Finding the branch points of $\log(\frac{z-1}{z+1})$

complex-analysissolution-verification

I want to determine if $z = 1$ is a branch point of $\log(\frac{z – 1}{z+1})$. I think it should be a branch point because we would have $\log(0)$ , yet somehow my work shows that it isn't, so I'm wondering if I've done something wrong.

Write $z – 1 = R_1e^{i \theta_1}$ and $z + 1 = R_2e^{i \theta_2}$. As we encircle the point $z = 1$, moving around the circle, $\theta_1$ changes by $2\pi$ and $\theta_2$ changes by $0$. Thus,
$$\log(\frac{z-1}{z+1}) = \log(\frac{R_1}{R_2}e^{i(\theta_1 – \theta_2)})$$
and at the start, we are at $\log(\frac{R_1}{R_2}e^0)$, while after encircling $z = 1$, we are at $\log(\frac{R_1}{R_2}e^{i(2\pi – 0)})$. $e^{i 2\pi} = \cos(2\pi) + i\sin(2 \pi) = 1$, so after encircling, the function value does not change, hence $z = 1$ is not a branch point.

Is this correct reasoning? if not, where was my error?

Best Answer

The source of the error is this sentence:

and at the start, we are at $\log(\frac{R_1}{R_2}e^0)$, while after encircling $z = 1$, we are at $\log(\frac{R_1}{R_2}e^{i(2\pi - 0)})$... so after encircling, the function value does not change.

It is not true that just because the interior terms do not change, the function value does not change. In fact, this is not true for any complex $z$, since $\log$ is multi-valued. At the start, the interior term is indeed at $\frac{R_1}{R_2}e^0$, and at the end, the interior term is indeed at $\frac{R_1}{R_2}e^{i(2\pi - 0)}$, but it says nothing about the behavior of the complex log at those points. Nothing concrete can be said about the entire function until we write

$$\log(\frac{R_1}{R_2}e^{i(\theta_1 - \theta_2)}) = \operatorname{Log}(\frac{R_1}{R_2}) + i(\theta_1 - \theta_2)$$

where $\operatorname{Log}$ denotes the real-valued logarithm, and now it is obvious that $z = 1$ would be a branch point.