I have got a question about the calculation of two specific integrals: \begin{align*} &\int_C z^n e^z \, dz \quad n\in\mathbb{N}_0 \\ &\int_{\vert z \vert = 1} z^{-n} e^z \, dz \quad n\in\mathbb{N} \end{align*} Where $C$ is the line from $0$ to $2 \pi i$ and $\vert z \vert = 1$ is the circle with radius $1$. Hint is to use cauchy's integral theorem. At this moment I tried solving the integrals using the line integrals but this leads to a lot of calculations which seem to be unsatisfying. Do you have an idea or parametrization for $C$ or $\vert z \vert =1$ which does 'work well'? Furthermore I am not quite sure whether we can use the CIT for the second integral because $\mathbb{C}-\{0\}$ is no star-domain.
Thank you in advance.
For the first integral just use $z=iy$,
$$ A_n = \int_C z^n e^z dz= \int_0^{2\pi} (iy)^n e^{iy} idy $$
$$= \int_0^{2\pi} (iy)^{n+1} e^{iy} dy $$
$$= (i)^{n+1}\int_0^{2\pi} y^n e^{iy} dy $$
This last integral can be obtained by using differentiation under the integral sign on,
$$ I(\alpha) = i\int_0^{2\pi} e^{i\alpha y}dy = i\frac{1}{i\alpha} \left( e^{2\pi i \alpha} - 1 \right),$$
Where
$$A_n = i\left(\frac{d}{d\alpha}\right)^n I(\alpha) \mid_{\alpha =1} $$
For the second integral over $\vert z \vert =1 $ we can use Cauchy's integral theorem in its generic form,
$$ \int_C \frac{f(z)}{(z-z_0)^{n+1}} dx = \frac{ 2\pi i}{n!} f^{(n)}(z_0), $$
Since the derivative of $e^z$ is itself and $z_0=0$ we can easily compute $\int_C z^{-n} e^z dz=2\pi i / (n-1)!$.