Determine the points of discontinuity of…
Determine the points of discontinuity of $\ f(z)=\dfrac{3z^2+4}{z^4-16}$.
Explanation of Method
A rational function is discontinuous exactly at the zeros of its denominator, unless the numerator also vanishes there (removable singularity). Factor $z^4-16$, locate its four roots, and check the numerator at each.
Step 1 · Factor the denominator
$$ z^4-16=(z^2-4)(z^2+4)=(z-2)(z+2)(z-2i)(z+2i) $$
so the zeros are $z=2,\,-2,\,2i,\,-2i$ — the fourth roots of $16$, i.e. $z_k=2e^{i\pi k/2}$ for $k=0,1,2,3$.
Step 2 · Check the numerator at each zero
$$ 3(2)^2+4=16\neq0,\qquad 3(-2)^2+4=16\neq0 $$
$$ 3(2i)^2+4=-12+4=-8\neq0,\qquad 3(-2i)^2+4=-12+4=-8\neq0 $$
The numerator is nonzero at every zero of the denominator, so none of the four singularities is removable.
Step 3 · Order of each pole
Each factor $(z-z_k)$ appears exactly once in the factorization, so each singularity is a simple pole (order 1). Near $z=2$, for instance, $$ f(z)=\frac{3z^2+4}{(z-2)(z+2)(z-2i)(z+2i)}\sim\frac{16}{(z-2)\cdot4\cdot(2-2i)(2+2i)} $$ confirming a simple pole there, and symmetrically at the other three points.
Geometric picture
All four poles lie at distance $2$ from the origin, at the vertices of a square: two on the real axis ($\pm2$) and two on the imaginary axis ($\pm2i$).
Result
Since $16=2^4$ has exactly four fourth roots equally spaced by $90^\circ$, the poles automatically form a square centered at the origin — a pattern that generalizes to any $z^n=c$.