COMPLEX NUMBERS EXAMPLES & SOLUTIONS

[Pages:12]COMPLEX NUMBERS EXAMPLES & SOLUTIONS

Department of Mathematics & Philosophy

The Open University of Sri Lanka

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Examples for Complex numbers

COMPLEX NUMBERS:EXAMPLES & SOLUTIONS

Question (01)

(i) Find the real values of x and y such that (1- i)x + 2i + (2 + 3i) y + i = - i

3-i

3+i

(ii) Find the real values of x and y are the complex numbers 3 - ix2 y and -x2 - y - 4i conjugate of each other.

(iii) Find the square roots of 4 + 4i

(iv) Find the complex number Z satisfying the equation Z -12 = 5 and Z - 4 = 1

Z - 8i 3

Z -8

(v) Find real such that 3 + 2i sin is 1- 2i sin

(a) real

(b) imaginary

Solution

(i) (1- i)x + 2i + (2 + 3i) y + i = - i

3-i

3+i

{(1- i)x + 2i}(3 + i) + {(2 + 3i) y + i}(3 - i) = - i

(3 - i)(3 + i)

(3 + i)(1- i)x + 6i + 2i2 + (2 + 3i)(3 - i) y + 3i - i2 9 -i2

=

-i

(3 + i - 3i - i2 )x + 6i + (6 + 9i - 2i - 3i2 ) y + 3i + i2 = - i 9 +1

(4 - 2i)x + (9 - 7i) y + 9i -1 = -10i

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[4x + 9 y -1] + i(19 - 2x - 7 y) = 0

4x + 9 y -1 = 0......................(1)

19 - 2x - 7 y = 0.....................(2)

By solving equations (1) and (2)

x = -82 y = 37

5

5

(ii) 3 - ix2 y = -x2 - y - 4i

3 - ix2 y = -(x2 + y) + 4i

-(x2 + y) = 3

-x2 y = 4

x2 = - 4 y

4y-y=3

y2 +3y - 4 = 0

( y + 4)( y -1) = 0 y = - 4 or y = 1

when y = - 4

x2 = - 4 = 1 x = ?1 -4

y =1

x2

=

-4 1

= -4

(Not real)

x = ?1 and y = - 4

(iii) Let z2 = 4 + 4i and z = (x + iy)

z = 4 + 4i z2 = (x + iy)(x + iy)

= x 2 +2ixy + i2 y2

z2 = (x2 - y2 ) + i2xy

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(x2 - y 2 ) + i2xy = 4 + 4i

x2 - y2 = 4 2xy = 4

y = 2x

{ } x2 -

2 x

2

=4

x 4 -4x2 + 4 = 0 (x 2 -2)2 = 0

x2 = 2

x=? 2

y =

2 ?

=? 2

2

z = 2 + 2 i or z = - 2 - 2 i

(iv) z -12 = 5 .................(1) z - 8i 3 z - 4 = 1 ....................(2) z -8

Let z = x + iy

z -12 = (x -12) + iy

z -12 = (x -12)2 + y2

z - 8i = x + i( y - 8)

z - 8i = x2 + ( y - 8)2

z - 4 = (x - 4) + iy

z - 4 = (x - 4)2 + y2

z - 8 = (x - 8) + iy

z - 8 = (x - 8)2 + y2

From (1); z -12 = 5 3 (x -12)2 + y2 = 5 x2 + ( y - 8)2 z - 8i 3

{ } 9 (x -12)2 + y2 = 25 x2 + ( y - 8)2

..............(1)

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From (2) z - 4 = 1 z -8

(x - 4)2 + y2 = (x - 8)2 + y2

(x - 4)2 + y2 = (x - 8)2 + y2

x2 - 8x +16 + y2 = x2 -16x + 64 + y2

8x = 48

x =6

form (1); 9 (6 -12)2 + y2 = 25 62 + ( y - 8)2

9 (36 + y2 ) = 25 36 + ( y - 8)2 16 y2 - 400 y + (2500 - 324) = 0

16 y2 - 400 y + 2176 = 0

y2 - 25y +136 = 0

( y -17) ( y - 8) = 0

y = 17 or y = 8

z = 6 +17i or z = 6 + 8i

(v)

z

=

3 + 2i sin 1- 2i sin

=

(3 + 2i sin )(1+ 2i sin ) (1- 2isin )(1+ 2i sin )

=

(3 - 1+

4sin2 4sin2

)

+

8i sin 1+ 4sin2

If

z

is

real

1

8 sin + 4sin2

=0

= 2

= n + (-1)n 2 n

If

z

is imaginary

3 - 4 sin2 1+ 4sin2

=0

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sin2

=

3 4

sin = ? 3 2

=? 3

= n ? 3

n

Question (02)

(i) Express the following complex numbers in the polar form

(a)

3 2

- +

i i

2

(b) (1- i) (1+ i)(1+ 3 i)

(c) 1+ 7i (2 - i)2

(d) 0 + 0i

(ii) Find the modulus and the principal value of the argument of the following complex numbers

(a) 1- 3i 1- 2i

(b) 1- i - 1+ i 1+i 1-i

(c)1+ cos + i sin (d) 1- (1- i)2

1+ 2i

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(iii)Find the modulus and the principal value of the argument of each of the following complex numbers

(a) (1- 3i)(1- i)

(b) (-3 + 3i) (4 - 4i)

(c) (-3 + 3i)(4 - 4i) (iii) Find the square roots of (a) 5 +12i (b)15 + 8i (c) 24 +10i

Solution

(2)(a)

z

=

3-i 2+i

2

=

(3

(2

- i)(2 - i) + i)(2 - i)

2

=

6

+

i2 5

-

5i

2

=

(1- i)2

=

2

1

-i

2

1 2 2

{ }2

= 2(cos(- 4)) + i sin(- 4))

= ( 2)2 {cos(- 4) + i sin (- 4)}{cos(- 4) + i sin(- 4)}

= 2[cos(- 4 - 4) + i sin(- 4 - 4)]

= 2[cos(- 2) + i sin(- 2)]

(b) z =

1-i

= (1- i)(1- i)(1- 3i)

(1+ i)(1+ 3i) (1+ i)(1- i)(1+ 3i)(1- 3i)

=

1+ i2 - 2i (1- 3 (1- i2 )(1- 3i2 )

i)

=

-2i(1 - 2.4

3i)

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= 1 (- 3 - i) 4

=

2 4

-

3 2

-

1 2

i

= 1 {cos(-5 6) + i sin(-5 6)}

2

(c). z

=

1+ 7i (2 - i)2

=

1+ 7i 22 + i2 - 4i

=

1+ 7i (3 - 4i)

=

(1+ 7i)(3 + 4i) (3 - 4i)(3 + 4i)

= 3 + 28i2 + 25i = -25 + 25i = (-1+ i)

25

25

= 2 -1 2 +1 2 i

= 2 [cos 3 4 + i sin 3 4]

(d) z = 0 + 0i is not possible to write in the form z = r {cos + i sin}

Where r > 0 - Since there is no value such that cos = 0 and sin = 0 There is no argument for Zero Complex number.

(iii)(a)

z

=

1- 3i 1+ 2i

=

(1- 3i) (1+ 2i)

(1 - (1 -

2i) 2i)

=

1+ 6i2 - 5i 1- 4i2

=

-5 - 5i 5

{ } = -1- i = 2 -1 2 + (-1 2)i

= 2 [cos(- 3 4) + i sin(-3 4)]

z = 2 and arg(z) = - 3 4

(b) z

= 1-i 1+ i

-

(1+ i) (1- i)

=

(1- i)2 - (1+ i)2 (1+ i)(1- i)

=

[1- i +1+ i][1- i -1- i]

1- i2

= 2(-2i) = -2i = 2[cos(- 2) + i sin(- 2)]

2

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