Uniform convergence of a series $\sum _{n=1}^{\infty}3^{-n}\sin (nz)$

Question

I'm solving the following problem: Show that $$\sum _{n=1}^{\infty}3^{-n}\sin (nz)$$ converges uniformly in the region of $$|Imz|<\ln3$$ At first, I tried to use a Weierstrass M-test and attempted to find a sequence $a_n$ such that $\sum _{n=1}^{\infty}a_n$ converges and $|3^{-n}\sin (nz)|\leq a_n$ for all n and $z$ such that $|Imz|<\ln3$. Using the definition of $\sin (z)$, I've obtained the following upper bound: $|3^{-n}\sin (nz)|\leq \frac {3^{n}+3^{-n}}{2\cdot 3^{n}}$. However, since RHS has a constant term $\frac 12$, the corresponding series doesn't converge at all! This upper bound is the best result I've got so far. If the region was $|Imz| \leq \ln(3-\epsilon)$, where $\epsilon$ is some small enough positive number, then it is easy to show the uniform convergence(It works with the upper bound I found). However, in the region of $|Imz|<\ln3$, it becomes quite difficult for me. Does the series really converge uniformly on the given region? If so, how can we show that? Any help would be appreciated!