Determine the points of discontinuity of f(z)=cot z

Determine the points of discontinuity of $\ f(z)=\cot z$.

Step 1 · Solve $\sin z=0$ in $\mathbb C$

Writing $z=x+iy$: $$ \sin z=\sin x\cosh y+i\cos x\sinh y=0 $$ requires both $\sin x\cosh y=0$ and $\cos x\sinh y=0$. Since $\cosh y>0$ always, the first forces $\sin x=0$, i.e. $x=n\pi$. Then $\cos(n\pi)=(-1)^n\neq0$, so the second forces $\sinh y=0$, i.e. $y=0$.

$$ \boxed{\sin z=0\iff z=n\pi,\ n\in\mathbb Z} $$

Step 2 · Order of the pole at $z=n\pi$

Let $w=z-n\pi$. Using the addition formula, $$ \sin z=\sin(n\pi)\cos w+\cos(n\pi)\sin w=(-1)^n\sin w\approx(-1)^nw $$ for small $w$, while $\cos z\to\cos(n\pi)=(-1)^n\neq0$. Hence $$ \cot z=\frac{\cos z}{\sin z}\approx\frac{(-1)^n}{(-1)^nw}=\frac1{z-n\pi} $$ so each $z=n\pi$ is a simple pole.

Step 3 · Residue at each pole

$$ \operatorname*{Res}_{z=n\pi}\cot z=\lim_{z\to n\pi}(z-n\pi)\cot z=\lim_{z\to n\pi}\frac{(z-n\pi)\cos z}{\sin z}\overset{\text{L'H}}{=}\frac{\cos(n\pi)}{\cos(n\pi)}=1 $$

Laurent series near a pole

$$ \cot z=\frac1{z-n\pi}-\frac{z-n\pi}3-\frac{(z-n\pi)^3}{45}-\cdots $$

Periodicity

Since $\cot(z+\pi)=\cot z$, the poles repeat exactly with period $\pi$, all lying on the real axis.