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?
On
Firstly we write $$f\left ( z \right )=\sqrt{1-z^2}=e^{\frac{1}{2}\ln\left ( 1-z^2 \right )}$$ We have branch points for all $z$ such that the argument of the log vanishes. We see that $$1-z^2=0\Leftrightarrow z_{1,2}=\pm1$$ We now need to investigate the point $z=\infty$. In that manner we need to see if if there is a branch point az $z=0$ for the function $$g\left ( z \right )=f\left ( \frac{1}{z} \right )=e^{\frac{1}{2}\ln\left ( z^2-1 \right )}e^{-\frac{1}{2}\ln z^2}$$ 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 $\infty$ and also $1$ and $\infty$. Note that this is actually one branch cut(in terms of the Riemann sphere)
$$f(z)=\sqrt{1-z^2}=\left(1-z^2\right)^{1/2}=e^{\log\left(1-z^2\right)/2}\,.$$
The branch cut of $\log$ runs from zero to negative infinity, therefore $f$ is discontinuous at point $z$ whenever $1-z^2\in{({{-\infty};{0}}]}$, hence the solution set of $$1-z^2\in{({{-\infty};{0}}]}$$ for $z$ over $\mathbb{C}$ will be the set of points of discontinuity of $f$.