There's also a simpler way of doing this. Observe that the function is holomorphic except at $z=a$ and $z=\dfrac{1}{a}$, then note your contour is the unit circle, as some other answers have discussed (if you're lost on this point, observe as a quick and easy method that generally contours with angles are circles or parts of circles, that we have $0$ to $2 \pi$ here, so clearly talking about one circle and you have $e^{it}$ which has modulus one so a unit radius). Now as $0 < a < 1$, $\frac{1}{a}$ is outside the circle so only need to look at $a$. You could use a Laurent series as previous answer has it, but this is quite messy (often one can avoid using Laurent's theorem). Now, there is a theorem (consult your textbook for a proof) that says a point is a pole of order m if $f(z)$ can be written as $\dfrac{\phi(z)}{{(z-z_0)}^m}$ where $\phi(z)$ is analytic and nonzero at $z_0$. Moreoever, the residue at $z_0$ is $\phi(z_0)$ if $m=1$ and $\dfrac{\phi^{m-1}(z_0)}{(m-1)!}$ if $m \geq 2$. So, here we note for $z=a$ we can set $\phi(z)=\dfrac{1}{z-1/a}$ and then we have that $\phi(a)=\dfrac{1}{a-1/a}$ and we note that $m=1$ in this case, so this is the residue and we are done.