Compute the following integrals of $z^i$ Compute the following integrals:


*

*$\int_{\gamma_1}z^idz$, where $\gamma_1(t)=e^{it}, \frac{-\pi}{2}\leq t \leq \frac{\pi}{2}$.

*$\int_{\gamma_2}z^idz$, where $\gamma_2(t)=e^{it}, \frac{\pi}{2}\leq t \leq \frac{3\pi}{2}$.
I think 1. is much easier than 2. since it does not cross the $Re(z)<0$. So for 1.
$$\int_{\gamma_1}z^idz=\int^{\frac{\pi}{2}}_{\frac{-\pi}{2}}(e^{it})^i ie^{it}dt=i\int^{\frac{\pi}{2}}_{\frac{-\pi}{2}}e^{it-t}dt=i\frac{1}{i-1}e^{it-t}|^{\frac{\pi}{2}}_{\frac{-\pi}{2}}=\frac{1-i}{2}\left(e^{-\frac{\pi}{2}+i\frac{\pi}{2}}-e^{\frac{\pi}{2}-i\frac{\pi}{2}}\right)$$
However, for 2. we cannot apply the same method directly... Any help? Thanks!
 A: This is a summary of the comments to the other answer. It seems that you are using this text book, and that this is exercise 4.19. In this chapter(Chapter 4) the integral $\int_\gamma f$ is defined only for functions $f$ that are continuous on $\gamma$, so it would seem that you should pick a branch cut of $\log(z)$ so that $z^i$ is continuous on $\gamma_2$. 
Without further instruction, we don't have a unique branch cut to use, so one could get different answers. But I suspect that it wants you to use a branch cut that coincides with the principal branch on $\{z\in \mathbb C:\Im z > 0\}$. This is the choice
$$\log z = \log |z| + i\arg(z), \quad \arg(z) \in [0,2\pi)$$
In fact, with this choice, $z^i := e^{i \log z}$ is an analytic function on an open neighbourhood of $\gamma_2$, so that with $\frac{d}{dz} \frac{z^{i+1}}{i+1} = z^i$,
\begin{align} \int_{\gamma_2} z^i dz  
&= \left. \frac1{i+1} e^{(i+1)\log z} \right|_{\exp(i\pi/2)}^{\exp(3i\pi/2)} \\
&= \left. \frac1{i+1} e^{(i+1)\log z} \right|_{i}^{-i} \\
&= \frac{e^{i(i+1)3\pi/2} - e^{i(i+1)\pi/2}}{i+1} \\
&= (1/2 - i/2) (-i e^{-3 π/2} - i e^{-π/2}) \\
&\approx -0.108 - 0.108i\end{align}
since $\log i = \operatorname{Log}i = i\pi/2, \log(-i) = i\operatorname{arg}(-i) = i3\pi/2. $ This answer is distinct from the answer obtained by splitting the region into two parts $\gamma_2 \cap \{\Im z<0\}, \gamma_2 \cap \{\Im z>0\}$ and then integrating the function given by the principal log on each curve, as in the other answer; the resulting answer there is approximately $9.03956 - 14.0579 i$.
A: Observe $$z^i = e^{iLn(z)} = e^{iln|z|-Arg(z)} = |z|^ie^{-Arg(z)}$$
$$\therefore, \int_{\gamma_2}z^idz = \int_{\frac{\pi}{2}}^{\frac{3\pi}{2}} |z|^ie^{-Arg(z)} ie^{it}dt = \int_{\frac{\pi}{2}}^{\frac{3\pi}{2}} |e^{it}|^ie^{-Arg(e^{it})} ie^{it}dt$$
$$ = \int_{\frac{\pi}{2}}^{\frac{3\pi}{2}} 1^ie^{-Arg(e^{it})} ie^{it}dt = \int_{\frac{\pi}{2}}^{\frac{3\pi}{2}} e^{-Arg(e^{it})} ie^{it}dt$$
$$ = \int_{\frac{\pi}{2}}^{\pi} e^{-Arg(e^{it})} ie^{it}dt + \int_{\pi}^{\frac{3\pi}{2}} e^{-Arg(e^{it})} ie^{it}dt$$
$$ = \int_{\frac{\pi}{2}}^{\pi} e^{-t} ie^{it}dt + \int_{\pi}^{\frac{3\pi}{2}} e^{-(t-2\pi)} ie^{it}dt$$
$$ = i\int_{\frac{\pi}{2}}^{\pi} e^{-t} e^{it}dt + i\int_{\pi}^{\frac{3\pi}{2}} e^{-(t-2\pi)} e^{it}dt$$
$$ = i\int_{\frac{\pi}{2}}^{\pi} e^{(i-1)t} + ie^{2\pi}\int_{\pi}^{\frac{3\pi}{2}} e^{(i-1)t} dt$$
$$ = i \frac{e^{(i-1)t}}{i-1}|_{\frac{\pi}{2}}^{\pi}  + ie^{2\pi} 
\frac{e^{(i-1)t}}{i-1}|_{\pi}^{\frac{3\pi}{2}}$$

Calvin Khor edit:
By considering a different branch of the logarithm where $Arg(z) \in [0,2\pi)$ instead of $Arg(z) \in (-\pi,\pi]$ s.t. $Ln$ will be differentiable for $Re(z)<0$,
[stuff to say]
