EQUALITY OF TWO COMPLEX NUMBERS EXAMPLES

Problem 1 :

Find real numbers x and y such that :

a) 2x + 3yi = -x - 6i 

b) x² + xi = 4 - 2i

c) (x + yi) (2 - i) = 8 + i 

d)  (3 + 2i) (x + yi) = -i

Solution :

a)  2x + 3yi = -x - 6i

b)

x² + xi = 4 - 2i

By equating the imaginary parts.

x = -2

c)  (x + yi) (2 - i) = 8 + i 

2x - ix + 2yi - yi² = 8 + i

2x - ix + 2yi + y = 8 + i

(2x + y) + i(2y - x) = 8 + i

1⋅(2) - (2)

4x + 2y - 2y + x = 16 - 1

5y = 15

y = 3

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

2(3) - x = 1

6 - x = 1

x = 6 - 1

x = 5

d)

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

3x + 3yi + 2ix + 2yi² = -i

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

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

(1) ⋅ 2 - (2) ⋅ 3

6x - 4y - 9y - 6x = 0 - (-3)

-13y = 3

y = -3/13

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

3x - 2(-3/13) = 0

3x = -6/13

x = -2/13

So, the real numbers x and y is -2/13 and -3/13.

Problem 2 :

Find x and y if x, y ϵ ℝ and :

a)  2(x + yi) = x - yi 

b)  (x + 2i) (y - i) = -4 - 7i

c)  (x + i) (3 - iy) = 1 + 13i

d)  (x + yi) (2 + i) = 2x - (y + 1)i

Solution :

a 

2(x + yi) = x - yi 

2x + 2yi =  x - yi

By equating the real and imaginary parts.

So, the real numbers x and y is 0 and 0.

b)  

(x + 2i) (y - i) = -4 - 7i 

xy - ix + 2iy - 2i² = -4 - 7i

xy - ix + 2iy + 2 = -4 - 7i

(xy + 2) + i(2y - x) = -4 - 7i

By equating the real and imaginary parts.

y = -6/x substitute the equation (2).

2(-6/x) - x = -7

-x² - 12 + 7x = 0

-(x² + 12 - 7x) = 0

x² - 7x + 12 = 0

(x - 3) (x - 4) = 0

x = 3 and x = 4

c 

(x + i) (3 - iy) = 1 + 13i 

3x - ixy + 3i - i²y = 1 + 13i

3x - ixy + 3i + y = 1 + 13i

(3x + y) + i(3 - xy) = 1 + 13i

By equating the real and imaginary parts.

y = -10/x substitute the equation (1).

3x + (-10/x) = 1

3x² - 10 = x

3x² - x - 10 = 0

(3x + 5) (x - 2) = 0

x = -5/3 and x = 2

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

y = 6 and y = -5

d)

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

2x + ix + 2iy + i²y = 2x - iy - i

2x + ix + 2iy - y = 2x - iy - i

(2x - y) + i(x + 2y) = 2x - iy - i

By applying the value of y, we get 

x = -1

Problem 3 :

Write z in the form a + bi where a, b ϵ ℝ and i = √-1, if the complex number z satisfies the equation 3z + 17i = iz + 11

Solution :

Given 3z + 17i = iz + 11

3z - iz = 11 - 17i

z(3 - i) = 11 - 17i

(a + ib)(3 - i) = 11 - 17i

3a - ia + 3bi - i²b = 11 - 17i

(3a + b) + i(-a + 3b) = 11 - 17i

(1) + (2)  ⋅ 3

3a + b - 3a + 9b = 11 - 51

10b = -40

b = -4

Applying the value of b, we get

3a = 11 + 4

3a = 15

a = 5

z = 5 - 4i

Problem 4 :

The complex number z is a solution of the equation 

√z = 4/(1 + i) + 7 - 2i.

Express z in the form a + bi where a and b ϵ ℤ.

Solution :

Let z = a + ib

2a = 142, then a = 71

-2b = 130, then b = -65.

Problem 5 :

Find the real values of m and n for which

3(m + ni) = n - 2mi - (1 - 2i).

Solution :

3(m + ni) = n - 2mi - (1 - 2i)

3m + 3ni = n - 2mi - 1 + 2i

3m + 3ni = (n - 1) + i(2 - 2m)

By equating real and imaginay parts.

3m = n - 1

n = 3m + 1 --- (1)

3n = 2 - 2m --- (2)

n = 3m + 1 substitute the equation (2).

3(3m + 1) = 2 - 2m

9m + 3 = 2 - 2m

9m + 2m = 2 - 3

11m = -1

m = -1/11

m = -1/11 substitute the equation (1).

n = 3(-1/11) + 1

n = -3/11 + 1

n = (-3 + 11)/11

n = 8/11

So, the real values m and n is -1/11 and 8/11.

Problem 6 :

Express z = 3i/(√2 - i) + 1 in the form a + bi where a, b ϵ ℝ, giving the exact values of the real and imaginary parts of z.

Solution :