Finding an upper bound for $|\int_{\gamma}e^{1/z}dz|$. Find the upper bound for $|\int_{\gamma}e^{1/z}dz|$ where $\gamma$ is the part of the circle $|z| = \sqrt{8}$ from $2+2i$ to $-\sqrt{8}$.  This is a question my professor went through quickly during class and so I am trying to fill in the blanks.
I know we will use the mL inequality here.  $L$ is pretty easy to find (just the length of the line), but I am struggling on how to find the maximum of $e^{1/z}$ on this interval.
This is where I am:
Using the fact that $|z|=\sqrt{8}$ and $x^2+y^2=8$, then the $e^{1/z}$ has a max of $e^{1/ \sqrt8}$.
But this is where I get stuck...  
 A: Your first step is right as by the estimation lemma: 
$$\left |\int_\gamma e^{\frac{1}{z}} \mathrm{d}z \right | \leq \mathrm{L}(\gamma) \cdot \max_{z \in \gamma}\left | e^{\frac{1}{z}} \right |$$
Where $\mathrm{L}(\gamma)$ is the length of the path.
Now we use the fact that $| e^z | = e^{\mathrm{Re}(z)}$ and that $\mathrm{Re}\left (\frac{1}{z}\right )=\frac{\mathrm{Re}(z)}{|z|^2}$ to obtain: $$\left |\int_\gamma e^{\frac{1}{z}} \mathrm{d}z \right | \leq \mathrm{L}(\gamma)\cdot \max_{z \in \gamma} e^{\frac{\mathrm{Re}(z)}{|z|^2}} = \underbrace{\sqrt 8 \cdot \frac{\color{green}3\pi}{4}}_{\mathrm{L}(\gamma)} \qquad \cdot \underbrace{\max_{t \in [\frac{\pi}{4},\ \color{green}{\pi}]} e^{\frac{\sqrt{8} \cos t}{(\sqrt8)^2}}}_{\text{Path parametrized in polar coordinates}}$$
or if the path is the other way round (through $\scriptsize{+} \normalsize\sqrt 8$): $$\underbrace{\sqrt 8 \cdot \frac{\color{green}5\pi}{4}}_{\mathrm{L}(\gamma)} \qquad \cdot \underbrace{\max_{t \in [\frac{\pi}{4},\ \color{green}{-\pi}]} e^{\frac{\sqrt{8} \cos t}{(\sqrt8)^2}}}_{\text{Path parametrized in polar coordinates}}$$
Take it from here :-)
