$\log(e^z - i)$ as a holomorphic function in $\mathbb{D}$

Question

I'm learning complex analysis, specifically holomorphic functions, and need help with the following exercise:

Examine if the function $\log(e^z - i)$ can be defined as a holomorphic function in the unit disk $\mathbb{D}$.

I wasn't able to do much except for this:

Let $f(x) = e^z - i$. We note that $e^z - i = 0 \iff e^xe^{yi} = e^{{\pi \over 2}i} \iff x = 0$ and $y = \frac{\pi}{2}$. The point $(0, \frac{\pi}{2}) \notin \mathbb{D}$ and therefore $f(z) \neq 0$ in $\mathbb{D}$.

This is as far as I got. I suspect the solution has to do with the existence of holomorphic logarithms on simply connected domains but I'm having difficulties understanding this concept.

Answer 1

You have already proved that $e^z-i$ is nowhere zero on $\mathbb{D}.$

$\textbf{Claim:} ~$If $f$ is analytic and nowhere zero in a simply connected domain $U$, then there exists a holomorphic branch of $\text{log}~f$. To prove this, you'll need to use the fact that a holomorphic branch of $\text{log}~f$ exists if and only if $\dfrac{f'}{f}$ has a primitive in $U$.

$\textbf{Explanation:}$ When we say a holomorphic branch of logarithm of a holomorphic function $f$ exists, we mean that $\exists$ a holomorphic function$~h$ such that $f=e^h$. Note that $\log f$ does not denote the usual branch of the logarithm on the image of $f$. It may not exist all the time, for example, consider $\exp:\mathbb{C}\to\mathbb{C_*}$.

Suppose $U$ is simply connected, $f:U\to \mathbb{C_*}, $ and a holomorphic branch of $\log f$ exists. Note that since $0 \not \in Im(f)$, $\dfrac{f'}{f}$ is well-defined and analytic on $U$. From $$f=e^h$$

we have, $$f'=h'e^h$$

or in other words,$$\dfrac{f'}{f}=h'$$

Which means that $\dfrac{f'}{f}$ has a primitive, namely $h$.

Now assume that $\dfrac{f'}{f}$ has primitive, say $g$. Since $g$ is holomorphic and $U$ is simply connected, $g$ has a primitive, say $h$ in $U$. Then, \begin{align*} &h'=g=\dfrac{f'}{f} \\ \implies&\bigg(fe^{-h}\biggl)^{'}=0\\ \implies&fe^{-h}=c\neq0\\ \implies&f=e^{h+k},~~~~\text{where} ~c=e^{k}. \end{align*}

Thus, $h+k$ is a holomorphic branch of the logarithm of $f$.

It only remains to prove that if $U$ is simply connected and $f$ is holomorphic in $U$, then $f$ has a primitive in $U$. This I leave as an exercise.

Answer 2

Since $$ e^{\rho e^{i\theta}}=e^{\rho\cos\theta}\cdot e^{i\rho\sin\theta}=e^{\rho\cos\theta}\cos(\rho\sin\theta)+i e^{\rho\cos\theta}\sin(\rho\sin\theta)$$ the map $z\to e^{z}$ sends the unit disk in a subset of $\text{Re}(z)\geq\color{red}{\large\frac{\cos(1)}{e}}$, and so does the map $z\to e^{z}-i$, whose range (depicted below) stays away from zero in the right half-plane.

The logarithmic map is regular over such a set, hence the original function is holomorphic over the unit disk.

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