Determine if $f(z) = \log(e^z+1)$ is analytic and where

Question

I tried to substitute $z$ with $x+iy$ and then write it down as $f(z) = log(e^xcis(iy)+1) $ but it looks worse than the starting point. on original branch

Answer 1

You're using the branch of logarithm that is analytic on the complement of the left half-line $B = (-\infty, 0]$. So your function is analytic at $z$ as long as $e^z+1 \notin B$, i.e. $e^z \notin (-\infty, -1]$.

Now if $z = x + i y$ with $x, y$ real, $e^z = e^x e^{i y}$. For this to be in $B$, you need $y = (2n+1) \pi$ with for integer $n$ (so that $e^{iy} = -1$) and $x \ge 0$ (so $|e^z| = e^x \ge 1$). Thus your function is analytic on the complement of the half-lines $y = (2n+1) \pi$, $x \ge 0$.

Answer 2

According to the determination of logarythm you saw, you should know on which domain the complex logarithm is holomorphic. The question would be : can the function exp(z) + 1 go outside of this domain ?

Answer 3

Summary of previous comments plus.

Convert $e^z+1$ to a form $re^{i\theta}$ Then $ln$ will be $ln(r)+i\theta=u+iv$ for Cauchy-Riemann equations to check analytics. $e^z+1=e^xcosy+1+ie^xsiny$. Therefore $r^2=(e^xcosy+1)^2+(e^xsiny)^2$, $\theta=arctan(\frac{siny}{cosy+e^{−x}})$.

For Cauchy-Riemann you could use $u=\frac{ln(r^2)}{2}$. Also note that although $\theta$ has two possible values, they differ by a constant ($\pi$), so it doesn't effect the analysis.