I would like to do the following exercise :
Let $f$ be a meromorphic function and $\mathcal{P}$ the set of its poles. We also assume that $f$ is even ($\forall z \in \mathbb{C}, \; f(z)=f(-z)$). Prove that $\mathcal{P}$ is symmetric with respect to $0$ and that, if $a \in \mathcal{P}$ and $\mathrm{Res}_{a}(f)$ denotes the residue of $f$ at $a$, then $\mathrm{Res}_{-a}(f) = - \mathrm{Res}_{a}(f)$.
My try : let $a \in \mathcal{P}$ a pole of $f$ of order $k \in \mathbb{N}^{\ast}$. Since $f$ is even, $f$ is not defined at $-a$ and to prove that $a$ is also a pole of $f$ of order $k$, I need to prove that $(z+a)^{k}f(z)$ has a finite limit when $z \to -a$. It follows from :
$$ \begin{align*} \lim \limits_{z \to -a}(z+a)^{k}f(z) &= {} \lim \limits_{z \to -a}(-1)^{k}(-z-a)^{k}f(-z) \\[2mm] &= (-1)^{k}\lim \limits_{z \to a}(z-a)^{k}f(z) \\ \end{align*} $$
And $\displaystyle \lim \limits_{z \to a} (z-a)^{k}f(z)$ exists because $a$ is a pole of order $k$ of $f$. As a consequence, $-a$ is also a pole of $f$. For the residue relation, we can note that if $a$ is a pole of order $k$ of $f$, then :
$$ \begin{align*} \mathrm{Res}_{-a}(f) &= {} \frac{1}{(k-1)!}\frac{d^{k-1}}{dz^{k-1}} \Big((z+a)^{k}f(z) \Big) \\[2mm] &= \frac{1}{(k-1)!} \frac{d^{k-1}}{dz^{k-1}} \Big( (-1)^{k} (-z-a)^{k}f(z) \Big) \\[2mm] &= (-1)^{k-1} \frac{1}{(k-1)!} \frac{d^{k-1}}{dz^{k-1}} (-1)^{k} \Big( (z-a)^{k}f(-z) \Big) \\[2mm] &= - \mathrm{Res}_{a}(f) \\ \end{align*} $$
Is this OK or can it be improved ? Thanks.
It was a joy to read.
Everything is correct except for you missing the limits in the last chain of equalities. As you noted $-a,a\not \in \text{dom}(f)$, so you need to to take the limits at this points. That is, you should have
$$ \begin{align*} \mathrm{Res} _{-a}(f) &= {} \frac{1}{(k-1)!} \lim \limits_{z\to -a}\left[\frac{d^{k-1}}{dz^{k-1}} \Big((z+a)^{k}f(z) \Big)\right] \\[2mm] &= \frac{1}{(k-1)!} \lim \limits_{z\to -a} \left[\frac{d^{k-1}}{dz^{k-1}} \Big( (-1)^{k} (-z-a)^{k}f(z) \Big)\right] \\[2mm] &= (-1)^{k-1} \frac{1}{(k-1)!} \lim \limits_{z\to -a}\left[\frac{d^{k-1}}{dz^{k-1}} (-1)^{k} \Big( (z-a)^{k}f(-z) \Big)\right] \\[2mm] &= - \mathrm{Res}_{a}(f). \\ \end{align*} $$