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