Singularities of $\frac{z^2}{(z^2+1)^2(z^2+2z+2)}$.

Question

Let $f(z) = \frac{z^2}{(z^2+1)^2(z^2+2z+2)}$.

Find singularities of this function. For each singularity determine if it is removable, a pole (if a pole determine it's order,) or essential.

I have worked out that the singularities of this function are $z=\pm i$, $z=-1\pm i$.

However, I'm struggling to determine whether these singularities are removable, a pole or essential. Is $z=\pm i$ a pole of order $2$ and $z=-1\pm i$ a simple pole?

Any help will be appreciated.

Answer 1

Hints:

• Rational functions only have poles because the zeros of the denominator are isolated.

• A singularity $z_0$ of a rational function is removable iff we can cancel $z-z_0$ enough times in both the numerator and the denominator to make it disappear in the denominator.

Answer 2

Let start with the poles: as your function is rational the pole are the zero of the denominator, so being $z^2+1=0 \iff z=\pm 1$ and $z^2+2z+2=0\iff z=-1\pm i$ we have $4$ singularities: $\{i,-i,-1+i,-1-i\}$.

Now note that $z^2+2z+2\neq 0$ if $z=\pm i$, and the same is true for $z^2$.

Let be $g(z)=\frac{1}{(z^2+1)^2}$, therefore $f(z)=\frac{z^2}{z^2+2z+2}g(z)$ and to study the singularities $\pm i$ we can study directly $g(z)$.

If we do a series expansion we get $g(z)=\sum_{n=-\infty}^{\infty}a_n(z-i)^n$ where $a_n=0$ if $n\leq 3$ and $a_n\neq 0$ if $n\geq 2$ therefore $-i$ is a singularity of order $2$.

The exact same rasonement is true for $-i$ and a very similar one for the other two poles.

I've gave to you the general procedure and a partial example, I'll leave to you the details of the calculations and the other cases.

You can also prove in a more general way (with the same procedure that I used in this answer) that the order of a pole in a rational function is exactly the multiplicity of the $0$ in the polynomial.