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

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

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  • $\begingroup$ 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. $\endgroup$ – Ron Gordon Apr 19 '13 at 12:13
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Write

$$\frac{d}{dw} \left [\frac{\phi(w)}{w-a} \right ]^n = n \left [\frac{\phi(w)}{w-a} \right ]^{n-1} \left [\frac{\phi'(w)}{w-a} - \frac{\phi(w)}{(w-a)^2}\right] = n [\phi(w)]^{n-1} \left [\frac{\phi'(w)}{(w-a)^n} - \frac{\phi(w)}{(w-a)^{n+1}} \right ]$$

The integral is then

$$\oint_C dw \, w \frac{d}{dw} \left( \frac{(\phi (w))^n}{(w-a)^n}\right) = i 2 \pi a n! \left [\frac{d^n}{dw^n} [\phi(w)^{n-1} \phi'(w)] \right ]_{w=a} \\+ i 2 \pi a n \cdot n! \left [\frac{d^{n+1}}{dw^{n+1}} \phi(w)^n \right ]_{w=a}$$

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  • $\begingroup$ ok ok thank you!! how does first term vanish?? I mean could we make it vanish?? $\endgroup$ – hasExams Apr 19 '13 at 12:15
  • $\begingroup$ If $\phi$ is linear, then both terms will vanish. If $\phi'(a)=0$, then the first term vanishes. Otherwise, not sure. $\endgroup$ – Ron Gordon Apr 19 '13 at 12:23
  • $\begingroup$ thanks anyway, i found the more generalized proof here $\endgroup$ – hasExams Apr 19 '13 at 23:53

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