Hey alphabeta.
Can you use the Cauchy-Formula and evaluate f(z) at alpha? (Since f(z) has no singularities in the interior of the contour)?
First, we need to check that there are not any singularities. The only place where there may be a singularity is where . But this is not possible since on the contour , for that denominator to be 0, , but we know . Now since is the unit circle oriented in the counterclockwise direction centred at the origin, we can parameterise it as with . Notice that and the integral becomes
Now let and note that and giving