Solve for z if 2z-z' = (3-5i)/(2-3i)

We'll consider the complex number put in rectangular
form:

z = x + i*y. It's conjugate is z' = x -
i*y

2z - z' = 2x + 2i*y - x +
i*y

2z - z' = x + 3i*y
(1)

We'll manage the right side and we'll multiply the
fraction by the conjugate of the denominator.

(3-5i)/(2-3i)
= (3-5i)(2+3i)/(2^2 + 3^2)

(3-5i)(2+3i)/(2^2 + 3^2) = (6 +
9i - 10i + 15)/13

(3-5i)/(2-3i) = (21-i)/13
(2)

We'll put (1)=(2):

x +
3i*y = (21-i)/13

13x + 39i*y = 21 -
i

Comparing, we'll get:

13x =
21

x = 21/13

y =
-1/39

The complex number z is: z = (21/13) -
(1/39)*i