Let $k$ be an integer and $\mathcal{H}$ denotes the upper half plane. Let $f: \mathcal{H}\to \mathbb{C}$ be a function satisfying the following conditions:
$1.$ $f$ is holomorphic on $\mathcal{H}$.
$2.$ $f(\frac{az+b}{cz+d})= (cz+d)^{k} f(z),$ for all $\begin{pmatrix} a & b \\ c & d \end{pmatrix} \in SL_{2}(\mathbb{Z})$ and $ z \in \mathcal{H}.$
$3.$ $f$ is given by a power series of the form, $$ f(z)= \sum_{ n \geq 0} a_{n} q^{n} (q= e^{2\pi i z}),$$ (Note that, the condition $2$ implies, $f$ is one periodic and hence has a Laurent series expansion.)
Can we say, that $f(z)$ is bounded as $\Im z\ \text{ to } \infty$ from the above three conditions?
I am unable to deduce this. If not, can someone tell me what should be the counterexample to this?