# evaluate: $\oint_c w \frac{d}{dw} \left( \frac{\phi (w)^n}{(w-a)^n}\right) dw$

If $\phi(w)$ is analytic in the circle with center $a$ then how to evaluate the integral? $$\oint_c w \frac{d}{dw} \left( \frac{(\phi (w))^n}{(w-a)^n}\right) dw$$

In second last line of the above picture, I don't understand how it is evaluated. Can anyone explain? Thanks in advance!!

• I solved the problem you originally posted. Please think hard before making such changes to your problem, as people are likely working hard on your original post. – Ron Gordon Apr 19 '13 at 12:13

$$\frac{d}{dw} \left [\frac{\phi(w)}{w-a} \right ]^n = n \left [\frac{\phi(w)}{w-a} \right ]^{n-1} \left [\frac{\phi'(w)}{w-a} - \frac{\phi(w)}{(w-a)^2}\right] = n [\phi(w)]^{n-1} \left [\frac{\phi'(w)}{(w-a)^n} - \frac{\phi(w)}{(w-a)^{n+1}} \right ]$$
$$\oint_C dw \, w \frac{d}{dw} \left( \frac{(\phi (w))^n}{(w-a)^n}\right) = i 2 \pi a n! \left [\frac{d^n}{dw^n} [\phi(w)^{n-1} \phi'(w)] \right ]_{w=a} \\+ i 2 \pi a n \cdot n! \left [\frac{d^{n+1}}{dw^{n+1}} \phi(w)^n \right ]_{w=a}$$
• If $\phi$ is linear, then both terms will vanish. If $\phi'(a)=0$, then the first term vanishes. Otherwise, not sure. – Ron Gordon Apr 19 '13 at 12:23