How to solve $e^z=2\sqrt3 - 2i$

Question

I have been working with this equation for a long time now, and I just can't figure it out.

Solve the following equation. $$e^z=2\sqrt3 - 2i.$$

There should be three solutions that have a positive imaginary part. I hope some of you, can help me. Thank you :-)

Answer 1

If $z=x+yi$, then $e^z=e^x\bigl(\cos y+i\sin y\bigr)$. So$$e^z=2\sqrt3-2i=4\left(\frac{\sqrt3}2-\frac12i\right)\iff e^x=4\wedge\cos y=\frac{\sqrt3}2\wedge\sin y=-\frac12.$$So, take $x=\log4$, and $y=-\frac\pi6+2k\pi$ ($k\in\mathbb Z$).

Answer 2

Let $z=x+yi$, where $\{x,y\}\subset\mathbb R$.

Thus, $$e^x(\cos{y}+i\sin{y})=2\sqrt3-2i,$$ which gives $$e^x\cos{y}=2\sqrt3$$ and $$e^x\sin{y}=-2.$$

Thus, $$e^{2x}=(2\sqrt3)^2+(-2)^2,$$ which gives $$e^{2x}=16$$ or $$x=2\ln2.$$ Hence, $\cos{y}=\frac{\sqrt3}{2}$ and $\sin{y}=-\frac{1}{2}$, which gives $$y=\frac{11\pi}{6}+2\pi k,$$ where $k\in\mathbb Z$.

Id est, we got the answer: $$\left\{2\ln2+\left(\frac{11\pi}{6}+2\pi k\right)i|k\in\mathbb Z\right\}.$$ If you wish three solutions with positive imaginary part then we obtain for example: $$\left\{2\ln2+\left(\frac{11\pi}{6}+2\pi k\right)i|k\in\{0,1,2\}\right\}.$$