So I need to compute
$$\int_{\vert z \vert = 1} ze^{2z}dz.$$
Attempt:
I used integration by parts to get
$$\frac{ze^{2z}}{2}\bigg\vert_0^{2\pi} - \int_0^{2\pi} \frac{1}{2}e^{2z}dz$$
which is
$$\pi e^{4\pi} - \bigg(\frac{1}{4}e^{2z} \bigg) \bigg\vert_0^{2\pi}$$
something seems off however, am I on the right track? Namely, are my bounds correct?
On
To add yet another perspective on this, we can compute this by explicitly finding an anti derivative for $f(z)=ze^{2z}$. One antiderivative is $F(z)=\frac14\left(2z-1\right)\mathrm{e}^{2z}$.
The contour integral of $f$ along a contour $\gamma$ beginning at $z_0$ and ending at $z_1$ is: $$\oint_\gamma z e^{2z}dz=F(z_1)-F(z_0)$$ Since in our case $\gamma$ is closed, $z_0$ and $z_1$ will be the same (say $z_0=z_1=0$) and the contour integral is zero.
If you wish to neglect application of Cauchy's Integral Theorem, and instead evaluate the integral directly, then we may proceed as follows.
On the unit circle, we note that $|z|=1$ and we may parameterize $z$ as $z=e^{i\phi}$, $\phi\in [0,2\pi]$. Therefore, we have
$$\begin{align} \oint_{|z|=1}ze^{2z}\,dz&=i\int_0^{2\pi}e^{i2\phi}e^{2e^{i\phi}}\,d\phi\\\\ &=\left.\left(\frac14e^{2e^{i\phi}}(2e^{i\phi}-1)\right)\right|_0^{2\pi}\\\\ &=0 \end{align}$$
as expected!