Attempting an old problem sheet question:
Use the Cauchy integral formula to compute (for any $a \in \mathbb{C}$), $$ \int_{|z|=1} \frac{\exp{az}}{2z^2 - 5z+ 2} dz $$
By following the example on wikipedia I arrive at
$$ \int_{|z|=1} \frac{\exp{az}}{2z^2 - 5z+ 2} dz = \frac{4 \pi i\exp{(\frac{1}{2} a)}}{3} $$
But nowhere did I have any consideration for what $a \in \mathbb{C}$ is. My thought is that it doesn't affect anything, since it doesn't change where the poles of $\frac{\exp{az}}{2z^2 - 5z+ 2}$ lie, but is this easily shown? And if it does affect it, how do we work out what affect it has (is it a pre- or post- integrating problem)?
When $f$ has a simple pole at a point $c$ the residue is given by $\lim_{z \to c} f(z)$. The poles in this case are at 2 and 1/2 and only 1/2 is inside the circle.. To find the residue at 1/2 use continuity of the exponential function: $\lim_{z \to 1/2} (z-1/2) \frac {e^{az}} {2z^{2}-5z+2}=e^{(1/2)a}\lim_{z \to 1/2} (z-1/2) \frac {e^{az}} {2z^{2}-5z+2}$ which you can compute by factoring the denominator.