Solutions of $e^z = \alpha$ in $\left\{z\in\mathbb C~|~0 \leq Im(z)<2\pi\right\}$

Question

I want to find the solution of the following equation:

$e^z = \alpha$

It only has one solution in $\left\{z\in\mathbb C~|~0 \leq Im(z)<2\pi\right\}$

I don't know how to begin. How can you find that solution of a complex function?

Answer 1

If you write $\alpha$ in polar coordinates, you can directly get the solution.

There exists $\varphi\in[0,2\pi)$ such that $\frac{\alpha}{|\alpha|}=e^{\varphi i}$. We get $$ \alpha=|\alpha|e^{\varphi i}=e^{\ln(|\alpha|)}e^{\varphi i}=e^{\ln(|\alpha|)+\varphi i}. $$ So you get $z=\ln(|\alpha|)+\varphi i$.

Answer 2

Assuming $\alpha=a+ib\in\Bbb{C}$. Writing $z=x+iy$,

$$e^z=e^x(\cos y + i\sin y)=a+ib.$$

Then equating real and imaginary parts, $e^x\cos y=a$ and $e^x\sin y=b$. Solve simultaneously.