How do I prove that there is a complex logarithm?

Question

I have the following problem:

Prove that there exists an analytic function $$f:\{z\in \Bbb{C}:|z-1|<1\}\rightarrow \Bbb{C}$$ such that $e^{f(z)}=z$ when $|z-1|<1$.

We have just shown this fact for the $l$-th root but somehow I did not see it now where to start. I know the analytic function which describes $e^z$ maybe this helps?

Can someone give me a hint?

Thanks for your help

Answer 1

One rather pedestrian way to do this is by constructing directly a sort of logarithm function. Here is one approach.

Given a real number $\theta_0$ and using polar coordinates, every $z\in\mathbb{C}\setminus\{0\}$ can be expressed uniquely in the form $$z=re^{i\theta}=r(\cos\theta + i \sin\theta)$$ where $r=|z|$ and $\theta\in[\theta_0,\theta_0+2\pi)$.

• The angle $\theta$ is called argument of $z$, which we denote by $\operatorname{arg}_{\theta_0}(z)$. Let $\ell_{\theta_0}=\{rx^{i\theta_0}:r\geq0\}$, $\Omega_{\theta_0}=\mathbb{C}\setminus\ell_{\theta_0}$. Define a function $L_{\theta_0}:\Omega_{\theta_0}\rightarrow\big(\mathbb{R}\setminus\{0\}\big)\times(\theta_0,\theta_0+2\pi)$ by \begin{align} z\mapsto \log(|z|) +i\operatorname{arg}_{\theta_0}(z), \end{align} where $\log$ is the usual logarithm function on the real line.
• $L_{\theta_0}$ is bijective function whose inverse is the exponential function restricted to $\big(\mathbb{R}\setminus\{0\}\big)\times(\theta_0,\theta_0+2\pi)$.
• Since $\exp\in H(\mathbb{C})$ and $\exp'=\exp\neq0$, $L_{\theta_0}\in H(\Omega_{\theta_0})$ by the inverse function theorem for holomorphic functions, and $L'_{\theta_0}(z)=\frac{1}{z}$ for $z\in\Omega_{\theta_0}$ (see remark below).
• The function $L_{\theta_0}$ is called the $\theta_0$--branch of logarithm. The branch $L_{-\pi}$ is called the principal branch of logarithm. When the branch of logarithm is clear from the context, we use $\log$ to denote the function $L_{\theta_0}$.

Notice that the ball $B(1;1)=\{z:|z-1|<1\}\subset \Omega_{-\pi}$ and so, $e^{L_{-\pi}(z)} = z$ for all $z\in B(1;1)$.

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Remark: The complex inverse function states that if $f:D\rightarrow U$, $D,U$ regions in $\mathbb{C}$, is analytic and bijective, and $f'\neq0$ on $D$, then $f^{-1}:U\rightarrow D$ is also analytic and $(f^{-1})'(w)=\frac{1}{f'(f^{-1}(w))}$. This can be proved by using methods of Calculus on the plane plus checking that the Cachy-Riemann equations are satisfied by $f^{-1}$.

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