UNIT 01 - Complex Numbers

8/12/2019 UNIT 01 - Complex Numbers

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PREPARED BY SITI HAWA BINTI AZIZ

1.1 UNDERSTAND COMPLEX NUMBERS IN CARTESIAN FORM ( .

A Complex Number is a combination of a Real Number and an Imaginary Number .

However, if the value of 1 is defined as equals to the value i, then i 2 = ( 1 )2 = -1.

You may remember that for a quadratic equation ax 2 + bx + c =0 , their roots can be found by using

the following formula:

1,12.38, √ ,-2, : real numbers 1 ,√ , : imaginary numbers

( when we squared, it cannot be stated in

a real number form)

aacbb

x2

42

UNIT 1 : COMPLEX NUMBERS

√

A Complex Number is a combination of a Real Number and an Imaginary

Number

Real part Imaginary part √

http://www.mathsisfun.com/numbers/real-numbers.html










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Example 1.2

Determine the values for the following complex numbers:

a) i 8

= i 2 (4)

= = 1

b) i 15

= i 2(7) i= (-1) 7 i= (-1) i= -i

c) 3 i 34 - i 13 +3

= 3 i 2(17) - i2(6) i +3= 3 (-1) 17 - (-1) 6 i +3= 3 (-1) – 1 i +3= - i

d) –2 i 3 + 2 i 18 - 3 i 50

= -2 i 2 i + 2 i 2(9) - 3 i 2(25)

= -2 (-1) i + 2 (-1) 9 - 3 (-1) 25 = 2 i + 2 (-1) – 3 (-1)= 2 i – 2 +3= 2 i + 1

1.2 ALGEBRA OPERATIONS ON COMPLEX NUMBERS

(a) Addition and Subtraction

Adding or subtracting two complex numbers is quite easy. If you are given two complex numbers,

say z = x + iy and w = u + iv , where x, y , u and v R , then

z + w = x + iy + u + iv

= ( x + u ) + ( y + v)

and z – w = x + iy – ( u + v

= (x – u) + ( y – v ) i



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Example 1.3

Solve the following expression :

a) ( 4 + 5i ) + ( 6 + 7i)= ( 4 + 6 ) + (5 + 7 )i= 10 + 12i

b) ( 4 + 5i ) – ( 6 + 7i)= ( 4 – 6 ) + ( 5 – 7 )i= – 2 – 2i

c) =

d) –

(b) Multiplication (Product)

Multiplying two complex numbers is a little bit different. You have to be careful with the

expression i 2 as it is equals to – 1 .

Example 1.4

Solve the following expression :

a) –√

b) –

c) ,

d)



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i 2 = 1

(c) Conjugate

Did you notice how the i 2 has been changed to -1?

Specifically, if you have z = x + iy and w = x – iy, where x and y R, then:

zw = ( x + iy) ( x – iy )

= x 2 - (yi)2

= x 2 + y 2 (real number)

Can you see that the answer is not a complex number anymore? This is a special case. The

complex number w is called the conjugate of z.

Example 1.5

Write the conjugate of:

a)

Therefore, the conjugate of 2 + 3i is 2–

3i.

b) –

Therefore, the conjugate of –3 + 4i is –3 – 4i.

(d) Division

In order to divide two complex numbers, first you have to convert the denominator to a real

number . This can be done by multiplying the denominator with its conjugate . Multiply both top

and bottom by the conjugate of the bottom .

A conjugate is where you change the sign inthe middle like this:

conjugate

http://www.mathsisfun.com/algebra/conjugate.html

http://www.mathsisfun.com/algebra/conjugate.html



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Example 1.6

Write each answer in the form a+bi .

a)

=

(We should then put the answer back into a + bi form)

b)

c) =

=



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(e) Equivalent Complex Numbers

Two complex numbers, z = x + iy and w = u + iv, can be considered as equivalent if and only if x = u

and y = v , then

z = w .and

x + iy = u + iv , so x = u and y = v

For example, if x + yi = 9 – 7i, you can solve this by comparing the real parts and the imaginary

parts.

Example 1.7

Find the value of a and b for each of the following problems :

a) b)

c)

d)



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Activity 1(a)

1) State the real and imaginary parts of each of the complex numbers below :

Complex Numbers Real part Imaginary part

a) -3-3i

b) 4+5i

c) √ - 4i

d) -4i

e) 5f) 3-2.5i

2) Express the following in the form of , where a and b are real numbers:

a) √ d) √

b) √ e) √ c) 6 √ f) √

3) Simplify the following Complex Numbers:

a) i 7 d) i 36

b) i 12 e) 7i 56 – i 3 5



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c) i 20 d) 8i 59 + 5 i 97

4) Solve the following equations :

a) x2 – 6x +10 = 0

b) x2 + 3x = -5

c) 6x - 5 = 5x2



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Feedback for Activity 1(a)

1)

a) -3-3i real part = -3 imaginary part = -3

b) 4+5i real part =4 imaginary part = 5

c) √ - 4i real part = √ imaginary part = -4

d) -4i real part r = 0 imaginary part = -4

e) 5 real part r = 5 imaginary part = 0

f) 3-2.5i real part r =3 imaginary part = -2.5

2)a)

b) √ c)

d)

e)

f)

3)

a) i 7 = i 2 (3) i = (-1)i = -i

b) i 12 = i 2 (6) = (-1) = 1

c) i 20 = i 2 (10) = (-1) = 1

d) i 36 = i 2 (18) = (-1) = 1

e) 6

f) -3i

4)

a) 3 + i , 3 - i c) , ,

b) ,



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Activity 1 (b)

1. State the following complex numbers in the form a + ib

a) √ b) 8 – 16

c) 9 d) √

e) √ + f) √ √

2. Simplify each of the following

a) ( 3 + 4i) + ( 5 – 2i) b) ( 7 + 6i) – (– 4 – 3i)

c) d)



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e) f)

g) – h)

3. State the following complex numbers in the form a + ib

a) i1

2

b) ii

212

c) d)



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e) f)

4. Determine the values of x and y.

a) b)

c) d) ( x + iy ) ( – 2 + 7i ) = – 11 – 4i



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e) x + iy = ( 3 + i )(2 – 3i) f)

5. Solve the following expressions.

a) b)

c) d) –

e) f)



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c) -3 - 4i d) 4 - 3i

1.4 ADDITION AND SUBTRACTION OF COMPLEX NUMBERS USING ARGAND’S DIAGRAM

Figure 1.2 : Addition of Complex Numbers Using Argand’s Diagram

OP3 = OP1 + OP2



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b)

c)

d)

(2, -6)

(8,2)



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Given that z = x + iy (x , y)

The modulus of z , R = z = 22 y x

Argument of z , arg z = tan– 1 ( y/x)

1.5 MODULUS AND ARGUMENT OF A COMPLEX NUMBER

In the Argand’s Diagram shown in Figure 2.3, the length of the line OP is kn own as its modulus

and is written as z = x + iy and it is always positive .

Notes: The value of or agument is measured within 0° < < 360 °.

O

x , real axis

P( x, y )

Imagianry axis, y

r

x

y

arg P , θ P = α

arg Q, θ Q = α + 180°, (- α )

arg R , θ R = α + 180°, (+ α )

arg S, θ S = + 360°, (- α )



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Example 1.10

Find the modulus and argument for the following complex numbers:a) b) c) d)

Example 1.11

1) Given that and .Calculate modulus and argument for :

a) b)

c) d)

||

()

a)

||

( )

b)

||

( )

c)

||

( )

d)

||

()

||

()



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||

( )

||

a)



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3) Given that z 1 = 3 + i and z 2 = – 2 + 4i, draw lines in an Argand’s Diagram to represent the

following: z 1 , z2 , z1 + z2 and z 1 – z2 .



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Activity 1(d)

1) Plot the Argand’s Diagram for each of the following complex numbers a nd use it to

determine its modulus and argument:

a) z = 3i b) z = 2 - 3i

c) z = -4 + i d) z = 1 + 2i

e) z = √ + 2i f) z = -1 - i



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2) Given that , and .Calculate modulus and argument

for and

3) Given that find the modulus and argument for and



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1.6 COMPLEX NUMBERS IN POLAR AND EXPONENTIAL FORMS

Complex numbers can be represented in 4 forms, which are :

Cartesian Form z = a + b i

Polar Form

z = R

( R is modulus and is argument of z in unit of

degree)

Trigonometric form R ( cos + i sin )

( R is modulus and is argument of z in unit ofdegree)

Exponential Form

Re i

( R is modulus and is argument of z in unit of

radian)

Example 1.12

1) Convert z = – 5 + 2i to the Polar and Exponential forms

2) Convert z = 2.5 (cos 189 + i sin 189 ) to the Cartesian, Polar and Exponential forms.



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Solutions :

a)

= =

=

=

b)

=

= = =

3) Given that a = 4+i , b = 2 10° and c= . Find, in polar form.

Solution :

a = 4 + i (4 , 1)

||

Polar form , a : 4.12

c = from this form we know that:R = 4

Polar form , c : 4 20.05



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=

=

=

=



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Activity 1(e)

1) Convert 6i--3z into polar form and exponential form.

2) Convert these complex numbers to the Polar and Exponential form.

a. z = 4 ( cos 54 + i sin 54 )

b. z = 15 ( cos 200 + i sin 200 )

3) Convert these complex numbers to the Polar and Exponential form.

a. 3 + 3i b. – 5 + 2i

c. – 3 – 3i

d. 5 – 2i

4) Given that )218sin218(cos10 00 i z . Express z in polar, exponential and cartesian

form.

5) Find the product of and √ . Then, convert into polar form.



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SELF ASSESSMENT 1 (a)

1) Solve the following quadratic equations

a) x2 + 6x + 13 = 0

b) 3x2 – 2x + 5 = 0

2) State the following complex numbers in a+bi form.

a)

√

b) √ 3) Simplify

a. i 3 b. i 9 c.2

2

i d.5

3i

4) Given

and Find :

a)

b)

c)

5) Simplify each of the following and state your answers in the form a+ib.

a) ( 7 – 5i ) + ( -4 – 2i )

b) ( -8 + 11i ) – ( 6 – 5i )c) ( 8 – 3i )( 7 + 4i )

d) ii23

9

6) Determine the real parts and the imaginary parts of the following.

a) i

i

32

3

b) iii

43)32)(1(



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SELF ASSESSMENT 1 (b)

1) Plot the following complex numbers in an Argand diagram and find its

modulus and argument.

a) 5i

b)

c) √ d)

√

2) Given and

a) Solve for and .

b) Sketch the answer in Argand’s diagram.

c) Find the modulus and argument based on the diagram in b).

3) Given and

a) Find and .

b) Sketch the answer in Argand’s diagram.

c) Find the modulus and argument based on the diagram in b).

4) Given that and . Calculate modulus and argument for

and .

5) Calculate modulus and argument for )



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SELF ASSESSMENT 1 (c)

1. Convert z = – 5 – 3i to the Polar form.

2. Given that z = 2 ( cos 25 o + i sin 25 o ) , state z 3 in Polar form.

3. If z1 = 12 ( cos 125 o + i sin 125 o ) and z2 = 3 ( cos 72 o + i sin 72 o),

find the values of :

a. z1 z2 b.2

1

z

z

Carry out your solution in Polar form.

4. Given that z = – 3 + 4i and w = 2 ( cos 30 o + i sin 30 o ) . Solve for zw and z/w in Polar

Forms.



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SOLUTION : SELF ASSESSMENT 1 (b)

1.

2.

a)

b)



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SOLUTION : SELF ASSESSMENT 1 (c)

1. 5.83 -149.04

2. 8 75

3. a. 36 197

b. 4 53

4. a. 10 – 23.1 o

b. 2.5 – 83.13 o

UNIT 01 - Complex Numbers

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