How to prove that $ \oint _{C(0,1/\pi)} \frac {e^{a/z}}{(z+2)^2} dz = \oint _{C(0,\pi)} \frac {e^{az}}{(2z+1)^2}dz $ I tried to parametrize each integral:
The first one with $h_1 (t) = \frac {1}{\pi} e^{it} $ from 0 to 2$\pi$
The second one with $h_2 (t) = \pi e^{it} $ from 0 to 2$\pi$
But it got complicated...
 A: The usual trick is using residues. For the first function use the Laurent series at $z = 0$ (the singularity at $z=-2$ is irrelevant). For the second function use the usual formula for a pole of order 2.
A: Make the change of variable $w=1/z$.  Now, $dw=-\frac1{z^2}\,dz$.
$$\begin{align}
dw=-\frac1{z^2}\,dz & \implies -z^2\,dw=dz\\
& \implies dz=-\frac1{w^2}\,dw
\end{align}$$
Let $\gamma(t)=\frac1\pi e^{it}$ for $t\in[0,2\pi]$ and $\gamma'(t)=\pi e^{it}$ for $t\in[0,2\pi]$.  If $f(z)=1/z$, then $f^{-1}(\gamma)=-\gamma'$. So,
$$\begin{align}
\int_\gamma\frac{e^{a/z}\,dz}{(z+2)^2} & = -\int_{-\gamma'}\frac{e^{aw}\,dw}{w^2(1/w+2)^2}\\
& = \int_{\gamma'}\frac{e^{aw}\,dw}{(2w+1)^2}.
\end{align}$$
A: Use residue theorem. For the first integral we have singularities at
$$z=-2,\quad z=0$$
but only $z=0$ lies inside the circle $C(0,1/\pi)$ so
$$\oint_{C(0,1/\pi)} \frac{e^{a/z}}{(z+2)^2}\,dz=2\pi i \operatorname*{Res}_{z=0}\frac{e^{a/z}}{(z+2)^2}$$
Now we have to turn it into series as the following:
$$\frac{e^{a/z}}{(z+2)^2}=
\sum_{n=0}^\infty \frac{n+1}{(-2)^{n+2}}z^n
\sum_{n=0}^\infty \frac{a^n}{n!z^n}$$
The residue is the coefficient of $z^{-1}$:
$$\operatorname*{Res}_{z=0}\frac{e^{a/z}}{(z+2)^2}=\sum_{n=0}^\infty\frac{n+1}{4(-2)^n}\cdot\frac{a^{n+1}}{(n+1)!}
=\frac{a}{4}\sum_{n=0}^\infty\frac{1}{n!}\left(-\frac{a}{2}\right)^n=\frac{ae^{-a/2}}{4}$$
The second integral has a singularity (a pole) at $z=-1/2$ so
$$\oint_{C(0,\pi)} \frac{e^{az}}{4(z+1/2)^2}\,dz=
2\pi i \operatorname*{Res}_{z=0}\frac{e^{az}}{4(z+1/2)^2}$$
Calculating the residue
$$\operatorname*{Res}_{z=0}\frac{e^{az}}{4(z+1/2)^2}=\lim_{z\to-1/2}\frac{d}{dz}\frac{e^{az}}{4}=\frac{ae^{-a/2}}{4}$$
we conclude that 
$$\oint_{C(0,1/\pi)} \frac{e^{a/z}}{(z+2)^2}\,dz=\oint_{C(0,\pi)} \frac{e^{az}}{(2z+1)^2}\,dz=\frac{i\pi a e^{-a/2}}{2}$$
A: Hint : try variable change : $u=\frac{1}{z}$
