Residues of p over q at a simple zero of q

Define $h(z):=\{\begin{array}{ll}{\displaystyle \frac{q(z)}{z-a}},& z\mathrm{\ne }a,\\ q^{\mathrm{\prime }}(a),& z=a\mathrm{.}\end{array}$ Because $q^{\mathrm{\prime }}(a)={\lim }_{z\to a}(q(z)-q(a))\mathrm{/}(z-a)$ and $q(a)=0$, the function $h$ is continuous at $a$ and holomorphic away from $a$; [L2] therefore makes $h$ holomorphic near $a$.

Step 1.1 gives $q(z)=(z-a)h(z)$ and $h(a)=q^{\mathrm{\prime }}(a)\mathrm{\ne }0$, so shrinking the neighbourhood if necessary makes $h$ nonzero there. Hence $k:=p\mathrm{/}h$ is holomorphic near $a$ by [L3], and on the punctured neighbourhood one has $(p\mathrm{/}q)(z)=k(z)\mathrm{/}(z-a)$.

If $p(a)=0$, define $s(z):=\{\begin{array}{ll}{\displaystyle \frac{p(z)}{z-a}},& z\mathrm{\ne }a,\\ p^{\mathrm{\prime }}(a),& z=a\mathrm{.}\end{array}$ The same argument as in step 1.1, using that $p$ is holomorphic and $p(a)=0$, shows that $s$ is holomorphic near $a$. Then step 1.1 gives $\frac{p(z)}{q(z)}=\frac{(z-a)s(z)}{(z-a)h(z)}=\frac{s(z)}{h(z)}$ on the punctured neighbourhood, so $p\mathrm{/}q$ is holomorphic at $a$ and its residue there is $0=p(a)\mathrm{/}{q}^{\mathrm{\prime }}(a)$.

If $p(a)\mathrm{\ne }0$, then $k(a)=p(a)\mathrm{/}h(a)\mathrm{\ne }0$, so step 2.1 makes $p\mathrm{/}q$ a simple pole at $a$. Applying [L1] gives $\mathrm{Res}\text{\hspace{0.17em}⁣}(\frac{p}{q},a)={\lim }_{z\to a}(z-a)\frac{p(z)}{q(z)}={\lim }_{z\to a}k(z)=k(a)=\frac{p(a)}{q^{\mathrm{\prime }}(a)}\mathrm{.}$

Steps 3.1 and 2.2 cover the cases $p(a)\mathrm{\ne }0$ and $p(a)=0$, so in all cases $\mathrm{Res}\text{\hspace{0.17em}⁣}(\frac{p}{q},a)=\frac{p(a)}{q^{\mathrm{\prime }}(a)}\mathrm{.}$