About branch of logarithm Let $U= \mathbb{C} - \{ it: t \leq 0 \text{ in } \mathbb{R}\}$. Let $L:U \rightarrow \mathbb{C}$ be the branch of logarithm given by $L(z)= \int _{\gamma} \frac{dw}{w}$, for $z\in U$, $\gamma: [a,b] \rightarrow U$ is any piecewise smooth curve with $\gamma(a)=1$ and $\gamma(b)=z$. Show that for $r>0$ and $-\frac{\pi}{2} < \theta < \frac{\pi}{2}$ we have $L(re^{i\theta})= \log r + i\theta$.
I see here that the branch cut is the lower $y$-axis.
In the usual definition, we already have  $L(re^{i\theta})= \int_{\gamma} \frac{1}{re^{i\theta}}d\theta=\log r + i\theta$.
Maybe I'm understanding what the question is asking. The part I'm confused about is curve $\gamma$. How does $\gamma$ fit in this question? Where do we need $\gamma$ in solving this question?
 A: You can choose any piecewise-smooth curve $\gamma$ (in $U$) from $1$ to $z$ to compute $L(z)$. Take $\gamma$ to be the straight line path from $1$ to $re^{i\theta}$. Let $\phi$ be the angle from the positive $x$-axis to the segment. Let $\mathscr{r} > 0$ such that $1 + \mathscr{r}e^{i\phi} = re^{i\theta}$. Then $\gamma$ is parametrized by $w = 1 + \rho e^{i\phi}$, $0 \le \rho \le \mathscr{r}$. Therefore \begin{align*}L(re^{i\theta}) &= \int_0^{\mathscr{r}} \frac{e^{i\phi}\, d\rho}{1 + \rho e^{i\phi}} = \int_0^\mathscr{r} \frac{d\rho}{e^{-i\phi} + \rho} \\&= \int_0^\mathscr{r} \frac{\rho + \cos \phi}{(\rho + \cos\phi)^2 + \sin^2\phi}\, d\rho + i\int_0^\mathscr{r} \frac{\sin \phi}{(\rho + \cos \phi)^2 + \sin^2\phi}\, d\rho\\
&= \frac{1}{2}\log[(\rho + \cos \phi)^2 + \sin^2\phi]\big|_{\rho = 0}^{\rho = \mathscr{r}} + i\arctan\left[\frac{\rho + \cos \phi}{\sin \phi}\right]\bigg|_{\rho = 0}^{\rho = \mathscr{r}}\\
&= \frac{1}{2}\log(\mathscr{r}^2 + 2\mathscr{r}\cos \phi + 1) + i\left\{\arctan\left[\frac{\mathscr{r} + \cos \phi}{\sin \phi}\right] - \arctan\left[\frac{\cos \phi}{\sin \phi}\right]\right\}\\
&= \frac{1}{2}\log|1 + \mathscr{r}e^{i\phi}|^2 + i\arctan\left[\frac{\mathscr{r}/\sin \phi}{1 + (\mathscr{r} + \cos\phi)\cos \phi/\sin^2\phi}\right]\\
&= \frac{1}{2}\log r^2 + i\arctan\left[\frac{\mathscr{r}\sin \phi}{1 + \mathscr{r}\cos\phi}\right]\\
&= \log r + i\arctan\left[\frac{r\sin \theta}{r\cos \theta}\right]\\
&= \log r + i\theta\end{align*}
