Find $\int_{\gamma} e^zz^n dz$ where $\gamma$ is the unit circle, using Cauchy's Integral Formula

I'm been banging my head against the wall trying to solve the following question which ask to solve the following integral using the Cauchy integral formula, and hence evaluating the corresponding real integrals.

$$\int_{\gamma} e^zz^n dz$$ where $$\gamma$$ is the unit circle {$$e^{i\theta}: -\pi \leq \theta \leq \pi$$} and $$n\in \mathbf{Z}$$.

To solve the question, I'm attempting to use the generalised form of the Cauchy integral formula. Although to use it, the $$z^n$$ is normally in the denominator not the numerator.

Any help will be much appreciated, thanks!!

• Hey. Is this from math2621? – user557493 Oct 7 '18 at 6:02
• @Bell yep it is! – Mr Bro Oct 7 '18 at 9:52
• I'm guessing you do math2221 as well. Are your group A or B? haha – user557493 Oct 7 '18 at 23:24

That integral is equal to $$0$$ if $$n\geqslant0$$. In fact,\begin{align}\int_\gamma e^zz^n\,\mathrm dz&=\int_\gamma\frac{e^zz^{n+1}}z\,\mathrm dz\\&=2\pi ie^00^{n+1}\\&=0.\end{align}On the other hand, if $$n<0$$, then\begin{align}\int_\gamma e^zz^n\,\mathrm dz&=\int_\gamma\frac{e^z}{z^{-n}}\,\mathrm dz\\&=\frac{2\pi i}{(-n-1)!}\exp^{(-n-1)}(0)\\&=\frac{2\pi i}{(-n-1)!}.\end{align}
• The question is for $n \in \mathbb Z$ not $n \in \mathbb N$. – Kavi Rama Murthy Sep 27 '18 at 8:46
Let $$f(z)=e^{z}$$. Then $$f^{(k)}(0)=\frac {k!} {2\pi i} \int_{\gamma} \frac {f(z)} {z^{k+1}}dz$$. Use this when $$n<0$$ in your integral.