Question

`find the general solution of e^2iz=4`​

Answer 1

Answer:

Let u = z^2 , then the given equation becomes

0 = (u^2 -2iu-1) + 5 ( by completing the square) = (u-i)^2 +5 = (u-i)^2 - 5i2 = (u-i-i sq(5)). (u-i+ i sq(5) = 0 which implies the two equations

(u-i-i sq(5) =0 or (u-i+ i sq(5) =0 . Now the last two equations are quadratic in z which can be solved