How can i find this complex integral?
$\int _{\Gamma} z^{i}dz$
where $\Gamma(t) =1-t^2+it, \quad -1 \leq t \leq 1$.
Thank you for your help.
How can i find this complex integral?
1
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complex-analysis
1 Answers
1
Note that $z^i$ is the multivalued function
$$\begin{align} z^i=e^{i\log(z)}&=e^{i(\log(|z|)+i\text{Arg}(z)+i2n\pi)}\\\\ &=e^{i\log(|z|)-\text{Arg}(z)-2n\pi} \end{align}$$
where $n$ is any integer and where $-\pi< \text{Arg}(z)\le \pi$ is the principal value of the argument of $z$. If we choose the principal branch of the logarithm to define $z^i$, then $z^i$ is holomorphic on $\mathbb{C}\setminus [-\infty,0)$ and we can write
$$\begin{align} \int_\Gamma z^i\,dz&=\left.\left(\frac{z^{i+1}}{i+1}\right)\right|_{-i}^{i}\\\\ &=\frac{i^{i+1}-(-i)^{i+1}}{i+1}\\\\ &=\frac{e^{(i+1)\log(i)}-e^{(i+1)\log(-i)}}{i+1}\\\\ &=\frac{e^{(i+1)i\pi/2}-e^{-(i+1)i\pi/2}}{i+1}\\\\ &=\frac{ie^{-\pi/2}+ie^{\pi/2}}{i+1}\\\\ &=(1+i)\cosh(\pi/2) \end{align}$$
and we are done!