Complex logarithm function

Question

I need some help understanding the logarithm function in complex plane. Let $w,z\in \mathbb{C}$. Define $$w=e^z$$ when $$z=\log(w).$$ Now I understand the representation of $$\log(z)=\log|z|+i(\arg(z)+2\pi k)$$ where $k\in \mathbb{Z}$. What I do not understand is how to tell when $\log(z)$ will be holomorphic? For example what is biggest open set that $\log(z^5+1)$ is holomorpic? My intuition tells me on $\mathbb{C}-\{z\in \mathbb{R} :z\le-1\} $ for $\log(z^5+1)$, but I am not sure how to prove this. What would the derivative be is it $$\log(z^5+1)=\frac{5z^4}{z^5+1}$$

Answer 1

The principal branch of the natural logarithm $\ln(z)=\ln(|z|)+i\arg(z)$ is analytic on the open disk, $D(1,1)$, centered at $z_0=1$ and of radius 1, and for $z\in D$ we have $\ln(z)=\sum_{n=1}^\infty (-1)^{n+1}\displaystyle\frac{(z-1)^n}{n}$. Hence $\ln(z)$ is holomorphic on this disk. Your function is holomorphic if $z^5+1$ belongs to the disc $D$, i.e. if $|z^5+1-1|<1$, which is equivalent to saying that $|z^5|<1$ or $|z|<1$. Hence the function $\ln {(z^5+1)}$ is holomorphic on the disk $D'(0,1)$, centered at $0$ and of radius 1. On this disk the derivative is computed as in the real case.