Extension of complex function Consider the function $$f(z) = \frac {z^2}{z^5 − 1}.$$
(a) Define an extension of f so that we have a holomorphic function
$f : C\cup \{ \infty \}  → C \cup \{ \infty \}.$
(b) Compute (1) the zeros and poles of f; (2) the order of the zero or pole of $f$ at $ z = \infty$; (3) the degree of $f$.
Could someone please tell me how to do (a). I have read some online slides saying that we can set whatever makes the denominator $0$, namely if $z$ is such that $z^5 -1=0$, then set $f(z) = \infty$ to extend such a function.
I am unsure how to proceed. I feel like there is an algorithm of some sort.
Thank you.
 A: When I do these sorts of problems, I like to introduce a new variable,
$$ w = \frac 1 z.$$
So $z = 0$ is the same as $w = \infty$, and $z = \infty$ is the same as $w = 0$.
Written in terms of the $w$ variable, the function $f$ is
$$ f(w) = \frac{w^{-2}}{w^{-5}-1} = \frac{w^3}{1-w^5}.$$
The zeroes and poles are:


*

*A double zero at $z = 0$.

*A triple zero at $w = 0$.

*Five single poles: one at each of the points $z = \exp(2\pi i n/5)$ for $n = 0,1,2,3,4$.
[Note that these poles are the same as the poles of $f(w)$ at $w = \exp(-2\pi i n/5)$ for $n = 0,1,2,3,4$. Be careful not to double-count!]
As a final check, observe that the total number of zeroes (counted with multiplicity) minus the total number of poles (counted with multiplicity) is
$$ 1\times 2 + 1 \times 3 - 5 \times 1 = 0.$$
This is how it should be. The number of zeroes minus the number of poles should always be zero, for any meromorphic functions on $\mathbb C \cup \{ \infty \}$. It is worth checking this every time you encounter a problem of this kind in the future.
