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Suppose a function $\eta (z)=log(\psi (z))$ where $$\psi (z)=\prod_{k=1}^{n} \left(z-z_k\right)$$ We know that $log(z)=log|z|+i(argz)$, this implies that $$log(\prod_{k=1}^{n} \left(z-z_k\right))=\sum_{k=1}^{n}log|z-z_k|+i\sum_{k=1}^{n}arg(z-z_k)=log|\psi (z)|+i\sum_{k=1}^{n}arg(z-z_k)$$ I am new to contour integration but I was told that for 2 complex numbers $\varepsilon_1$ and $\varepsilon_2$ $log(\varepsilon _1,\varepsilon _2)\neq log(\varepsilon _1)+log(\varepsilon _2)$. how do we choose the arguments of $\left(z-z_k\right)$ so that we can expand the log of products of complex numbers as sum of their individual logs and vice versa when doing complex analysis?

how do we choose the arguments of $\left(z-z_k\right)$ for every individual $z_k$ so that $$\sum_{k=1}^{n}arg(z-z_k)\in(-\pi,\pi]$$ or the principle branch?

Madhav Asthana
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  • we cannot do this in general- what we can do is to make branch cuts and choose a continuous hence analytic version of $\log \psi$ but its imaginary parts will be in $[-n \pi, n \pi]$ – Conrad Feb 20 '24 at 17:51
  • how do we do that? many times when working with residues that involve logarithms and multiple branch cuts, I chose different sets of arguments for the branch points and each one led to different answers – Madhav Asthana Feb 20 '24 at 18:30
  • you need a simple connected domain that avoids all the $z_k$ (for example you can draw horizontal lines through all of them to the left or right, but other choices work as well, so depends on particular values and particular problems) and then fixing a value of the $\log$ at a point gives you the above; for example if say $\psi(z)=(z-1)(z-3)$ a good choice would be cutting out $[1, \infty]$ and defining $\log \psi(0)= \log 3$ the usual real one, defines a unique $\psi$ on the cut plane; at $2+i$ you pass through a negative number so will go beyond $\pi$ – Conrad Feb 20 '24 at 18:51
  • how would one go doing the same with $ln(z^{4}+1)$ or when $\psi (z)=(z-e^{\frac {\pi i}{4}})(z-e^{\frac {3\pi i}{4}})(z-e^{\frac {-3\pi i}{4}})(z-e^{\frac {-\pi i}{4}})$? There are 4 branch points in this case and you draw branch cuts along the lines $\Re(z) \pm \Im(z)=0$ from $z_k$ to $\infty$ along these 2 lines. But then how do we chose the arguments for each so that $\sum_{k=1}^{4}arg(z-z_k)$ can be bounded and we can avoid the need of taking the sum after finding the arguments of $z-z_k$ individually? – Madhav Asthana Feb 20 '24 at 19:27
  • not sure exactly what the question is - once you choose the cut, the branch just depends on the initial value at some point in the domain and all branches differ by $2\pi ik$; in other words there is nothing to choose really then - the choices are essentially in the cuts only – Conrad Feb 20 '24 at 21:27
  • we can choose $arg(z-e^{\frac{i\pi}{4}})$ to be $(\frac{\pi}{4},\frac{9\pi }{4}]$ or even $(\frac{-7\pi}{4},\frac{\pi }{4}]$ my doubt is how do you choose a set of arguments for all the the branch points so that you can skip finding and summing the arguments of a complex no. $z$ from each branch point, and instead take the argument directly from a set of bound arguments. eg: https://math.stackexchange.com/q/4862653/1289545 in this problem, I chose $arg(z+i)\in (\frac {-5\pi}{2}, \frac{-\pi}{2}]$ and $arg(z-i)\in (\frac {\pi}{2}, \frac{5\pi}{2}]$ and in the end $arg(z)\in (-\pi,\pi]$ – Madhav Asthana Feb 20 '24 at 22:03
  • that set of arguments for the branch points also allowed me to open $log(z_1,z_2)=log(z_1)+log(z_2)$ I am curious if there is a way or method to select a range of arguments for multiple branch points that permits that allows us to restrict the sum of arguments to the principle branch? – Madhav Asthana Feb 20 '24 at 22:09

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