Prove that there exists a complex number $z$ such that $|z|>1$ and $z^{135} + (2+3i)z -100=0$.

Prove that there exists a complex number $z$ such that $|z|>1$ and $z^{135} + (2+3i)z -100=0$.

I understand that we can express $z$ in terms of $r(\cos \theta + i\sin \theta) = re^{i\theta}$ but I honestly don't see how this helps in solving this problem. I have considered representing $z$ as $a+bi$ and using the Binomial Theorem to state that $z^{135} = \sum_{k=0}^{135}\begin{pmatrix}135\\k\\\end{pmatrix}a^k(bi)^{135-k}$, but I also don't think this is helpful. Am I approaching this all wrong? What should I do?

(As was also explained in lulu's comment:) The product of the roots of a polynomial over $\mathbb{C}$ gives the constant term of that polynomial, up to a sign. This follows by just factoring any polynomial as $p(z) = \lambda (z - \alpha_1) \cdots (z - \alpha_n)$ and then expanding. In particular, in your case $\prod_{j = 1}^{135} \lvert \alpha_j \rvert = 100$. If $\lvert \alpha_j \rvert \leq 1$ for all $j$ then this couldn't happen, so some $\alpha_j$ satisfying your requirements must exist.