Bruce Sherman

2022-10-03

Branch points and natural maximal domain of $\mathrm{log}\frac{1+z}{1-z}$

So I have a function $\mathrm{log}\frac{1+z}{1-z}$ and I'm supposed to find its branch points, natural maximal domain.

So far I converted $z$ to $x+iy$ but no real good. Can't check holomorphicity using this. I know that $\mathrm{log}$ is holomorphic after a branch cut. How do I proceed?

I also conclude that $\frac{1+z}{1-z}$ is holomorphic as it doesn't contain $\overline{z}$

So I have a function $\mathrm{log}\frac{1+z}{1-z}$ and I'm supposed to find its branch points, natural maximal domain.

So far I converted $z$ to $x+iy$ but no real good. Can't check holomorphicity using this. I know that $\mathrm{log}$ is holomorphic after a branch cut. How do I proceed?

I also conclude that $\frac{1+z}{1-z}$ is holomorphic as it doesn't contain $\overline{z}$

Houston Ellis

Beginner2022-10-04Added 4 answers

We can't allow $0$ or $\mathrm{\infty}$ in $\mathrm{log}$, so the points $z=\pm 1$ are out.

Also: the complex logarithm is multi-valued, acquiring a multiple of $2\pi i$ as its argument moves along a closed curve separating $0$ from $\mathrm{\infty}$. If you don't want this to happen, you must make sure that your domain has no closed curves separating $1$ from $-1$There is no canonical way of making this happen. Cutting out the line segment from −1 to 1 is one natural choice; there is another one with cuts along the real axis (I leave it for you to find it), and plenty of others, which are less nice.

Also: the complex logarithm is multi-valued, acquiring a multiple of $2\pi i$ as its argument moves along a closed curve separating $0$ from $\mathrm{\infty}$. If you don't want this to happen, you must make sure that your domain has no closed curves separating $1$ from $-1$There is no canonical way of making this happen. Cutting out the line segment from −1 to 1 is one natural choice; there is another one with cuts along the real axis (I leave it for you to find it), and plenty of others, which are less nice.

ivybeibeidn

Beginner2022-10-05Added 2 answers

We can define

$$\begin{array}{}\text{(1)}& \mathrm{log}\left(\frac{1+z}{1-z}\right)=\frac{\pi}{2}i+{\int}_{i}^{z}(\frac{1}{w+1}-\frac{1}{w-1})\phantom{\rule{thinmathspace}{0ex}}\mathrm{d}w\end{array}$$

Since the residues at $+1$ and $-1$ cancel, the integral along any closed path that circles them both an equal number of times will be $0$. The smallest branch cut which forces any closed path to circle both an equal number of times is the segment $[-1,+1]$. Thus the largest natural domain would be $\mathbb{C}\setminus [-1,+1]$

$$\begin{array}{}\text{(1)}& \mathrm{log}\left(\frac{1+z}{1-z}\right)=\frac{\pi}{2}i+{\int}_{i}^{z}(\frac{1}{w+1}-\frac{1}{w-1})\phantom{\rule{thinmathspace}{0ex}}\mathrm{d}w\end{array}$$

Since the residues at $+1$ and $-1$ cancel, the integral along any closed path that circles them both an equal number of times will be $0$. The smallest branch cut which forces any closed path to circle both an equal number of times is the segment $[-1,+1]$. Thus the largest natural domain would be $\mathbb{C}\setminus [-1,+1]$

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