Question 464116
Note that


*[tex \LARGE 2i = 2e^{i\frac{\pi}{2}} = e^{\ln{2}} e^{i\frac{\pi}{2}} = e^{\ln{2} + i\frac{\pi}{2}}]


Hence, all we need to solve is


*[tex \LARGE e^{-z + 5} = e^{\ln{2} + i\frac{\pi}{2}}]


*[tex \LARGE -z + 5 = \ln{2} + i\frac{\pi}{2}]


*[tex \LARGE z = 5 - \ln{2} - i\frac{\pi}{2}]


Note that we can replace pi/2 with 5pi/2, 9pi/2, ... because e^{ix} is periodic with period 2pi. The general solution is


*[tex \LARGE z = 5 - \ln{2} - i(2n\pi + \frac{\pi}{2})] where n is any integer.