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## determination of logarithm (complex)

The question is this:

Consider p(z) a polynomial and C a closed path containing all the zeroes of p in its interior. Compute

$$\frac{1}{2\pi i}\int_C z\frac{p'(z)}{p(z)}dz$$

The solution given by the manual starts by saying that

$$\frac{p'(z)}{p(z)}=(log(p(z)))'$$.

But there is no determination of log(p(z)) on C. Isn't that a problem?
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 Recognitions: Gold Member Science Advisor Staff Emeritus I imagine it would depend on what you do next.
 Recognitions: Gold Member Homework Help Science Advisor Next we differentiate log(p(z)), it gives $$z\frac{p'(z)}{p(z)} = \sum_{j=1}^k \frac{zn_j}{z-zj}$$ where zj are the zeros of p(z) and nj their relative order of multiplicity. Then we use the residue theorem on the integral ansd use the fact that each zj is a simple pole for $\sum_{j=1}^k \frac{zn_j}{z-z_j}$ such that $$\sum_{j=1}^k Res(z\frac{p'(z)}{p(z)},z_j) = \sum_{j=1}^k \lim_{z\rightarrow z_j}(z-z_j)\frac{zn_j}{z-z_j} = \sum_{j=1}^k z_jn_j$$ to get the result.

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