could you help me with this question? I am a beginner and I struggling with it
Consider a curve $\alpha$ connecting the points $1$ and $-1$ in which is found above the axis X, except in the extreme. Be the function
$f(z)= z^i:=e^{i ln z}$
where $lnz$ is the main log of z
a. Calculate $\int _{\alpha} f(z)dz $, assuming that $\alpha$ is the circuference $\alpha (t) =e^{it}$, $t$ $belongs$ $[-\pi, \pi]$
the answer given is $(\frac{-1}{2} + \frac{1}{2}i)(e^{-\pi} +1)$
b. Why is there not the antiderivative of $f$ in {$ z: -\pi < arg z\leq \pi $}
c. Verify that if $g(z)=e^{iLn(z)}$ , where $Ln (z)$ is the branch of the log defined in {$ \frac{-\pi}{2} < arg z\leq \frac{3\pi}{2} $} so $g$ and $f$ coincide and $\alpha$
d. Find the primitive of g
the answer given is $ \frac{z^{i+1}}{i+1}$
e. Use (c) and (d) to computer $\int _{ \alpha} z^i: $
the answer given is $ \frac{1+e^{-\pi}}{2} (1-i) $
For the first, I did substitute $\alpha(t)=e^{it}$ so \begin{align} \int_\alpha z^i\,\mathrm{dz} &= \int_{-\pi}^{\pi}e^{iit}ie^{it}\,\mathrm{dz} \\ &= i\int_{-\pi}^{\pi}e^{t(i-1)}\,\mathrm{dz} \\ &= \dfrac{i}{i-1}\left[e^{\pi(i-1)}-e^{-\pi(i-1)}\right] \hspace{1cm};\hspace{1cm}e^{\pi}=e^{-\pi}=-1\\ &= \dfrac{1-i}{2}\left[e^{\pi}-e^{-\pi}\right] \\ &= (1-i)\sinh\pi \end{align} but it seems different from yours!