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It is rather easy to see that the function $$f(z) = \ln \left( \frac{z-a}{z-b} \right)$$ has branch points at $z=a$ and $z=b$, My question is why considering a branch cut "connecting" $a$ and $b$ makes this function single-valued ? For simplicity suppose $a=1$ and $b=-1$ are both real. a real number $c=2$ which is bigger than $a$ and $b$ , hence, does not lie on the branch cut, therefore it is not a singular value of $f(z)$ ! (I can not understand this, because $z=2$ can be represented in so many ways for $\theta_1 := \arg (z-1)$ and $\theta_2 = \arg (z+1)$) Is this onkay, only because we can define sheets in a way that, on each sheet everything works nicely...?

Let me try that, suppose we consider a branch cut which connects $1$ to $-1$ and consider the following sheets for the Riemann surface of $f(z)=\ln \left( \frac{z-1}{z+1} \right)$:

first sheet : $ - \pi < \theta_1 \leq \pi$ and $ 0 \leq \theta_2 < 2\pi$

second sheet : $ \pi < \theta_1 \leq 3\pi$ and $ 2 \pi \leq \theta_2 < 4\pi$

and so on ...

If I consider $z=2$ on the first sheet : $\theta_1 = \theta_2 = 0$, so $\arg(\frac{2-1}{2+1}) = 0-0 =0$

If I consider $z=2$ on the second sheet : $\theta_1 = \theta_2 = 2\pi$, so $\arg(\frac{2-1}{2+1}) = 2\pi-2\pi =0$

Therefore for a very special way of defining the sheets $z=2$ is a regular point for $f(z)$, right ?

I would be very thankful if you correct any incorrect statement/conclusion in what I wrote above.

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  • $\begingroup$ It doesn't make the function "single-valued", $\ln \frac{z-a}{z-b}$ still has infinitely many possible values for every $z$ outside the branch cut, since $\exp$ is periodic. But, the point is that you have a branch [and hence infinitely many, all differing by integer multiples of $2\pi i$] defined and holomorphic on the whole complement of the branch cut. $\endgroup$ – Daniel Fischer Jan 24 '15 at 0:19

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