Post on 01-Jun-2018
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)
8/9/2019 4-Chapter Complex Number
2/35
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)
8/9/2019 4-Chapter Complex Number
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
8/9/2019 4-Chapter Complex Number
4/35
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)
8/9/2019 4-Chapter Complex Number
5/35
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-=-
8/9/2019 4-Chapter Complex Number
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
8/9/2019 4-Chapter Complex Number
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
Solve following problems with the help of above text and
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
8/9/2019 4-Chapter Complex Number
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 ι= 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
8/9/2019 4-Chapter Complex Number
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
8/9/2019 4-Chapter Complex Number
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
8/9/2019 4-Chapter Complex Number
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
8/9/2019 4-Chapter Complex Number
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)
8/9/2019 4-Chapter Complex Number
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.
8/9/2019 4-Chapter Complex Number
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
8/9/2019 4-Chapter Complex 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
8/9/2019 4-Chapter Complex Number
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.
8/9/2019 4-Chapter Complex Number
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
8/9/2019 4-Chapter Complex Number
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
8/9/2019 4-Chapter Complex Number
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
8/9/2019 4-Chapter Complex Number
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)
8/9/2019 4-Chapter Complex Number
21/35
Laws of Motion 21
4.5
Solve following problems with the help of above text and
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 ++££
++-
8/9/2019 4-Chapter Complex Number
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)
8/9/2019 4-Chapter Complex Number
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
8/9/2019 4-Chapter Complex Number
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)
8/9/2019 4-Chapter Complex Number
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
- +
+ -
8/9/2019 4-Chapter Complex Number
26/35
8/9/2019 4-Chapter Complex Number
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
8/9/2019 4-Chapter Complex Number
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
8/9/2019 4-Chapter Complex Number
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