Find all complex numbers z such that 4 2iz 8 2iz 2 10i

Find all complex numbers z such that (4 + 2i)z + (8 - 2i)z\' = -2 + 10i, where z\' is the complex conjugate of z.

Solution

Let z = a + bi where a and b are real numbers. The complex conjugate z\' is written in terms of a and b as follows: z\'= a - bi. Substitute z and z\' in the given equation

(4 + 2i)(a + bi) + (8 - 2i)(a - bi) = -2 + 10i

Expand and separate real and imaginary parts.

(4a - 2b + 8a - 2b) + (4b + 2a - 8b - 2a )i = -2 + 10i

Two complex numbers are equal if their real parts and imaginary parts are equal. Group like terms.

12a - 4b = -2 and - 4b = 10

Solve the system of the unknown a and b to find:

b = -5/2 and a = -1

z = -1 - (5/2)i