have 222211

22

211 )zzz||zzz(| --+-+

|zzz|2|zzz||zzz| 2221

21

222

211

222

211 +-+--+-+=

22

222

21

21 |z|2|zz||z|2 +úû

ùêëé -+=

= |zz|2)|z||z(|222

21

22

21 -++

= |zz||zz|2|zz||zz| 21212

212

21 -++-++= 22121 |)zz||zz(| -++

\ |zzz||zzz| 22211

22

211 --+-+

|zz||zz| 2121 -++= Answer (d)

8. The smallest positive integer n for which 1i – 1

i1n

=÷ ø

öçè

æ + is

(a) 4 (b) 3 (c) 2 (d) 1

Sol. i2

i2

i – 1

i2i1

i1

i1

i – 1

i1

i – 1

i12

2

==++

=++

´+

=+

1i1i – 1

i1 nn

=Þ=÷ ø

öçè

æ +\

Clearly the smallest value of n is 4. Answer (a)

9. If .1iba =+ the simplified form ofai b1

ai b1

-+++

is

(a) b + ai (b) a + bi(c) (1 + b)2 + a2 (d) ai

Sol. As 1 ba1iba 22 =+Þ=+

=)ai b1)(ai – b1(

)ai b1(

ai – b1

ai ba 2

+++++

=+

++ 22

22

a) b1(

ai) b1(2a – ) b1(

++

+++=

= b2) ba(1

abi2ai2 b2 b)a – 1(22

22

+++

++++

b211

abi2ai2 b2 b b 22

++

++++=

= ai b b1

i) b1(a) b1( b b1

abiai b b2 +=+

+++=+

+++

Answer (a)

10. The argument of the complex number a-a+ cosisin1 is

(a)42

p-

a(b)

2

a(c)

42

p+

a(d)

22

p+

a

Sol. Let 1+ sina – icosa = r(cosq + isinq)Then r cos q = 1 + sina and r sinq = – cosa

r 2 = (1 + sina)2 + (–cosa)2 = 2 + 2 sina =2

2sin

2cos2 ÷

ø

öçè

æ a+

a

r =÷

ø

öçè

æ a+

a

2sin

2cos2

or ÷ ø

ö

çè

æ a-

p

24cos2


7/35

Laws of Motion 7

Also, tan q =÷ ø

öçè

æ a-p

+

÷ ø

öçè

æ a-p

-=

a+

a-

2cos1

2sin

sin1

cos [Note the step]

=÷

ø

öçè

æ a-p

÷ ø öç

è æ a-p÷

ø öç

è æ a-p-

24cos2

24cos

24sin2

2 ÷

ø

öçè

æ p-

a=÷

ø

öçè

æ a-

p-=

42tan

24tan

\ q =42

p-

a

Hence, Modulus = ÷ ø

öçè

æ a-

p

24cos2 and argument

=42

p-

a.

Answer (a)

11. Convert the complex number16

1 3i

-

+ into polar form.

Sol. The given complex number16

1 3i

-

+ =

16

1 3i

-

+ ×

1 3

1 3

i

i

-

-

=2

16(1 3)

1 ( 3)

i

i

- -

- =

16(1 3)

1 3

i- -+

= – 4 (1 3)i-

= – 4 + i4 3

q

O

Y

X

Y'

X'

P( 4,4 3)-

Let – 4 = r cos q, 4 3 = r sin qBy squaring and adding, we get

16 + 48 = r 2 (cos2 q + sin2 q)which gives r 2 = 64, i.e., r = 8

Hence, cos q = –

1

2 , sin q =

3

2

q = p –3

p

=2

3

p

Thus, the required polar form is 82 2

cos sin3 3

ip pæ ö

+ç ÷è ø

12. If | z2 – 1| = | z |2 + 1, then show that z lies on imaginaryaxis.

Sol. Let z = x + iy. Then | z2 – 1| = | z |2 + 1Þ | x2 – y2 – 1 + i2 xy | = | x + iy |2 + 1Þ ( x2 – y2 – 1)2 + 4 x2 y2 = ( x2 + y2 + 1)2

Þ 4 x2

= 0 i.e., x = 0Hence z lies on y-axis.

4.2


examples.

1. For any two complex number z1, z

2

221

221 |zz||zz1| --- is equal to :

(a) )|z|1()|z|1(2

22

1 ++ (b) )|z|1()|z|1(2

22

1 --

(c) )|z|1()|z|1( 222

1 -+ (d) )|z|1()|z|1(2

22

1 +-

2. If z1, z

2 and z

3, z

4 are two pairs of conjugate complex

numbers, then arg ÷÷ ø

öççè

æ +

÷÷

ø

ö

çç

è

æ

3

2

4 z

zarg

z

z1 equals

(a) 0 (b)2

p

(c)

2

3p(d) p

3. If |b| = 1, then ba-a-b

1 is equal to

(a) 0 (b) ½ (c) 1 (d) 2

4. Let z1 and z

2 be complex numbers such that 21 zz ¹ and

|z1| = |z

2|. If z

1 has positive real part and z

2 has negative

imaginary part, then21

21

zz

zz

-

+ may be

(a) real and positive(b) zero(c) real and negative(d) either zero or purely imaginary

5. If1

2

z7

z5 is purely imaginary number then

21

21

z3z2

z3z2

-

+ is

equal to

(a) 7

5

(b) 9

7

(c) 49

25

(d) None


8/35

8 Phys ics

6. If z1 and z

2 be two non-zero complex numbers such that

|zz||zz| 2121 -=+ then arg (z1) – arg (z2)=

(a)

2

p(b) 0

(c)2

p- (d)

4

p

7. The Argument of the complex number z =)3i – 1(i4

)3i1( 2+ is

(a)6

p(b)

4

p

(c)2

p(d) None of these

8. The modulus of the complex number

z =)sini – (cos)i – 1(2)sini(cos)3i – 1(

qq q+q is

(a)2

1(b)

22

1

(c)

3

1(d) None of these

9. If (a + ib) (c + id) (e + if) (g + ih) = A + iB, then

( 2a + 2 b ) (2c + 2d ) ( 2e + 2f ) (

2g + 2h ) =

(a) 2A – 2B (b) 2A + 2B

(c) 44 BA + (d) 44 B – A

10. If q is real, then the modulus ofq+q+ sini)cos1(

1 is

(a)2

1sec

2

q(b)

2

1cos

2

q

(c) sec2

q(d) cos

2

q

DE MOIVRE THEOREM

These are two statements of De Moivre Theorem

1. (cosq+i sinq)n = cos nq + i sin nq, IÎn

2. If QÎn ÷÷ ø

öççè

æ Î¹= Iq , p,0q ,

q

pn then cos nq+i sinnq

is one of the values of (cosq+ i sinq)n

1. Take care that n(sin icos ) (sin n icos n )q+ q ¹ q + q in fact

n)cosi(sin q+q =

n

2sini

2cos ú

û

ùêë

é÷ ø

öçè

æ q-

p+÷

ø

öçè

æ q-

p

= ÷ ø

öçè

æ q-

p+÷

ø

öçè

æ q-

pn

2

nsinin

2

ncos

2. (cos i sin )(cos isin )(cos i sin )....a + a b + b g + g

= cos( .....) i sin( .....)a + b + g + + a + b + g +

3. )sin(i)cos(sinicos

sinicosb-a+b-a=

b+b

a+a

ROOTS OF A COMPLEX NUMBER

If )sini(cosr z q+q= and n is a positive integer, then

n

1

n

1

)]sini(cosr [)z( q+q=

[ ] n1

n )k 2sin(i)k 2cos(r q+p+q+p=1

k = 0, 1, 2, ......, n – 1

ANSWER KEY

1. (b) 2. (a) 3. (c) 4. (d) 5. (d) 6. (a) 7. (c) 8. (a) 9. (b) 10. (a)

úû

ùêë

é÷ ø

öçè

æ q+p+÷

ø

öçè

æ q+p=\

n

k 2sini

n

k 2cosr )z( n

1

n

1

Putting k = 0, 1, 2, ....., n – 1, we get n values which representnth roots of complx number z

PROPERTIES

1. These n roots always form a G. P. with common ratio ei 2p/

n

2. Their images on argand diagram lie on a circle of radiusn/1r and centre origin.

3. The points representing these roots from the vertices of aregular polygon of n sides

y¢

P1x

O

y

P2P3

Pn

q

2q

CUBE ROOTS OF UNITY

3

k 2sini

3

k 2cos]k 2sinik 2[cos)1(z 3/13/1

p+

p=p+p== , k = 0, 1, 2

So, three cube roots of unity are 1,3

2sini

3

2cos

p+

p and

3

4sini

3

4cos

p+

p or 1,

2

3i1+- and

2

3i1--.

As ÷÷ ø

ö

ççè

æ --=÷÷ ø

ö

ççè

æ +-2

3i1

2

3i12

and

21 i 3 1 i 3

2 2

æ ö æ ö- - - +=ç ÷ ç ÷ç ÷ ç ÷è ø è ø


9/35

Laws of Motion 9

So, we denote the non-real roots by w and w2 we write mostly

2

3i1+-=w and

2

3i12 --=w

ALGEBRAIC METHOD

Let z = (1)1/3 01z3 =-Þ 0)1zz)(1z(2 =++-Þ , which gives

the roots of z as z = 1, the real root and2

3i1z

±-= ,the non-

real roots , i.e., w and w2. Clearly, we can always write

)z)(z)(1z()1z( 23 w-w--=-

1. 01 2 =w+w+

2. 2w=w and w=w )( 2 , so 13 =w=ww

3. 1|||| 2 =w=w

4.3

2)arg(

p=w and

2 4arg( )3

pw = or

3

2p-

5. For nay positive integer k, ,1k 3 =w w=w +1k 3 and

22k 3 w=w +

6. For any real a, b, c; 0c ba 2 =w+w+ c ba ==Þ7. The cube roots of unity lie on a unit circle and divide the

circumference into three equal parts8. The points represented by cube roots of unity form the

vertices of an equilateral triangle.9. |z||z||||z| =w=w

10. )zarg(3

2)zarg( +

p=w

–1 1

w

w2

i

–i

xx¢

y

y¢

(i) )iyx)(iyx(yx 22 -+=+

(ii) )yx)(yx)(yx(yx 233 w+w++=+

)yx)(yx)(yx(22

w+ww+w+=(iii) )y – x)(y – x)(yx(yx 233 ww-=-

(iv) x2 + xy + y2 = (x – yw) (x – yw2), in particular,x2 + x + 1 = (x – w) (x – w2)

(v) x2 – xy + y2 = (x + yw) (x + yw2), in particular,x2 – x + 1 = (x + w) (x + w2)

(vi) x2 + y2 + z2 – xy – xz – yz = (x + yw + zw2)(x + yw2 + zw)

(vii) xyz3 – zyx 333 ++

)zx – yz – xy – zyx)(zyx( 222 ++++=

= )zyx)(zyx)(zyx( 22 w+w+w+w+++

nTH ROOTS OF UNITY

Since 0sini0cos1 += , therefore

n/1n/1 )0sini0(cos)1( += 1/ n(cos 2k isin 2k )= p + p

2 k 2 k cos isinn np p= + ; k = 0, 1, 2, ........, (n – 1)

2k

ine

p

= ; k = 0, 1, 2, ...., (n – 1)

]n/)1n(2i[)n/4i()n/2i( e,.....,e,e,1 p-pp=

)n/2i(1n32 ewhere,....,,,,1 p- =aaaaa=

SQUARE ROOTS OF A COMPLEX NUMBER

Let z = x + iy and let the square root of z be the complex number a + ib. Then

ibaiyx +=+ or i)ab2() ba()iba()iyx( 222 +-=+=+Equating real and imaginary part, we get

22 bax -= …(1)

and y = 2ab …(2)

Now, 222222222 yx ba4) ba( ba +=+-=+ …(3)

Solving the equations (1) and (3), we get

÷÷÷

ø

ö

ççç

è

æ ++±=

2

xyxa

22

;÷÷÷

ø

ö

ççç

è

æ -+±=

2

xyx b

22

From (2), we can determine the sign of ab. If ab > 0, then a and b will have same sign. Thus

úúú

û

ù

êêê

ë

é

÷÷÷

ø

ö

ççç

è

æ -++

÷÷÷

ø

ö

ççç

è

æ ++±=+

2

xyxi

2

xyxiyx

2222

If ab < 0, then

úúú

û

ù

êêê

ë

é

÷÷÷

ø

ö

ççç

è

æ -+-

÷÷÷

ø

ö

ççç

è

æ ++±=+

2

xyxi

2

xyxiyx

2222

Thus, square roots of z = a + ib are :

úúû

ù

êêë

é -+

+±

2

a|z|i

2

a|z|for b > 0

and úúû

ù

êêë

é --

+±

2

a|z|i

2

a|z| for b < 0

For example :

(i) Square root of i is ÷÷ ø

öççè

æ +±

2

i1, as x = 0, y = 1 > 0 and

| i | = 1


10/35

10 Phys ics

(ii) Square root of -i is ÷÷ ø

öççè

æ -±

2

i1, as x = 0, y = -1 < 0 and | i | =

1

(iii) Square root of -3 + 4i is÷÷ ø

öççè

æ --+-±2

)3(5i

2

35 ,

or )i21( +± , As x = –3 and y = 4 > 0.

(iv) Square root of i31- is ÷÷ ø

öççè

æ --

+±

2

12i

2

12 or

÷÷ ø

öççè

æ -±

2

i3; As 03y


11/35

Laws of Motion 11

17. If sin 5x = å=

5

0r

r r xsina , then

(a) å= =5

0r r 1a (b) a1 = a3 = a5

(c) a0 = a2 = a4 (d) a1+a3+a5=a0+a2+a4Sol. Solve as above, expand (cos x + i sinx)5 using two theorems

and equate imaginary parts. Answer (a, c)

18. If sin6x = å=

6

0r r rxcosa . Then

(a) a0 = 0 (b) a1 = a3 = a5

(c) a2 + a6 = 2

1

(d) 2a0 + a1 + 3a2 + a3 + a4 + a5 + a6 = 0

Sol. To express cosn

x or sinn

x as a series of multiple anglesof cos and sin, we use Euler's representation cosx =

2

ee ixix -+ and sinx =

i2

ee ixix -- and expand cosnx =

nixix

2

ee÷÷

ø

ö

çç

è

æ + - by binomial theorem etc. as for the above

example.

sin6x =

6ixix

i2

ee÷÷

ø

ö

çç

è

æ - -

[

]ix666ix5ix56ix4ix2

46ix3ix3

36

ix2ix42

6ixix51

6ix60

666

eCeeC

eeCeeC

eeCe.eCeCi2

1

--

--

--

+-

+-

+-=

= [

]36ix2ix226

ix4ix41

6ix6ix60

66

C)ee(C

)ee(C)ee(C2

1

-++

+-+-

-

--

]CC[ r nn

r n

-=Q

= [ ]20x2cos215x4cos26x6cos264

1-´+´--

= 8

5

x2cos32

15

x4cos16

3

x6cos32

1

+-+- = å=

6

0r r rxcosa

We get a0 = 8

5, a1 = 0, a2 = 32

15- , a3 = 0, a4 = 16

3, a5 = 0

and a6 = 32

1- .

Only (d) is satisfied with these values. Answer (d)

19. If a, b, g are the cube roots of a negative number p then

for any three real numbers x, y, z, the value of a+g+b

g+b+a

zyx

zyx

is

(a)2

i31--(b) (x + y + z)i

(c) ip (d) pi2

y

2

x÷

ø

öçè

æ +

Sol. Let q = –p, q > 0

Cube root of p = p1/3 = (–q)1/3 = –q 1/3. (1)1/3

Cube roots of p are – q 1/3, –q 1/3w, –q 1/3w2

Let a = –q 1/3, b = –q 1/3w, g = –q 1/3w2

\ )q (z)wq (y)wq (x

)wq (z)wq (y)q (x

zyx

zyx3/123/13/1

23/13/13/1

-+-+-

-+-+-=

a+g+bg+b+a

=2

i31w

zywxw

)zywxw(w

zywxw

zwywx 22

22

2

2 --==

++

++=

++

++

1 3ior 2

æ ö- +ç ÷è ø

Answer (a)

4.3

Solve following problems with the help of above text and examples.

1. If z =50

17

)i – 1(

)i3( +, then amp (z) is equal to

(a)3

2p(b) –

3

p(c)

3

2p- (d) None

2. Arg2

6

)3i – 1(4

)i3(i + is equal to

(a) –6

p(b)

6

p

(c)3

10

p(d)

10

5p


12/35

12 Phys ics

3. For z = cos q + i sin q, then the value of n2

n2

z1

z1

+

- is

(a) –i tan n q (b) i tan n q (c) tan n q (d) i4. If x = a + b, y = aw + bw2 and z = aw2 + bw, then x3+y3+z3

=

(a) 3 (a3 + b3) (b) (a + b)3

(c) a3 + b3 – a2 b – ab2 (d) None of these

5. If 1, w, w2 are the cube roots of unity then(1 – w + w2)5 + ( 1 + w – w2)5 =(a) 32 (b) 0

(c) 32w (d) 16w2

6. If 1, w, w2 are the cube roots of unity then(3 + 3w + 5w2)6 – (2 + 6w + 2w2)3 =(a) 0 (b) 64 (c) 36 (d) – 36

7. RÎx , then square roots of 1xxix 24 +++ are

(a) ÷ ø ö

çè æ +-+++± 1xxi1xx

2

1 22

(b) ÷ ø ö

çè æ +--++± 1xxi1xx

2

1 22

(c) ÷ ø ö

çè æ ++++-± 1xxi1xx

2

1 22

(d) ÷ ø

ö

çè

æ ++-+-± 1xxi1xx2

1 22

8. If ¹w 1 is a cube root of unity, then the roots of theequation (x+aw)3 + a3 = 0 are :(a) a, aw2, – 2aw (b) a, –2aw2, aw(c) –2a, aw2, aw (d) a, – 2aw2, –aw

9. Let ,zz,zz 22

121 w+w=b+=a 212

zz w+w=g wherew is a complex cube root of unity, then :

(a) 21222 zz8=g+b+a

(b) 21zz3=ga+bg+ab

(c) )zz(3 323

1333 +=g+b+a

(d) )zz(2 323

1 +=abg

10. The polynomial 2k 31n3m3 xxx ++ ++ , is exactly divisible

by 1xx 2 ++ if (a) m n, k are rational(b) m, n, k are integers(c) m, n, k are positive integers(d) none of these.

11. If ,1i -= then334 365

1 i 3 1 i 34 5 3

2 2 2 2

æ ö æ ö+ - + + - +ç ÷ ç ÷ç ÷ ç ÷

è ø è ø is

equal to

(a) 3i1- (b) 3i1+-

(c) 3i (d) 3i-

12. If )1(¹w be a cube root of unity and ,BA)1( 7 w+=w+

then A and B are respectively the numbers(a) 0, 1 (b) 1, 1 (c) 1, 0 (d) –1, 1

GEOMETRY OF A COMPLEX NUMBER

O

Y

X

P(z)

As in vectors, we represent a point by the position vector of the

poi nt r OP = with respect toorgin O. Similarly the point P can be rep resen ted by a comp lexnumber z, such that length

OP= z

and )zarg(XOP =Ð . The point P is called the IMAGE of thecomplex number z and z is said to be AFFIX or complexcoordinate of the point P.

DISTANCE BETWEEN TWO POINTS

If two points P and Q have affices z1and z2 respectively then

12 z – zPQ = == 12 z – zPQ Affix of Q – Affix of P..

SECTION FORMULA

n

m

QR

PR =

P (z )1

Q (z )2

R (z)

If a point R (z) divides the join of two points P (z1) and Q (z2) in ratio m:n, then

nm

nzmzz 12

+

+= (internally) and

nm

nzmzz 12

-

-= (externally)

ØMid Point of PQ is given by 2

zz 21 +

ANSWER KEY

1. (c) 2. (b) 3. (a) 4. (a) 5. (a) 6. (a) 7. (a) 8. (a) 9. (c) 10. (b) 11. (c) 12. (b)


13/35

Laws of Motion 13

ANGLE BETWEEN TWO LINES(CONCEPT OF ROTATION)

)z(R3

)z(P 1

)z(Q 2

1q

q

2q

(i) Consider three points P(z1), Q(z2) and R(z3)Then angle between PQ an PR (counter clockwise)

12 q-q=q )PQarg( – )PR arg(=

)z – zarg( – )z – zarg( 1213=

12

13

z – z

z – zarg=q\

If P(z1), Q(z2) and R(z3) be collinear Points, then q = 0 or p,

i.e.12

13

z – z

z – z= 0

1zz

1zz

1zz

z – z

z – z

33

22

11

12

13 =Þ

(A complex number z is purely real if zz = )

If PR is perpendcular to PQ.Then

arg12

13

12

13

z – z

z – z

2z – z

z – zÞ

p±=÷÷

ø

öççè

æ is puerly imaginary

That is,12

13

12

13

z – z

z – z

z – z

z – z-=

(A Complex number z is purely imaginary if zz -= )

(ii))z(R 3 )z(P 1

)z(Q 2 )z(S 4

q

Consider Four Points P, Q, R and S with affices respectively

321 z,z,z and z4. As in (i) the angle between SR and QP..

÷÷ ø

öççè

æ =q

21

43

z – z

z – zarg

If SR and QP be perpendicular, 2

p±=q

Then21

43

z – z

z – z is purely imaginary..

21

43

21

43

z – z

z – z –

z – z

z – z=Þ

or alternatively {0}k somefor ik,z – z

z – z

21

43 -Î= R

)z – z(ik z – z 2143 =Þ(iii) Multiplying a complex number z by i is equivalent to

rotating the image of z in Argand plane by 90º about origin,

anticlockwise, as |z| = |iz| and arg2

)i(argz

iz p==÷

ø

öçè

æ

O

P (z)

Q (iz)

p / 2

(iv) Multiplying a complex number z by w is equivalent torotating the image of z in Argand plane by 120º (or 240°)about origin anticlockwise, for |z| = |wz| (|w| = 1) and

argwz 2 4

arg (w) or

z 3 3

p pæ ö æ ö= = ç ÷ç ÷

è øè ø

O

P (z)Q (wz)

3

2p

ANGLES OF A TRIANGLE

)z(C 3

)z( A 1

)z(B 2

If z1 z2 and z3 be the affices of vertices A, B and C of atringle ABC described in counterclockwire sense. Then

)AsiniA(cosBA

CA

z – z

z – z

12

13 += or iA

12

13 eBA

CA

z – z

z – z=

Similarly relations with other vertical angles can be given.


14/35

14 Phys ics

If z1, z2 , z3 are the vertices of a triangle then

1. centroid z is givn by3

zzzz 321

++=

2. Incentre I (z) of the DABC is given byc ba

cz bzazz 321

++++

=

3. Circumcentre O(z) of theDABC is given by

C2sinB2sinA2sin

)C2(sinz)B2(sinz)A2(sinz

1zz

1zz

1zz

1z|z|

1z|z|

1z|z|

z 321

32

3

22

2

12

1

32

3

22

2

12

1

++

++==

4. Orthocentre H(z) of DABC is given by

1zz

1zz

1zz

1z|z|

1z|z|

1z|z|

1zz

1zz

1zz

z

32

3

22

2

12

1

32

3

22

2

12

1

32

3

22

2

12

1

+

= OR

1 2 3

1 2 3

(tan A)z (tan B)z (tan C)zz

tan A tan B tan C(a sec A)z (b sec B)z (c sec C)z

a sec A bsec B csec C

+ +=

+ ++ +=

+ +

5. The centroid G lies on the segment joining the orthocentreH and the circumcentre O of the triangle and divides

internally in ratio 2 : 1, i.e1

2

OG

HG=

H

OG

6. Ar ea of t he DABC is given by the modulus of

1 1

2 2

3 3

z z 11z z 1

4 z z 1

7. Triangle ABC is equilateral if and only if

0zz

1

zz

1

zz

1

211332

=-

+-

+-

21133223

22

21 zzzzzzzzz ++=++Û

0

zz1

zz1

zz1

21

13

32

=Û

EQUATION OF STRAIGHT LINE THROUGH TWO

POINTS Z1 AND Z2Let variable point z be a point on this line then

01zz1zz

1zz

22

11 = or 0zzzz)zz(z)zz(z 21211221 =-+-+-

0)zzzz(ii)zz(zi)zz(z 21211221 =-+-+-Þ

(i) Let ,ai)zz( 12 =- a constant complex number then

ai)zz( 12 =-- Also )zzzz(i 2121 - is purely real constant

number, say b then the above equation is 0 bzaza =++It is called the general equation of a straight line

(ii) The complex equation 0 bazaz =++ represents a straightline in complex plane where 'a' is a complex number and 'b' is a real number. The complex slope of the line is given

bya

a- .

(iii) The equation of the perpendicular bisector of the linesegment joining the points A(z1) and B(z2) is

22

212121 |z||z|)zz(z)zz(z -=-+-

1. The complex slope of line joing points A (z1) and B(z2)

is define as 21

21

zz

zz

-

-=m

2. Two lines with complex slopes 1m and 2m are parallel

if 21 m=m and perpendicular if 021 =m+m

3. The length of perpendicular from a point )(A a to the

line 0 bzaza =++ is given by|a|2

| baa| p

+a+a=

EQUATION OF A CIRCLE :

P (z)

r

)z(C 0

(1) Consider a circle with centre C having affix z0 and radiusr.For any point P(z) on this circle CP = r

i.e. r z – z 0 = .....(1)

where, r is a positive real number


15/35

Laws of Motion 15

14

tan76y12x14yx

36y622

=p

=+--+

-Þ

Þ x2 + y2 – 14x – 18y + 112 = 0 ..... (i)which is a circle with centre (7, 9)

and radius = 18112814911297 22 =-+=-+ Answer (b)

Note : The equation (i) may be converted to complex form asfollowing :x2 –14 + y2 – 18y + 112 = 0

Þ (x2 – 14x + 49) + (y2 – 18y + 81) + 112 – 49 – 81 = 0Þ (x – 7)2 + (y – 9)2 = 18 Þ |(x – 7) + i (y – 9)|2 = 18Þ |(x + iy) – (7 + 9i)|2 = 18

Þ |z – (7 + 9i)|2 = 18 Þ |z – (7 + 9i)| = 18 , which is equivalent to |z – z0| = r

Hence centre of the circle is z0 = 7 + 9i = (7, 9) radius of

the circle is r = 18

21. Consider the complex number z satisfying |z – 5i| £ 3, then(a) Value of z having the least modulus is z =2i(b) Value of z having the greatest modulus is z = 8i(c) Value of z having the least positive argument is

z = )i43(5

4+

(d) Value of z having the greatest positive argument is

z = )i43(5

4+-

Sol. The inequation | z – 5i | £ 3 represents all the complexnumbers lying inside or on the circle| z – 5i | = 3 ...(1)Clearly the circle (1) has centre at (0, 5) and radius = 3 Letus plot this circle on the xy – plane.The given inequation represents the points inside or on thecircle shown in figure.

O

BD

C

E

y

x

A

(5i)

ab

f

Note that the modulus of a complex number is represented by the distance of the image of the complex number fromorigin. Clearly the point at the least distance from origin isD and the point at the greatest distance from origin is E.

Hence the affices of D and E give us the complex numbersof the least and the greatest moduli respectively.

Ø r z – z 0 < represents the region inside the circle given by

(1)

Ø r z – z 0 > represents the region outside the circle given

by (1).(2) General equation of a circle

Consider the equation of a circle r |zz| 0 =-

0r |z|zzzzzzr )zz)(zz( 220002

00 =-+--Þ=--Þ

Let ,az0 =- a constant complex number and

, br |a|r |z| 22220 =-=- a constant real number then the

above equation becomes 0 bzazazz =+++ ,It is called the general equation of circle .

Hence the complex equation 0 bzazazz =+++reprectents a circle in complex plane where 'a' is a complex

number and 'b' is a real number. The centre of the circle is

a point with affix '–a' and the radius is given by b – a2

.

For the existence of the circle .0 b – a2

>

(3) 2 11 2(z – z )(z – z ) (z – z )(z – z ) 0+ = represents a circle in

the complex plane which is described on a line as diameter having extremities z1 and z2.

20. Let z1 = 10 + 6i and z2 = 4 + 6i. If z is any complex number

such that the argument of2

1

zz

zz

-

- is

4

p. Then z must lie on

a circle, which has(a) Centre (0, 0); radius 6

(b) Centre (7, 9); radius 18

(c) Centre ( 2 , 1); radius 18

(d) Centre (7, 9); radius 6Sol. Let z = x + iy

Then i)6y()4x(

i)6y()10x(

)i64()iyx(

)i610()iyx(

zz

zz

2

1

-+-

-+-=

+-+

+-+=

-

-

i)6y()4x(

i)6y()4x(

xi)6y()4x(

i)6y()10x(

---

---

-+-

-+-

=

52y12x8yx

i)36y6(

52y12x8yx

76y12x14yx2222

22

+--+

-+

+--+

+--+=

(on simplifying)

Now given that arg 4zz

zz

2

1 p=÷÷ ø

öççè

æ

-

-;

Hence476y12x14yx

36y6tan

221 p=

úúû

ù

êêë

é

+--+

--

úûùêë

é =+ -xytan)iyx(arg 1Q


16/35

16 Phys ics

Now OC = 5, DC = 3, Hence OD = 2 and OE = 8\ Point D, i.e. complex number with least modulus is 2iPoint E, i.e. complex number with maximum modulus is 8iFurthermore the argument of a complex number in Argand diagram is given by the angle that the line joining the origin

to its image forms with positive x–axis. Clearly the twotangents drawn on the circle from origin, OA and OBrepresent the least and the greatest values of this anglerespectively for any point on or inside the circle. Hence,the affices of points A and B give us the complex numberswith least and greatest arguments respectively.

Now Ð XOA = a = AOC,2

Ð=ff-p

and ÐXOB = b = f+p2

Clearly cos f = OCOA = OCACOC

22

- = 54

535

22

=-

[Where 3radiusAC == ] and sin f =5

3

Affix of point A is r (cos a+ i sin a)

r = OA = 4, cos a = cos ÷ ø

öçè

æ f-

p2

= sin f =5

3 ;

sin a = sin ÷ ø

öçè

æ f-p

2 = cos f =

5

4. \ A is )i43(

5

4+

Also affix of point B is r(cos b + i sinb)

r = OB = OA = 4, cos b = cos ÷ ø

öçè

æ f+

p

2 = – sin f = –

5

3

sin b = sin ÷ ø

öçè

æ f+

p2

= cos f =5

4

\ B is )i43(5

4+-

Hence, the complex number with least argument is

)i43(5

4+ and the complex number with greatest argument

is )i43(5

4+- .

All options (a), (b), (c) and (d) are correct

Note : Student may feel that the solution of this example islengthy on the contrary the solution is quite simple and directfrom the figure only.

22. Suppose that z1, z2, z3 represent the vertices of a triangletaken in order. The triangle is equilateral if and only if

(a) 0zz

1

zz

1

zz

1

133221

=-

+-

+-

(b) 0zz1

zz1

zz1

133221=-----

(c)3221 zz

1

zz

1

-=

-

(d) 0zzzzzzzzz 13322123

22

21 =---++

Sol. Let a = 1332,21 zzand zzzz -=g-=b-Then a + b + g = 0 ....(1)

Clearly g+b+a = 0 ....(2)

Let the triangle be equilateral, then| z1 – z2 | = |z2 – z3 | = | z3 – z1 | = l sayThat is | a | = | b | = | g | = l

Nowal

=aÞl=a=aa2

22||

Similarly,

g

l=g

b

l=b

22

&

Hence, from (2)g

l+

b

l+

a

l 222 = 0

g+

b+

aÞ

111= 0

133221 zz

1

zz

1

zz

1

-+

-+

-Þ = 0

Conversely let133221 zz

1

zz

1

zz

1

-+

-+

- = 0

bga

=bg

g+b-=

aÞ=

g+

b+

aÞ

10

111 from (1)

||||||||||||y 322 gba=aÞbg=aÞb=a\

Similarly ||||||||3 gba=b and |||||||| 3 gba=g

Hence |zz||zz||zz||||||| 133221 -=-=-Þg=b=a .That is the triangle is equilateral.

Also note that 0zz

1

zz

1

zz

1

133221

=-

--

+-

0zzzzzzzzz 13322123

22

21 =---++Þ Answer (a, d)

23. The centre of a regular hexagon has affix i. The affix of one

vertex is 2 + i. The affix z of adjacent vertices are

(a) 1 + i )31( ± (b) 31i ±+

(c) )31(i2 ±+ (d) i21±

Sol.


17/35

Laws of Motion 17

Let A be the image of complex number 2 + i, ÐAOB =3

p

Let B (or F) be z, then ÐAOB = arg i)i2(iz

-+

-

(refer to angle between two lines in the text)

3

p±=

÷ ø

öçè

æ p±

p=

-+-

\3

sini3

cosAO

BO

i)i2(

iz

= cos3

sini3

p±

p[ ]BOAO =Q

( )1 3

z i 2 i z i 1 i 3 1 i 1 32 2

æ ö\ - = ± Þ = + ± = + ±ç ÷ç ÷

è øAnswer (a)

24. Let RÎ b,a , such that 0 3q (b) P2 < 3q (c) p 2 = 3q (d) p = 3q

Sol. Given z1 and z

2 are roots of z2 + pz + q = 0

Hence, z1 + z2 = – p – (i), z1z2 = q –(ii)We know that if z

1, z

2, z

3 are vertices of an equilateral

triangle

then 0zzzzzzzzz 13322123

22

21 =---++

(see example 16)

Here z3 = 0, then 0zzzz 21

22

21 =-+

0q 3 p0zz3)zz( 2212

21 =-Þ=-+Þ [from (i) & (ii)]

Answer (c)

26. If the imaginary part of2 1

1

z

iz

+

+

is – 2, then show that the

locus of the point representing z in the argand plane is astraight line.

Sol. Let z = x + iy. Then

2 1

1

z

iz

+

+ =

2( ) 1

( ) 1

x iy

i x iy

+ +

+ + =

(2 1) 2

(1 )

x i y

y ix

+ +

- +

={(2 1) 2 }

{(1 ) }

x i y

y ix

+ +

- + ×

{(1 ) }

{(1 ) }

y ix

y ix

- -

- -

=2 2

2 2

(2 1 ) (2 2 2 )

1 2

x y i y y x x

y y x

+ - + - - -

+ - +

Thus, Im2 1

1

z

iz

æ ö+ç ÷+è ø

=2 2

2 2

2 2 2

1 2

y y x x

y y x

- - -

+ - +

But Im2 1

1

z

iz

æ ö+ç ÷+è ø

= – 2 (Given)

So,2 2

2 2

2 2 22

1 2

y y x x

y y x

- - -= -

+ - +

Þ 2 y – 2 y2 – 2 x2 – x = – 2 – 2 y2 + 4 y – 2 x2

i.e., x + 2 y – 2 = 0, which is the equation of a line.

27. Let z1 and z2 be two complex numbers such that

1 2 0 z i z+ = and arg ( z1 z2) = p. Then find arg ( z1).

Sol. Given that 1 2 0 z i z+ =Þ z1 = iz2, i.e., z2 = – iz1Thus, arg ( z1 z2) = arg z1 + arg (– iz1) = pÞ arg (– iz1

2) = pÞ arg (– i) + arg ( z1

2) = pÞ arg (– i) + 2 arg ( z1) = p

Þ2

- p + 2 arg ( z1) = p

Þ arg ( z1) = 34p


18/35

18 Phys ics

4.4Solve following problems with the help of above text and

examples.1. The points z

1, z

2, z

3, z

4, are the vertices of a parallelogram

(in a complex plane) taken in order if and only if

(a) 3241 zzzz +=+ (b) 4231 zzzz +=+

(c) 4321 zzzz +=+ (d) None2. A square ABCD has its centre at the origin. If A be z

1 then

the centroid of triangle ABC is

(a)3

iz1± (b)3

z1±

(c) ÷ ø

öçè

æ p±

p÷

ø

öçè

æ 3

sini3

cos3

z1(d) ÷

ø

öçè

æ p±

p÷

ø

öçè

æ 6

sini6

cos3

z1

3. If z1 = 1+2i, z2 = 2 + 3i, z3= 3 + 4i then z1, z2 and z3 representthe vertices of a/an(a) equilateral triangle (b) right angled triangle(c) isoceles triangle (d) none of these

4. Circumcentre of an equilateral triangle is at the origin and one of the vertex is r cis a then the other two vertices are at(a) )2(cisr ),(cisr aa-

(b) aa 3cisr ,2cisr

(c) ÷ ø

öçè

æ p+a÷

ø

öçè

æ p+a

3

4cisr ,

3

2cisr

(d) ÷ ø

öçè

æ p

+a÷ ø

öçè

æ p

+a 32

cisr ,3cisr

5. Let z be a complex number satisfying £i5 – z 1 such that

amp z is minimum. Then z is equal to

(a)5

62 +

5

i24(b)

5

24 +

5

i62

(c)5

62 –

5

i24(d) None of these

6. If i25 – z £ 15, then )z(amp.min – )z(amp.max =

(a) ÷ ø

öçè

æ 5

3cos 1 – (b) p – 2 ÷

ø

öçè

æ 5

3cos 1 –

(c)2

p + ÷

ø

öçè

æ 5

3cos 1 – (d) ÷

ø

öçè

æ 5

3sin 1 – – ÷

ø

öçè

æ 5

3cos 1 –

7. In the Argand plane, the area in square units of the triangleformed by the points 1 + i, 1 – i, 2i is

(a)2

1(b) 1 (c) 2 (d) 2

8.If the complex numbers 1z , 2z , 3z are in A.P. , then theylie on a

(a) circle (b) parabola

(c) line (d) ellipse.

9. Let the complex numbers 1z , 2z and 3z be the vertices

of an equilateral triangle. Let 0z the circumcentre of the

triangle. Then 21z +2

2z +2

3z – 32

0z equals

(a) 1 (b) – 1

(c) 0 (d) None of these

10. The complex numbers z = x + iy which satisfy the equation

i5z

i5 – z

+ = 1 lie on

(a) the x-axis

(b) the straight line y = 5

(c) a circle passing through the origin

(d) None of these.

ANSWER KEY

1. (b) 2. (a) 3. (d) 4. (c) 5. (a) 6. (b)

7. (b) 8. (c) 9. (c) 10. (a)

SOME LOCI IN COMPLEX PLANE :

1. arg (z) = RÎaa, represents a line starting from the origin(excluding origin) and making an angle a with the realaxis.

a

z

Y

XO

2. R ,)zzarg( 0 Îaa=- represents a line starting from the

point z0 (excluding the point z0) and making an angle awith the real axis.

az

Y

XO

z03. If z

1 and z

2 are two fixed points then locus of z, satisfying


19/35

Laws of Motion 19

|zz||zz| 21 -=- is the perpendicular bisector of the

segment joining )z(A 1 and )z(B 2

4. The complex equation a2zzzz 21 =-+- where

,zza2 21 -> 'a' is positive real number represents anellipse in complex plane, z1 and z2 are affices of two foci

of ellipse. If |,zz|a2 21 -= then a2|zz||zz| 21 =-+-represents the line segment joining z1 and z2If |,zz|a2 21 -< then the equation does not represent anycurve

5. The complex equation 1 1 2z z – z z 2a- - = .where

21 zza2 -< , and 'a' is positive real number, represents ahyperbola in complex plane, z1 and z2 are affices of twofoci of hyperbola.

If |zz|a2 21 -= , a2|zz||zz| 21 =--- represents thestraight line joining A(z1) and B(z2) but excluding thesegment AB

A(z )1 B(z )2

6. The complex equation K z – z

z – z

2

1 = represents a circle if

1K ¹ and a straight line if K=1.

7. The complex equation K z – zz – z2

22

1 =+ represents

a circle if K ,z – z2

1

K

2

21³ a real number..8. Let z1 and z2 be two fixed points and a be a real number

such that p£a£0 then

(a) ,zz

zzarg

2

1 a=÷÷ ø

öççè

æ

-

-

2,0

p¹ap


20/35

20 Phys ics

Given arg zz

zz

2zz

zz

2

1

2

1

-

-Þ

p=÷÷

ø

öççè

æ

-

- is purely imaginary

Or Re 0zz

zz

2

1 =÷÷ ø

ö

ççè

æ

-

-

0)3y()1y()x4()6x( =+----Þ

021y2x10yx 22 =++-+Þ

Which is a circle with centre (5, – 1) and radius = 5

The equivalent form in complex number is

|z – (5 – i)| = 5 Answer (b)

29. The complex number z in the argand diagram satisfying

| z – a | = 2 and arg (z) = tan –1

3

4 where a = 3 + 4i, is

(a) )i43(5

7or )i43(

5

3++ (b) )i43(

7

5or )i43(

3

5++

(c) 3 + 4i and 3 – 4 i (d) None of these.

Sol.

| z – a | = 2 is the circle with centre at a = (3, 4) and radius= 2.

Clearly3

4tanXOC 1 – =q=Ð

Since arg (z) =3

4tan 1 – , z must lie along OC and on the

circle. It must therefore be either point A or B. Now, OA = OC – AC = 5 – 2 = 3

OB = OC + CB = 5 + 2 = 7

\ A is 3 (cos q +i sin q ) = )i43(53

+

B is 7 (cos q +i sin q ) = )i43(57

+ Answer (a)

30. If z = x + iy such that | z + 1| = | z – 1| and amp41z

1z p=

+

-

then

(a) 0y,12x =+= (b) 12y,0x +==

(c) 12y,0x -== (d) 0y,12x =-=

Sol.

Given that | z + 1 | = | z – 1 | Þ (x +1)2 + y2 = (x – 1)2 + y2

Þ x = 0 ...(1)

41z

1zarg

p=÷

ø

öçè

æ +-

Þ arg 4y)1x(

]iy1x[]iy)1x[(22

p=

ïþ

ïýü

ïî

ïíì

++

-++-

4y)1x(

iy2)1yx(22

22 p=

ïþ

ïýü

ïî

ïíì

++

+-+Þ

11yx

y2,01yx,0y

2222 =

-+>-+>Þ

01yx,0y,01y2yx 2222 >-+>=--+ÞTherefore the locus represented by the equation

arg41z

1z p=÷

ø

öçè

æ +-

is the arc ABC of the circle x2 + y2 – 2y – 1 = 0. Solvingwith x = 0, we get

y = 21y,0y,21

2

82+=\>±=

±Answer (b)

31. In argand plane the locus of 1z ¹ such that

arg 3

2

2zz3

3z5z22

2 p=

úúû

ù

êêë

é

--

+- is

(a) the straight line joining the points z = 3/2, z = –2/3(b) the straight line joining the points z = –3/2, z = 2/3(c) a segment of a circle passing through z = 3/2, z = –2/3(d) a segment of a circle passing through z = –3/2, z = 2/3

Sol .)1z)(2z3(

)1z)(3z2(

2zz3

3z5z22

2

-+--

=--

+-= .1zas

3/2z

2/3z.

3

2

2z3

3z2¹

+-

=+-

\ The given condition reduces to arg3

2

3/2z

2/3z p=÷ ø öç

è æ

+-

This implies that the line joining the points z =2

3. and

z =3

2- subtend a constant angle

3

2p at the point z. Thus

z describes the segment of a circle through z =2

3 and

z =3

2- at which the chord subtends an angle

3

2p.

Answer (c)


21/35

Laws of Motion 21

4.5


examples.

1. If z = x + iy and w =iz

iz1

-

-, then |w| = 1 implies that, in the

complex plane.(a) z lies on the imaginary axis(b) z lies on the real axis(c) z lies on the unit circle(d) None of these.

2. The Points representing the complex numbers z for which| z + 4 |2 – | z – 4 |2 =8 lie on(a) a straight line parallel to x – axis(b) a straight line parallel to y – axis(c) a circle with centre as origin(d) a circle with centre other than the origin

3. If 1z , 2z , 3z are in H.P. they lie on a:

(a) circle (b) sphere(c) straight line (d) None of these

4. The locus of the point Z in the Argand plane for which

21Z+ +

21 – Z = 4 is a

(a) straight line (b) pair of straight lines(c) circle (d) parabola

5. If the imaginary part of

1iz

1z2

+

+ is – 2, then the locus of z in

the complex plane is(a) a circle (b) a straight line(c) a parabola (d) None of these

6. If the real part of i – z 2z + is 4, then the locus of the point

representing z in the complex plane is a/an(a) circle (b) parabola(c) hyperbola (d) ellipse

7. The complex number iyxz += which satisfy the

equation 1i5z

i5z=

+-

lie on

(a) the x -axis (b) the line y = 5(c) a circle through the origin (d) None of these

8. If 11z

iz=+

-, then locus of z is

(a) x-axis (b) y-axis(c) x = 1 (d) x + y = 0

9. If iyxz += and ‘a’ is a real number such that

|aiz| - = |aiz| + , then locus of z is

(a) x-axis (b) axisy -

(c) yx = (d) 1yx 22 =+

10. If ,8|zz||zz| =-++ then z lies on(a) a circle (b) a straight line

(c) a square (d) None of these11. If z is a complex number satisfying

|z – i Re(z)| = |z – Im(z)| then z lies on

(a) xy = (b) xy -=

(c) 1xy += (d) 1xy +-=

ANSWER KEY

1. (b) 2. (b) 3. (a) 4. (c) 5. (b) 6. (a) 7. (a) 8. (d) 9. (a) 10. (c) 11. (a, b)

1. 21 z| – |z is the least value of | z1 + z2 | and 21 zz + is the greatest value of 21 zz +

Thus 222121 zzzzzz +£+£-

2. |z|)zRe(|z| ££-

3. |z|)zIm(|z| ££-

4. | z | < | Re(z) | + | Im(z) | < |z|2

5. If ,az

1z =+ a is positive real number then

2

4aa|z|

2

4aa 22 ++££

++-


22/35

22 Phys ics

1. If the complex number z satisfies the equation

iz3 + z2 – z + i = 0, then

(a) z lies on a unit circle

(b) Re(z) = Im (z)

(c) | z | = 1

(d) None of these

Sol. iz3 + z2 – z + i = 0 Þ i2z3 + iz2 – iz + i2 = 0(Multiply with i)

Þ – z3 + iz2 – iz + i2 = 0 (i2 = –1)Þ –z(z2 + i) + i (z2 + i) = 0 Þ (z2 + i) (z – i) = 0

Either z = i Þ |z| = 1 or z2 = –i Þ |z2| = 1 Þ |z| = 1Hence, |z| = 1, implies that z lies on unit circle.

Answer (a, c)

2. All non zero complex numbers z satisfying the equation

2izz = are(a) i (b) –i

(c) i2

1

2

3- (d) i

2

1

2

3--

Sol. Let z = x + iy, x, y Î R[Note that the problems with lower powers of z can easily be solved using z = x + iy]

Then 2izz = Þ x – iy = i (x + iy)2

Þ x – iy = i (x2 – y2 + 2ixy) = i (x2 – y2) – 2xyEquating real and imaginary parts x = –2xy and

–y = x2 –y2

x = – 2xy Þ x (1 + 2y) = 0 Þ x = 0 or y =2

1-

If x = 0, from second equation, y (y – 1) = 0 Þ y = 0 or 1

If y =2

1- from second equation

2

3x

4

3x

4

1x

2

1 22 ±=Þ=Þ-=

Discording the solution x = 0, y = 0 (for it is z = 0)

We get the solution z = 0 + 1i = i,

z = i2

1

2

3zand i

2

1

2

3-

-=- Answer (a, c & d)

3. The minimum value of |z| + | z – i| is

(a) 0 (b) 1 (c) 2 (d) none

Sol.

Using the result |z1+z

2| £ |z

1| + |z

2|, we get |z| + |z – i| = |z|

+ |i – z| [since |z| = | – z|]£ | z + i – z | = |i| = 1\ minimum value of |z| + |z–i| is 1

ALTERNATE :We may obtain the above answer using geometricalrepresentation. Consider a triangle which has vertices O(origin) P (z) and Q (z – i) thenOP = |z|. OQ = | z – i| and PQ | z + i–z| = |i| = 1 Now in a triangle sum of two sides ³ third sideThat is OP + OQ ³ PQ[The equality holds, when O, P and Q are collinear]Thus | z | + | z – i| ³ 1 Answer (b)

4. If the roots of 0i2izz23

=++ represent the vertices of aDABC in the Argand plane, then the area of the triangle is

(a)2

73(b)

4

73(c) 2 (d) None

Sol. 0)2iz2z)(iz(i2izz 223 =-+-=++

i – 1 – ,i1,iz -=ÞLet A = (0,1), B = (1,–1), C=(–1, –1),

2|22|2

1

111

111

110

2

1ABC =--=

--

-=D\ Answer (c)

5. The complex number z satisfying the equations

0|i5z||iz|4|z| =+--=- , is

(a) i3 - (b) i232 - (c) i232 -- (d) 0

Sol. We have two equations

04|z| =- and 0|i5z||iz| =+--

Putting ,iyxz += these equations become

4|iyx| =+ i.e. 16yx 22 =+ ...(1)

and |i5iyx||iiyx| ++=-+

or 2222 )5y(x)1y(x ++=-+ i.e y = – 2 ...(2)


23/35

Laws of Motion 23

Hence the complex numbers z satisfying the given

equations are )2,32(z1 -= and )2,32(z2 --=

that is, i232z,i232z 21 --=-= Answer (b, c)

6. If nn ii)n(S -+= , where 1i -= and n is a positiveinteger, then the total number of distinct values of )n(S is

(a) 1 (b) 2(c) 3 (d) 4

Sol. We have, n

n2

nnnn

i

1i

i

1iii)n(S

+=+=+= -

,...4,3,2,1n,i

1)1(n

n

=+-

=

\ values of )n(S are 0, –2, 0, 2, 0, –2, ....

\ Total number of distinct values of )n(S is 3.Answer (c)

7. If z1, z2 , z3, z4 are represented by the vertices of a rhombustaken in the anticlock wise order, then

(a) 4321 zzzz +=+ (b) 0zzzz 4321 =-+-

(c)2 4

1 3

z zamp

z z 2

- p=

- (d) 2zz

zzamp

43

21 p=-

-

Sol. Since diagonals of a rhombus bisect each other

)say(z2

zz

2

zz0

4231 =+

=+

\

0zzzz 4321 =-+-ÞAlso, since diagonals of a rhombus are at right angles

z2 z1

z3 z4

z0O

2zz

zzamp

01

02 p=-

-\

2

2

zzz

2

zzz

amp31

1

422 p

=+

-

+-

Þ2zz

zzamp

31

42 p=-

-Þ

Answers (b), (c)

8. If z in any complex number satisfying 1|1z| =- , thenwhich of the following is correct ?

(a) zarg2)1zarg( =-

(b) )zzarg(3

2)zarg(2 2 -=

(c) )1zarg()1zarg( +=-

(d) )1zarg(2zarg +=

Sol. Since 1|1z| =-

,e1z iq=-\ where q=- |1z|arg

q+q+=\ sinicos1z

2/ie.2

cos22

sini2

cos2

cos2 qq

=úû

ùêë

é q+

qq=

2cos

2sini2

2cos2 2 qq+q= )1zarg(212zarg -=q=\

Thus, .zarg2)1zarg( =- Answer (a)9. Let z be a complex number having the argument

20,

p


24/35

24 Phys ics

\ Real part of tan (a + ib) is )ee(2cos2

2sin222 b-b ++a

a

11. If |z| ³ 3, then find the least value ofz

1z +

Sol. : Let z = r (cos q + i sin q ). Then | z | = r ³ 3

Let t =z

1z +

t2 =

2

)sini(cosr

1)sini(cosr q-q+q+q

=

2

sinr

1r icos

r

1r q÷

ø

öçè

æ -+q÷

ø

öçè

æ +

= q÷ ø

öçè

æ -+q÷ ø

öçè

æ + 22

22

sinr

1r cos

r

1r

= q+÷÷ ø

öççè

æ + 2cos2

r

1r

22

t2 is least if cos2 q = – 1 (occurs when q =2

p)

\ for a given value of r

(t2)least

=

2

22

r

1r 2

r

1r ÷

ø

öç

è

æ -=-+ ÷

ø

öç

è

æ -=\

r

1r )t( least

Now3

8

3

13t3r least =-³Þ³ .

3

8

z

1ztof ValueLeast =+=\

12. Find the set of exhaustive values of real number a for

which the equation 0i2|1z|z =+-a+ has a solution

Sol. Let iyxz +=

We have, 0i2|1z|z =+-a+

0y)1x()2y(ix 22 =+-a+++Þ

Equating real and imaginary parts

02y =+ 2y -=\ and 04)1x(x 2 =+-a+

\ )5x2x(x 222 +-a= or 2 2 2 2(1 )x 2 x 5 0- a + a - a =

Since x is real, 0AC4BD 2 ³-=\ (For 2 1,a = the

equation already has solution)

0)1(204 224 ³a-a+aÞ 054 24 ³a+a-Þ

04

54 22 £÷

ø

öçè

æ -aaÞ

Now a being real implies 2a is +ve and hence we

conclude that4

52 -a is –ve orúúû

ù

êêë

é-a

úúû

ù

êêë

é+a

2

5

2

5 is -ve

2

5

2

5£a£-Þ

13. If w is complex cube root of unity and a, b, c are three realnumbers such that

22c

1

b

1

a

1w=

w++

w++

w+

and w=w+

+w+

+w+

2c

1

b

1

a

1222

Then show that

1c

1

1 b

1

1a

1

++

++

+ = 2

(a) 1 (b) 2 (c) 3 (d) None

Sol. Sincew

=w12 and

2

1

w=w the given relation may be

rewritten as w=w++w++w+

2

c

1

b

1

a

1

and2222

2

c

1

b

1

a

1

w=

w++

w++

w+Clearly w and w2 are the roots of

x

2

xc

1

x b

1

xa

1=

++

++

+...(1)

or x

2

)xc)(x b)(xa(

)x b)(xa()xc)(xa()xc)(x b(=

+++

++++++++

or ]abca bcx)c ba(2x3[x 2 ++++++

= ]xx)c ba(x)abca bc(abc[2 32 +++++++

0abc2x)abca bc(x3 =-++-Þ

Now if a is the third root of this equation then sum of the

roots, 102 =aÞ=w+w+aHence, 1 is the root of equation (1) we get

21c

1

1 b

1

1a

1=

++

++

+Answer (b)


25/35

Laws of Motion 25

Fill in the Blanks

1. If | z | = 2 and arg ( z) =4

p then z = .............

2. The greatest and the least absolute value of z + 1 where

| z + 4 | £ 3 is ............. and .............3. In the Argand plane, the vector z = 4 – 3i is turned in the

clockwise sense through 180° and stretched 3 times. Thecomplex number represented by the new vector is .............

4. The value of | |a , where a is a non-real cube root of unity is .............

5. For any two complex numbers

| az1 – bz2 |2 + | bz1 + az2 |

2 .............

6. If x + iy =a ib

c id

+

+, then ( x2 + y2)2 = .............

7. The real value of ‘a’ for which 3i3 – 2ai2 + (1 – a)i + 5 isreal is .............

8. If zn

= cos(2 1) (2 3)n n

æ öp

ç ÷+ +è ø + sin

(2 1) (2 3)n n

æ öp

ç ÷+ +è ø,

then limn ®¥

( z1 z2 ...... z3) is equal to .............

9. The real value of ‘a’ for which 3i3 – 2ai2 + (1 – a) i + 5 isreal is .............

10. If (2 + i) (2 + 2i) (2 + 3i) .... (2 + ni) = x + iy then 4. 8. 13....... (4 + n2) = .............

11. The value of 4 3( 1) n-- - , where n Î N, is .....

True or False

12. Multiplication of a non-zero complex number by i rotates

it through a right angle in the anti-clockwise direction.13. The complex number cos q + i sin q can be zero for some

q.14. If a complex number coincides with its conjugate, then the

number must lie on imaginary axis.15. The argument of the complex number

(1 3)(1 ) (cos sin ) z i i i= + + q + q is7

12

p+ q .

16. The points representing the complex number z for which| z + 1 | < | z – 1 | lies in the interior of a circle.

17. If three complex numbers z1, z2 and z3 are in A.P., thenthey lie on a circle in the complex plane.

18. If n is a positive integer, then the value of in + (i)n+1 +(i)n+2 + (i)n+3 is 0.

19. The points having affixes z1, z2, z3 form an equilateral

triangle iff1 2 2 3 3 1

1 1 10

z z z z z z+ + =

- - -.

20. Multiplication of a non-zero complex number by – i rotatesthe vector through a right angle in counter clockwise sense.

21. If z1, z2, z3 respectively are affixes of points A, B, C then,

1 2

3 2

CBA z z

z z

æ ö-Ð = ç ÷

-è ø

22. If three complex numbers are in A.P., then they lie on a

circle in the complex plane.23. The trigonometric form of the complex number

z = 1 + i tan a where – p < a < p, a ¹ ± p / 2, is1

cosα

(cos a + i sin a).24. The points representing the complex numbers z for which

| z + 1 | < | z – i | lie on a circle.25. If a complex number coincides with its conjugate, then the

number must lie on the imaginary axis.26. The fourth roots of – 1, if plotted, would lie at the vertices

of a square.

27. If the complex numbers z1, z2, z3 represent the vertices of an equilateral triangle such that | z1 | = | z2 | = | z3 |, then

z1 + z2 + z3 ¹ 028. If in the Argand plane z1, z2, z3, z4 are four points such that

| z1 | = | z2 | = | z3 | = | z4 |, then the four points are thevertices of a square.

Short nswer Questions

29. If 4 x + i (3 x – y) = 3 + i (– 6), where x and y are realnumbers then find the values of x and y.

30. Express the following in the form x + iy

(i)5 2

1 2

i

i

+

- (ii)

31

2 3 i

æ ö

- -ç ÷è ø

(iii) i –35 + i – 40 (iv)2

2

(1 2 )

i

i

-

+

31. If ( ) x iy a ib+ = ± + then show that ( ) x iy b ia- - = ± -

32. If Re8

6

z i

z

æ ö-ç ÷+è ø

= 0 then show that z lies on the curve

x2 + y2 + 6 x – 8 y = 0

33. Find the conjugate of(3 2 )(2 3 )

(1 2 ) (2 )

i i

i i

- +

+ -


26/35


27/35

Laws of Motion 27

8. If yixz -= and ,iq pz 31

+= then )q p(q

y

p

x 22 +÷÷ ø

öççè

æ +

is equal to [AIEEE 2004](a) –2 (b) –1

(c) 2 (d) 1

9. If ,1|z||1z|22 +=- then z lies on [AIEEE 2004]

(a) an ellipse (b) the imaginary axis(c) a circle (d) the real axis

10. If the cube roots of unity are 1, w , 2w then the roots of

the equation 3)1 – x( + 8 = 0, are [AIEEE 2005]

(a) –1, –1 + 2 w , – 1 – 2 2w (b) –1, – 1, – 1

(c) – 1, 1 – 2 w , 1 – 2 2w (d)– 1, 1 + 2 w , 1 + 22w

11. If 1z and 2z are two non- zero complex numbers suchthat |zz| 21 + = |z| 1 + |z| 2 , then argg 1z – argg 2z is equalto [AIEEE 2005]

(a)2

p(b) – p

(c) 0 (d)2

p-

12. If w =i

3

1z

z

- and | w | = 1, then z lies on

[AIEEE 2005]

(a) an ellipse (b) a circle(c) a straight line (d) a parabola

13. The value of å=

÷ ø

öçè

æ p+

p10

1k 11

k 2cosi

11

k 2sin is

[AIEEE 2006]

(a) i (b) 1(c) – 1 (d) – i

14. If z2 + z + 1 = 0, where z is complex number, then the valueof

2 222 3

2 3

1 1 1z z z

z z z

æ ö æ öæ ö+ + + + +ç ÷ ç ÷ ç ÷è ø è ø è ø

+ . . . . . . . . . +

26

6

1z

z

æ ö+ç ÷

è ø is [AIEEE 2006]

(a) 18 (b) 54(c) 6 (d) 12

15. If | z + 4 | £ 3, then the maximum value of | z + 1 | is[AIEEE 2007]

(a) 6 (b) 0

(c) 4 (d) 10

16. The conjugate of a complex number is1

i –1then that

complex number is [AIEEE 2008]

(a) –1 –1i (b) 11i +

(c) –1

1i +(d)

1

–1i

17. The number of complex numbers z such that

| z – 1| = | z + 1| = | z – i| equals [AIEEE 2010]

(a) 1 (b) 2

(c) ¥ (d) 018. Let a, b be real and z be a complex number. If z2 + a z + b

= 0 has two distinct roots on the line Re z =1, then it is

necessary that : [AIEEE 2011]

(a) ( 1,0)b Î - (b) 1b =

(c) (1, )b Î ¥ (d) (0,1)b Î

19. If ( 1)w ¹ is a cube root of unity, and ( )7

1 .+ w = + w A B

Then ( A, B) equals [AIEEE 2011]

(a) (1, 1) (b) (1, 0)

(c) (–1, 1) (d) (0, 1)

20. If w ¹ 1is the complex cube root of unity and matrix

ω 0

0 ω H

é ù= ê ú

ë û

, then H70 is equal to – [AIEEE 2011]

(a) 0 (b) –H

(c) H2 (d) H

21. If z ¹ 1 and2

1

z

z -is real, then the point represented by the

complex number z lies : [AIEEE 2012]

(a) either on the real axis or on a circle passing through

the origin.

(b) on a circle with centre at the origin

(c) either on the real axis or on a circle not passing throughthe origin.

(d) on the imaginary axis.22. If z is a complex number of unit modulus and

argument q, then arg1

1

z

z

+æ öç ÷è ø+

equals: [JEE M 2013]

(a) – q (b)2

p – q

(c) q (d) p – q23. If the cube roots of unity are 1, w, w2, then the roots of the

equation ( x – 1)3 + 8 = 0 are [IIT 1979]

(a) – 1, 1 + 2w, 1 + 2w2 (b) – 1, 1 – 2w, 1 – 2w2

(c) – 1, – 1, – 1 (d) None of these


28/35

28 Phys ics

24. The smallest positive integer n for which [IIT 1980]

11

1

+æ ö=ç ÷è ø-

ni

i is

(a) n = 8 (b) n = 16(c) n = 12 (d) none of these

25. The complex numbers = + z x iy which satisfy the

equation – 5

15

=+

z i

z i lie on [IIT 1981]

(a) the x-axis(b) the straight line y = 5(c) a circle passing through the origin(d) none of these

26. If

5 53 3

2 2 2 2

æ ö æ ö= + + -ç ÷ ç ÷

è ø è ø

i i z , then [IIT 1982]

(a) Re( z) = 0 (b) Im( z) = 0(c) Re( z) > 0, Im ( z) > 0 (d) Re( z) > 0, Im ( z) < 0

27. The inequality – 4 – 2 0 (d) none of these

28. If z = x + iy and w = (1–iz)/ (z–1) then |w| = 1 implies that,in the complex plane, [IIT 1983](a) z lies on the imaginary axis(b) z lies on the real axis

(c) z lies on the unit circle(d) None of these

29. The points z1, z2, z3, z4 in the complex plane are the verticesof a parallelogram taken in order if and only if

[IIT 1983]

(a) z1 + z4 = z2 + z3 (b) z1 + z3 = z2 + z4(c) z1 + z2 = z3 + z4 (d) None of these

30. If a, b, c and u, v, w are complex numbers representing thevertices of two triangles such that c = (1 – r) a + rb and w = (1 – r)u + rv, where r is a complex number, then thetwo triangles [IIT 1985](a) have the same area (b) are similar

(c) are congruent (d) none of these31. If w (¹ 1) is a cube root of unity and (1 + w)7 = A + Bwthen A and B are respectively [IIT 1995S](a) 0, 1 (b) 1, 1(c) 1, 0 (d) – 1, 1

32. Let z and w be two non zero complex numbers such that| z | = | w | and Arg z + Argg w = p, then z equals(a) w (b) - w [IIT 1995S]

(c) w (d) - w

33. Let z and w be two complex numbers such that | z | £ 1,

| w | £ 1 and | z + i w | = | z – i w | = 2 then z equals(a) 1 or i (b) i or – i [IIT 1995S]

(c) 1 or – 1 (d) i or – 1

34. For positive integers n1, n

2 the value of the expression

1 1 2 23 5 7(1 ) (1 ) (1 ) (1 )+ + + + + + +n n n ni i i i , where

i = 1 – is a real number if and only if [IIT 1996]

(a) n1= n

2 +1 (b) n

1= n

2 –1

(c) n1= n

2(d) n

1> 0, n

2> 0

35. If i = 1- , then 4 + 5

334 3651 3 1 3

32 2 2 2

æ ö æ ö- + + - +ç ÷ ç ÷

è ø è ø

i i

is equal to [IIT 1999]

(a) 1 3- i (b) – 1 3+ i

(c) 3i (d) 3- i

36. If arg( z) < 0, then arg (- z) - arg( z) = [IIT 2000S](a) p (b) p-

(c)2p- (d)

2p

37. If z1, z2 and z3 are complex numbers such that[IIT 2000S]

1 2 3 1 2 31 2 3

1 1 11, then z= = = + + = + + z z z z z

z z z

is(a) equal to 1 (b) less than 1(c) greater than 3 (d) equal to 3

38. Let z1 and z2 be nth roots of unity which subtend a right

angle at the origin. Then n must be of the form

[IIT 2001S](a) 4k + 1 (b) 4k +2(c) 4k + 3 (d) 4k

39. The complex numbers z1, z2 and z3 satisfying

1 3

2 3

1 3

2

- -=

-

z z i

z z are the vertices of a triangle which is

(a) of area zero [IIT 2001S](b) right-angled isosceles(c) equilateral(d) obtuse-angled isosceles

40. For all complex numbers z1, z2 satisfying | z1|=12 and

| z2-3-4i | = 5, the minimum value of | z1- z2| is[IIT 2002S](a) 0 (b) 2(c) 7 (d) 17

41. Let1 3

2 2w = - + i , then the value of the det.

42

22

1

11

111

ww

ww-- is [IIT 2002]

(a) w3 (b) )1(3 -ww

(c) 23w (d) )1(3 w-w


29/35

Laws of Motion 29

42. If1

1 and ( where 1)1

-= w = ¹ -

+ z

z z z

, then Re( w ) is

(a) 0 (b) 21

1

-

+ z

[IIT 2003S]

(c) 21

.1 1+ +

z

z z(d) 2

4-Chapter Complex Number

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Transcript of 4-Chapter Complex Number