Find $\int_C f(z)dz$, where $f(z)$ is the principal branch of $z^i$,$|z|>0$ and $-\pi<Argz<\pi$ and $C$ is the semicircle $z=e^{i\theta}$ with $0\leq\theta\leq \pi$
My try:$z^i=e^{i\log z}=e^{i(\ln|z|+iArg z)}$.Then how to proceed to get $f(z)$ ,so that we can solve this problem.Thank you
Just substitute $z = e^{i\theta}$ immediately. $$ \int_Cf(z) dz = \int_0^\pi \left(e^{i\theta}\right)^i(ie^{i\theta}d\theta) = i\int_0^\pi e^{(i-1)\theta}d\theta = \left.\frac{e^{(i-1)\theta}}{1+i}\right|^\pi_0 = -\frac{1+e^{-\pi}}{1+i} $$