Determine the solutions to $e^z - ie^{\pi} = 0$ which absolute value is smaller than $2\pi$

Question

Given the equation $$e^z-ie^{\pi}=0 $$ solve the equation for which $|z|<2\pi$.

My attempt

Manipulating the equation $$e^z=ie^{\pi} $$ Let's take the natural log $$\ln(e^z)=\ln(i)\cdot \ln(e^{\pi}) $$ $$z = \frac{i\pi}{2}+\pi $$ So I guess this is the only solution, but there is something bothering me. Why would they explicit state that they want us to find the solution, for which $\vert z \vert<2\pi$. I feel I have solved in a "wrong" way, because for this exercise I am not supposed to know what $\ln(i)$ is, but I just looked it up.

Is there different way to solve this, without using $\ln(i)$?

Answer 1

Note that $ie^\pi=e^{\pi i/2}e^\pi=e^{\pi+\pi i/2}$. So\begin{align}e^z=ie^\pi&\iff e^z=e^{\pi+\pi i/2}\\&\iff z=\pi+\frac{\pi i}2+2n\pi i\\&\iff z=\pi\left(1+\left(\frac12+2n\right)i\right)\end{align}for some $n\in\Bbb Z$. Can you take it from here?

Answer 2

Hint:

$$e^\pi i = e^\pi e^{\frac{\pi}{2}i}$$

Answer 3

As an alternative

$$e^z-ie^{\pi}=0 \iff e^z=ie^{\pi}\iff e^x\cdot e^{iy}=e^{\pi}\cdot i$$

which requires

• $x=\pi$
• $y=\frac \pi 2+2k\pi$