Let $f$ be a meromorphic function on the region $Im(z)>0$, $v_p(f)$ be the order of $p$. (The number $n$ such that $\frac{f(z)}{(z-p)^n}$ is holomorphic and non-zero at $p$.)
Moreover, assume $f$ is a $modular$ $function$ for $SL(\mathbb Z)$ of weight $k$.
Then we have the following application of $residue$ $theorem$.
Why is the last limitation true? The argument principle only tell us the integral over a circle.
$f$ need not be modular, it holds for all meromorphic $f$. You can see it by writing
$$f(z) = (z-\rho)^{v_\rho(f)}\cdot g(z)$$
with $g$ holomorphic and nonzero in a neighbourhood of $\rho$. Then you compute
$$\frac{df}{f}(z) = \frac{f'(z)}{f(z)}\,dz = \frac{v_\rho(f)(z-\rho)^{v_\rho(f)-1}g(z) + (z-\rho)^{v_\rho(f)}g'(z)}{(z-\rho)^{v_\rho(f)}g(z)}\,dz = \biggl(\frac{v_\rho(f)}{z-\rho} + \frac{g'(z)}{g(z)}\biggr)dz,$$
and since $\frac{g'}{g}$ is bounded in a neighbourhood of $\rho$,
$$\int_{B}^{B'} \frac{g'(z)}{g(z)}\,dz \to 0$$
as the radius of the circular arc tends to $0$. So what remains is the limit of
$$\int_B^{B'} \frac{v_\rho(f)}{z-\rho}\,dz.$$
But a parametrisation $z = \rho + r\cdot e^{i(\pi/2-t)}$ shows that that last integral is
$$-iv_\rho(f)\cdot \alpha(r),$$
where $\alpha(r)$ is the angle of the circular arc of radius $r$. Here we have $\lim\limits_{r\to 0} \alpha(r) = \frac{2\pi}{6}$, so the result follows.