# EQUALITY OF TWO COMPLEX NUMBERS

Problem 1 :

Find real numbers a and b such that :

a)  a + ib = 4

b) (1 – 2i) (a + bi) = -5 – 10i

c)  (a + 2i) (1 + bi) = 17 – 19i

Solution :

a)  a + ib = 4

By equating the real and imaginary parts.

a = 4 and b = 0

b)  (1 – 2i) (a + bi) = -5 – 10i

By multiplying the binomials, we get

a + bi - 2ia - 2bi2 = -5 - 10i

Here the value of i2 is -1.

a + bi - 2ia + 2b = -5 - 10i

(a + 2b) + i(b - 2a) = -5 - 10i

By equating the real and imaginary parts.

 a + 2b = -5 ---(1) b - 2a = -10-2a + b = -10  ----(2)

(1) ⋅ 2 + (2)

2a + 4b - 2a + b = -10 - 10

5b = -20

b = -4

Applying the value of b in (1), we get

a - 8 = -5

a = 3

c.

(a + 2i) (1 + bi) = 17 – 19i

By multiplying two complex numbers on the left side, we get

a + abi + 2i + 2bi2 = 17 - 19i

Here the value of i2 is -1.

a + abi + 2i - 2b = 17 - 19i

(a - 2b) + i(ab + 2) = 17 - 19i

By equating the real and imaginary parts.

 a - 2b = 17a = 17 + 2 b --- (1) ab + 2 = -19ab = -21  --- (2)

Applying the value of a in (2), we get

(17 + 2b)b = -21

17b + 2b2 = -21

2b2 + 17b + 21 = 0

(b + 7)(2b + 3) = 0

b = -7, b =  -3/2

 When b = -7a = -21/(-7)a = 3 When b = -3/2a = -21/(-3/2)a = 14

Problem 2 :

Find a complex number z such that 2z -1 = iz – i. Write your answer in the form z = a + bi where a, b ϵ ℝ.

Solution :

2z -1 = iz – i

2z - iz = -i + 1

z(2 - i) = -i + 1

Dividing 2 - i on each sides.

Problem 3 :

if z = 4 + i and w = 3 – 2i find 2w* - iz.

Solution :

Given, z = 4 + i and w = 3 - 2i

To find 2w* - iz :

w* = 3 + 2i

= 2(3  +2i) - i(4 + i)

= 6 + 4i - 4i - i2

= 6 + 1

= 7

Problem 4 :

Find rationals a and b such that

(2 – 3i)/(2a + bi) = 3 + 2i.

Solution :

Given (2 – 3i)/(2a + bi) = 3 + 2i

Multiplying 2a + bi on each sides.

2 – 3i = (3 + 2i) (2a + bi)

2 - 3i = 6a + 3bi + 4ai + 2bi2

2 - 3i = 6a + 3bi + 4ai - 2b

2 - 3i = 6a - 2b + i(3b  + 4a)

Equating real and imaginary parts.

 6a - 2b = 23a - b = 1 --- (1) 4a + 3b = -3 -----(2)

(1) ⋅ 3 + (2)

9a - 3b + 4a + 3b = 3 - 3

13a = 0

a = 0

Applying the value of a in (1), we get

3(0) - b = 1

-b = 1

b = -1

Problem 5 :

a + ai is a root of x2 – 6x + b = 0 where a, b ϵ ℝ. Explain why b has two possible values. Find a in each case.

Solution :

Since one of the root is in the form of complex number, other root will be its conjugate.

So, the roots are a + ai and a - ai

 Sum of the roots := (a + ai) + (a - ai)=   a + ai + a - ai= 2a Products of the roots := (a + ai) (a - ai)= a2 + a2= 2a2

x2 - (sum of the roots)x + products of the roots = 0

x2 - 2ax + 2a2 = 0

Given x2 - 6x + b = 0

x2 - 2ax + 2a2 = x2 - 6x + b

 Equating coefficients of x-2a = -6a = -6/-2a = 3 Equating constants terms2a2 = b2(3)2 = b18 = b

Problem 6 :

Find real numbers x and y such that

(3x + 2yi) (1 - i) = (3y + 1)i – x.

Solution :

(3x + 2yi) (1 - i) = (3y + 1)i – x

3x - 3xi + 2yi - 2yi2 = 3yi + i - x

3x - 3xi + 2yi + 2y = 3yi + i - x

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

By equating the real and imaginary parts.

 3x + 2y = -x2y = -x - 3x-4x - 2y = 02x + y = 0 ---(1) 2y - 3x = 3y + 1-3x - y = 1--- (2)

(1) + (2)

2x + y - 3x - y = 0 + 1

-x = 1

x = -1

By applying the value of x in (1), we get

-2 + y = 0

y = 2

Problem 7 :

Find real x and y such that :

a) x + iy = 0

b) (3 – 2i) (x + i) = 17 + yi

c) (x + iy)2 = x – iy

Solution :

a)

x + iy = 0

Equating real and imaginary parts.

x = 0 and y = 0

b)

(3 – 2i) (x + i) = 17 + yi

3x + 3i - 2ix + 2 = 17 + yi

(3x + 2) + i(3 - 2x) = 17 + yi

Equating real and imaginary parts.

3x + 2 = 17 --- (1)

x = 5

3 - 2x = y --- (2)

x = 5 substitute the equation (2).

3 - 2(5) = y

3 - 10 = y

-7 = y

So, the real numbers x and y is 5 and -7.

c)

(x + iy)2 = x – iy

x2 + i2y2 + 2(x)(iy) = x - iy

x2 - y2 + 2xiy = x - iy

Equating real and imaginary parts.

x2 - y2 = x --- (1)

2xy = -y --- (2)

2x = (-y) (1/y)

2x = -1

x = -1/2

x = -1/2 substitute the equation (1).

Problem 8 :

Find z if √z = 2/(3 – 2i) + 2 + 5i

Solution :

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