Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

$f,g$ are holomorphic in $D(0,1)$. $P_1,P_2,...,P_k$ are roots of $f$ in $D(0,1)$. their orders are $n_1,...,n_k$. Compute $$\frac{1}{2\pi i}\oint_\gamma\frac{f'(z)}{f(z)}\cdot g(z)dz.$$

Using residue theorem I have $$ \frac{1}{2\pi i}\oint_\gamma\frac{f'(z)}{f(z)}\cdot g(z)dz=\sum_{j=1}^k\frac{1}{(n_j-1)!}\left(\frac{\partial}{\partial z}\right)^{n_j-1}\left[(z-P_j)^{n_j}\frac{f'(z)}{f(z)}g(z)\right]_{z=P_j}. $$

But someone says $$ \frac{1}{2\pi i}\oint_\gamma\frac{f'(z)}{f(z)}\cdot g(z)dz=\sum_{j=1}^kn_jg(P_j). $$ I don't know how to get it. Any hint is appreciated.

share|improve this question
1  
Have you actually tried taking the derivative in your formula? –  Andrew Nov 30 '12 at 20:52
1  
Can I suggest you to give an on topic title to the question? –  Pragabhava Nov 30 '12 at 20:55
add comment

1 Answer 1

up vote 2 down vote accepted

You should use the argument principle: $$\frac{1}{2\pi i}\oint_\gamma\frac{f'(z)}{f(z)}\cdot g(z)dz=\sum_{\mbox{zeroes of } f}n(\gamma,a)g(a)-\sum_{\mbox{poles of } f}n(\gamma,b)g(b)=\sum_{j=1}^kn_jg(P_j)$$

share|improve this answer
    
Thank you very much, @Dennis. Your answer and link are very helpful. –  Sam Nov 30 '12 at 21:11
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.