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

Question

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?

Actually, all roots must be outside the unit circle since $100=\left|z^{135}+(2+3i)z\right|$ $\le |z|^{135} + \sqrt{13}, |z|,$. — dxiv, Nov 30 '21 at 00:58

Keeley Hoek · Accepted Answer · 2021-11-30T00:20:10.950

(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.

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

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