# Math Help - Need proof of theorem about log branch-cuts

1. ## Need proof of theorem about log branch-cuts

I'm going to need the following theorem for a problem in here, and I was wondering if anyone could help me prove it?

Theorem:

For $p$ and $q$ polynomials, there exists a holomorphic extension $\text{Log}_h(p/q)=\ln|p/q|+i\arg(p/q)$ to the entire complex plane except along branch-cuts extending from each of the zeros and poles of the expression $p/q$ with $\arg(p/q)$ a continuous and analytic function in this domain. Additionally, the difference across each branch-cut is determined by the order of the branch point such that traveling in the positive sense around the branch point, this difference is $2n\pi i$ for a zero of order n and $-2m\pi i$ for a pole of order m.

I believe I have a proof but not sure it's correct.

2. Setting...

$\lambda (z) = \ln \frac{p(z)}{q(z)} = \ln p(z) - \ln q(z)$ (1)

... is...

$\lambda^{'} (z) = \frac{p^{'}(z)}{p(z)} - \frac{q^{'}(z)}{q(z)}$ (2)

From (2) we observe that in all z where is $p(z) \ne 0$ and $q(z) \ne 0$ the derivative of $\lambda (*)$ does exist and therefore $\lambda (*)$ is holomorphic...

Kind regards

$\chi$ $\sigma$

3. Originally Posted by chisigma
Setting...

$\lambda (z) = \ln \frac{p(z)}{q(z)} = \ln p(z) - \ln q(z)$ (1)

... is...

$\lambda^{'} (z) = \frac{p^{'}(z)}{p(z)} - \frac{q^{'}(z)}{q(z)}$ (2)

From (2) we observe that in all z where is $p(z) \ne 0$ and $q(z) \ne 0$ the derivative of $\lambda (*)$ does exist and therefore $\lambda (*)$ is holomorphic...

Kind regards

$\chi$ $\sigma$
Hi Chisigma. I'm extending the principal branch logarithm $\text{Log}$ to the entire complex plane. At each zero and pole, we have to define a branch-cut over which the function is not analytic so it's more than just at the zeros of p and q the function is non-analytic. Looks to me anyway. It's these branch-cuts that I wish to determine how the function value changes across, and I believe that is only dependent on the order of the zeros and poles of p/q.