Determine the points of discontinuity of f(z)=(tanh z)/(z^2+1)
Determine the points of discontinuity of $\ f(z)=\dfrac{\tanh z}{z^2+1}$.
Explanation of Method
Write $\tanh z=\sinh z/\cosh z$, so $f(z)=\sinh z\big/\!\left[\cosh z\,(z^2+1)\right]$ fails to be defined where $\cosh z=0$ or $z^2+1=0$. Solve both, then check whether $\sinh z$ vanishes at any of those points (which would make the singularity removable).
Step 1 · Zeros of $\cosh z$
$$ \cosh z=0\iff e^{2z}=-1=e^{i\pi}\iff z=\frac{i\pi(2n+1)}2,\quad n\in\mathbb Z $$
i.e. $z\in\left\{\ldots,-\tfrac{3i\pi}2,-\tfrac{i\pi}2,\tfrac{i\pi}2,\tfrac{3i\pi}2,\ldots\right\}$, all purely imaginary.
Step 2 · Zeros of $z^2+1$
$$ z^2+1=0\iff z=\pm i $$
Step 3 · Check the numerator at the $\cosh z$ zeros
At $z_n=i\pi(n+\tfrac12)$, using $e^{i\pi(n+1/2)}=(-1)^ni$: $$ \sinh z_n=\frac{e^{z_n}-e^{-z_n}}2=\frac{(-1)^ni-(-1)^n(-i)}2=(-1)^ni\neq0 $$ so none of these poles is removable.
Step 4 · Check the numerator at $z=\pm i$
$$ \sinh(i)=\frac{e^i-e^{-i}}2=i\sin1\neq0,\qquad \sinh(-i)=-i\sin1\neq0 $$
Verified numerically: $\sinh(i)\approx0.8415i=i\sin(1)$. Neither is removable.
Step 5 · No overlap between the two zero-sets
If $i=\frac{i\pi(2n+1)}2$ for some integer $n$, then $2n+1=\frac2\pi\approx0.637$, not an odd integer — contradiction. So $\pm i$ never coincide with a zero of $\cosh z$, and all poles listed below are genuinely distinct, each simple (order 1, since every zero found is simple in both factors).
Result
The pole set splits cleanly into two independent families — the two isolated poles contributed by the algebraic factor $z^2+1$, and the infinite arithmetic progression along the imaginary axis contributed by $\cosh z=0$ — and Step 5 confirms they never collide.