Show holomorphic branch of $\log z$ on $\Omega=\mathbb{C}\backslash \{z\text{ real } : z\leq 0\}$ satisfying $\log(1)=0$.

Question

Show holomorphic branch of $\log z$ on $\Omega=\mathbb{C}\backslash \{z\text{ real } : z\leq 0\}$ satisfying $\log(1)=0$. Also satisfies $|Im\log(z)|<\pi$.

Proof: Recall the polar form of the Cauchy-Riemann equations: $u_r=\dfrac{1}{r}v_\theta$ and $u_\theta=-\dfrac{1}{r}v_r$ (If a proof for these equations is necessary, one can prove these equations by using the multi-variable chain rule on $z=x+iy=r\cos\theta+ir\sin\theta$ and the regular Cauchy-Riemann equations.)

If $\log(z)$ is holomorphic, then it must satisfy the Cauchy-Riemann equations. Define $\log(z)=\ln r + i(\theta +2\pi k)$, where $r=|z|$, $\theta=arg(z)$ and $k\in\mathbb{Z}$. Then $u_r=\dfrac{1}{r}$, $v_r=0$, $u_\theta=0$, and $v_\theta=1$.

Furthermore, \begin{gather*} \dfrac{1}{r}=u_r=\dfrac{1}{r}v_\theta=\dfrac{1}{r}\cdot 1 \\ 0 = v_r =-\dfrac{1}{r}u_\theta=0 \end{gather*} i.e. The Cauchy-Riemann equations (in polar form) are satisfied.

But note that $\log(z)$ isn't a function because for every $z$ there are infinite solutions. So, we would like to restrict it, then $\log(z)$ is a function. Look at the following $$ |Im \log(z)|=|arg(z)|<\pi $$ This is true because $[0,2\pi]$ can be defined as $[-\pi,\pi]$.

So $\log(1)=\ln 1+i(0+2\pi k)=0+2\pi ki$. By what we just showed, $k=0$, which implies $\log(1)=0$.

Questions:

1. Do I need to show that $\log z$ is holomorphic?
2. Is my argument for $|Im \log z|<\pi$ correct?
3. Same question for $\log(1)=0$.

Answer 1

My interpretation is that you are asked to prove a uniqueness result: the principal branch of logarithm satisfies the given condition and you have to show that there is no other branch: suppose f and g are an analytic branches with $f(1)=g(1)=0$. Then $e^{f(z)} =e^{g(z)}$ for all z. This implies that $f(z)=g(z)+2n(z)\pi i$ for some integer $n(z)$. However f and g are continuous, so $z \to n(z)$ is a continuous integer valued function on a connected set, so is is constant, say $n$. Now put $z=1$ in $f(z)=g(z)+2n \pi i$ to see that $n=0$. Hence $f=g$ and the principal branch is the only branch that vanishes at 1. Since $|\Im f(z)| < \pi /2$ for the principal branch we are through.