# EQUALITY OF COMPLEX NUMBERS

Two complex numbers are equal if and only if their real parts are equal and their imaginary parts are equal. I.e.,

a+bi = c+di

if and only if a = c, and b = d.

Find real numbers x and y such that :

Problem 1 :

(x + 2i) (1 - i) = 5 + yi

Solution :

(x + 2i) (1 - i) = 5 + yi

x - xi + 2i - 2i2 = 5 + yi

Here the value of i2 is -1.

x - xi + 2i  + 2 = 5 + yi

By equating the real and imaginary parts.

(x + 2) + i(-x + 2) = 5 + yi

 x + 2 = 5 ---(1)x = 5 - 2x = 3 -x + 2 = y ---(2)-3 + 2 = y-1 = yy = -1

Problem 2 :

2x + 3yi = -x - 6i

Solution :

2x + 3yi = -x - 6i

Adding x - yi on each sides.

2x + 3yi + x + 6i = 0

3x + i(3y + 6) = 0

 3x = 0x = 0 3y + 6 = 0y = -6/3y = -2

Problem 3 :

x2 + xi = 4 - 2i

Solution :

x2 + xi = 4 - 2i

Subtracting 4- 2i on each sides.

x2 + xi - 4 + 2i = 0

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

By equating the real and imaginary parts.

x2 - 4 = 0

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

x = 2 and x = -2

So, the real number x is -2.

Problem 4 :

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

Solution :

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

2x - ix + 2yi -i2y = 8 + i

Here the value of i2 is -1.

2x - ix + 2yi  + y = 8 + i

(2x + y) + i(2y - x) =  8 + i

By equating the real and imaginary parts.

 2x + y = 8 ---(1) 2y - x = 1-x + 2y = 1---(2)

(1) + 2(2)

2x + y - 2x + 4y = 8 + 2

5y = 10

y = 2

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

-x + 2(2) = 1

-x = 1 - 4

x = 3

Problem 5 :

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

Solution :

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

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

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

Here the value of i2 is -1.

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

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

By equating the real and imaginary parts.

3x - 2y = 0 ---(1)

-2y = -3x

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

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

Applying the value of y in the third equation.

Applying the value of x in the second equation.

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

Problem 6 :

2(x + yi) = x - yi

Solution :

2(x + yi) = x - yi

2x + 2yi = x - yi

Subtracting x - yi on each sides.

2x + 2yi - x + yi = 0

By equating the real and imaginary parts.

(2x - x) + i(2y + y) = 0

x + 3iy = 0

x = 0 and y = 0

Problem 7 :

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

Solution :

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

xy - xi + 2iy - 2i2 = -4 - 7i

Here the value of i2 is -1.

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

By equating the real and imaginary parts.

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

xy + 2 = -4 ---(1)

xy = -4 - 2

xy = -6

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

2y - x = -7 ---(3)

Applying the value of y in the third equation.

By using algebraic expression.

x2 - 4x - 3x + 12 = 0

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

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

x - 4 = 0 and x - 3 = 0

x = 4 and x = 3

Applying the value of x in the second equation.

y = -6/4

y = -3/2

So, the values of x and y is 4, -3/2.

Problem 8 :

(x + i) (3 - iy) = 1 + 13i

Solution :

(x + i) (3 - iy) = 1 + 13i

3x - ixy + 3i - i2y = 1 + 13i

Here the value of i2 is -1.

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

By equating the real and imaginary parts.

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

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

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

3 - xy = 13 ---(3)

Applying the value of y in the third equation.

3 - x(1 - 3x) = 13

3 - x + 3x2 = 13

3x2 - x + 3 - 13 = 0

3x2 - x - 10 = 0

By using algebraic expression.

3x2 - 6x + 5x - 10  = 0

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

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

3x = -5 and x = 2

x = -5/3

Applying the value of x in the second equation.

y = 1 - 3(-5/3)

y = 1 + 5

y = 6

Problem 9 :

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

Solution :

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

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

2x + ix + 2iy + i2y = 2x - (iy + i)

Here the value of i2 is -1.

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

By equating the real and imaginary parts.

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

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

-y = 2x - 2x

-y = 0

y = 0 ---(2)

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

Applying the value of y in the second equation.

x + 2(0) = 0 - 1

x = -1

So, the values of x and y is -1, 0.

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