I'm trying to solve exercise $1.1.9$ from Diamond's A First Course in Modular Forms, in which we must prove the $j$-invariant $j:\mathcal{H}\to\mathbb{C}$ with $j(\tau)=1728\frac{g_2(\tau)^3}{\Delta(\tau)}$ is surjective.
The author gives the following suggestion: suppose $\exists\,c\in\mathbb{C}$ with $j^{-1}(c)=\emptyset$. Consider the integral $I=\frac{1}{2\pi i}\int_\gamma \frac{j'(\tau)}{j(\tau)-c}dz$, where $\gamma$ is described below :
(the lower segment is an arc of the unit circle and the upper horizontal line has height $>1$)
By the argument principle, $I=0$. Exploring the $\text{SL}_2(\mathbb{Z})$-invariance of $j$, we conclude that the contribution of the two vertical lines cancel each other out (by invariance under $\left(\begin{matrix} 1 & 1 \\0 & 1 \end{matrix}\right)$). Furthermore, the contribution of left and right sides of the lower arc also cancel each other (by invariance under $\left(\begin{matrix} 0 & -1 \\1 & 0 \end{matrix}\right)$).
For the horizontal piece, he suggests us the change variables $q=e^{2\pi i\tau}$, remembering that $j(\tau)=\frac{1}{q}+h(q)$, where $h$ is holomorphic, to conclude the integral equals $1$.
The only part I'm struggling with is the last paragraph. By my computations, $dq =(2\pi i)q\,d\tau$, so: \begin{align*} \frac{j'(\tau)}{j(\tau)-c}d\tau &=\frac{-\frac{2\pi i}{q}+(2\pi i)q\,h'(q)}{-c+\frac{1}{q}+h(q)}\cdot \frac{1}{2\pi i}\frac{dq}{q}\\ &=\frac{-\frac{1}{q^2}+h'(q)}{-c+\frac{1}{q}+h(q)}dq\\ &=\left(-\frac{1}{q}+\widetilde{h}(q)\right)dq,\,\,\text{ where }\widetilde{h}\text{ is holomorphic} \end{align*}
I don't know what to do with this and how I'm supposed to get $1$ from the integral.