Domianpv

2022-09-30

Analyze the Complex Function by using the Principal log Branch

I am trying to analyze the function $\sqrt{1-{z}^{2}}$, where the square root function is defined by the principal branch of the log function. I want to locate the the discontinuities.

I know the discontinuities will lie on the negative real axis but I cannot figure out for which values of $z$ this will occur. I first tried rewriting the function as $(1-{e}^{2i\theta})$. But this seems to be getting me nowhere. Any thoughts?

I am trying to analyze the function $\sqrt{1-{z}^{2}}$, where the square root function is defined by the principal branch of the log function. I want to locate the the discontinuities.

I know the discontinuities will lie on the negative real axis but I cannot figure out for which values of $z$ this will occur. I first tried rewriting the function as $(1-{e}^{2i\theta})$. But this seems to be getting me nowhere. Any thoughts?

Corbin Hanson

Beginner2022-10-01Added 10 answers

Firstly we write

$$f\left(z\right)=\sqrt{1-{z}^{2}}={e}^{\frac{1}{2}\mathrm{ln}(1-{z}^{2})}$$

We have branch points for all $z$ such that the argument of the log vanishes. We see that

$$1-{z}^{2}=0\iff {z}_{1,2}=\pm 1$$

We now need to investigate the point $z=\mathrm{\infty}$. In that manner we need to see if if there is a branch point az $z=0$ for the function

$$g\left(z\right)=f\left(\frac{1}{z}\right)={e}^{\frac{1}{2}\mathrm{ln}({z}^{2}-1)}{e}^{-\frac{1}{2}\mathrm{ln}{z}^{2}}$$

We can now conclude that the original function has two finite branch points, and one branch point at infinity. We chose a branch cut such that it connects $-1$ and $\mathrm{\infty}$ and also $1$ and $\mathrm{\infty}$. Note that this is actually one branch cut(in terms of the Riemann sphere)

$$f\left(z\right)=\sqrt{1-{z}^{2}}={e}^{\frac{1}{2}\mathrm{ln}(1-{z}^{2})}$$

We have branch points for all $z$ such that the argument of the log vanishes. We see that

$$1-{z}^{2}=0\iff {z}_{1,2}=\pm 1$$

We now need to investigate the point $z=\mathrm{\infty}$. In that manner we need to see if if there is a branch point az $z=0$ for the function

$$g\left(z\right)=f\left(\frac{1}{z}\right)={e}^{\frac{1}{2}\mathrm{ln}({z}^{2}-1)}{e}^{-\frac{1}{2}\mathrm{ln}{z}^{2}}$$

We can now conclude that the original function has two finite branch points, and one branch point at infinity. We chose a branch cut such that it connects $-1$ and $\mathrm{\infty}$ and also $1$ and $\mathrm{\infty}$. Note that this is actually one branch cut(in terms of the Riemann sphere)

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