Compute $ \int _{|z|=1} z^{-1+i} dz$ using the branch of log defined in {${z:0 <\arg (z) < 2\pi}$}

Question

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Compute $ \oint _{|z|=1} z^{-1+i} \,dz$ using the branch of log defined in {${z:0 \le \arg (z) < 2\pi}$}

The solution given is $i(1-e^{-2\pi})$

Answer 1

With the branch cut chosen along the positive real axis we write $z=e^{i\phi}$, $\phi \in [0,2\pi]$. Then, we have

$$\begin{align} \oint_{|z|=1}z^{-1+i}\,dz&=\int_0^{2\pi} \frac{1}{e^{i\phi}}e^{i\log(e^{i\phi})}\,ie^{i\phi}\,d\phi\\\\ &=i\int_0^{2\pi}e^{-\phi}\,d\phi\\\\ &=i(1-e^{-2\pi}) \end{align}$$

as expected!