Consider the function ($PV$ denotes the principal value) $$PV(z-1)^{\frac{1}{2}}PV(z+1)^{\frac{1}{2}}, \ \ \forall z\in\mathbb{C}.$$ Find where $k$ is continuous and differentiable, giving reasons.
We first recall that $f(w)=w^{\frac{1}{2}}$ is not continuous when $w<0$. Using this, we can conclude that $PV(z-1)^{\frac{1}{2}}$ is not continuous when $z<1$ and $PV(z+1)^{\frac{1}{2}}$ is not continuous when $z<-1$. Hence, $k(z)$ is not continuous when $z<1$.
Is this sufficient to determine that $k(z)$ is not differentiable on the interval $(-\infty, 1)$, as $k(z)$ is not continuous for $z<1$ and not differentiable at $z=1$ (by inspection)?
Carefully rewrite $f(z)=\text{PV}(z-1)^{1/2}(z+1)^{1/2}$ as $\exp\left(\frac{\ln(z-1)+\ln(z+1)}2\right)$.
Consider $g(z)=\frac{\ln(z-1)+\ln(z+1)}2$, $$\lim_{z\to x,\Re z>0}g(z)=g(x),x<1$$ $$\lim_{z\to x,\Re z<0}g(z)=g(x)-\pi i,-1<x<1$$ $$\lim_{z\to x,\Re z<0}g(z)=g(x)-2\pi i,x<-1$$ Hence, $$\lim_{z\to x,\Re z>0}f(z)=f(x),x<1$$ $$\lim_{z\to x,\Re z<0}f(z)=-f(x),-1<x<1$$ $$\lim_{z\to x,\Re z<0}f(z)=f(x),x<-1$$ Therefore, $f(z)$ is discontinuous at $-1<x<1$.
One can prove $$\lim_{z\to1}f(z)=\lim_{z\to-1}f(z)=0,$$ hence it is continuous at $z=1$ and $-1$, but not continuous at neighborhood of $z=1$ and $-1$.