Show that $ \int_{-1}^{1} z^{i} dz= \frac{1 + e^{- \pi}}{2} (1 -i)$

Question

The problem says: Show that $ \int_{-1}^{1} z^{i} dz = \frac{1 + e^{- \pi}}{2} (1 -i)$ with $z^{i}$ the main branch $z^{i} = exp(i log z)$ $(|z|>0, -\pi < Arg z < \pi)$ and where the path of integration is any contour from $z=-1$ to $z=1$.

hint: use the antiderivative branch $z^{i} = exp(i log z)$ $(|z|>0, -\frac{\pi}{2} < Arg z < \frac{3\pi}{2})$.

I need to understand the branch of a complex function. Any book suggestion that talk about branches.

Answer 1

My solution

$\int_{-1}^{1} z^{i} dz = \frac{z^{i+1}}{i + 1} = \frac{(1)^{i+1}}{i+1} - \frac{(-1)^{i+1}}{i+1}$ by the fundamental theorem of calculus.

Then we use the principal branch of $z^{i}$ $->$ $z^{i} = e^{i log z}$.

$\int_{-1}^{1} z^{i} dz = \frac{1}{i+1} (e^{(i+1) log (1)} - e^{(i+1) log (-1)})$

remembered that $log z = ln |z| + arg z$. Then,

$\int_{-1}^{1} z^{i} dz = \frac{(i+1)[1 + e^{-\pi}]}{2}$

Answer 2

By definition, $z^i=e^{i\log z}$ where $\log z=\ln |z|+i\theta: -\pi\le \theta \le \pi.$ Now then,

$\int _{[-1,1]}z^idz=\frac{z\cdot z^i}{i+1}\bigg | _{-1}^1=\frac{1}{i+1}(1+e^{-\pi})=\frac{1 + e^{- \pi}}{2} (1 -i).$