Determination of logarithm (complex)

[quasar987]

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?

 

[Hurkyl]

I imagine it would depend on what you do next.

 

[quasar987]

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.

 

[Hurkyl]

So do you really need to specify a branch of log on the entire curve? You're only using it to prove an algebraic identity.

 

[quasar987]

I hear you!