How to prove there exists a solution? Guillemin Pollack Prove there exists a complex number $z$ such that 
$$
z^7+\cos(|z^2|)(1+93z^4)=0.
$$
(For heaven's sake don't try to compute it!)

Although the answers above are correct ones, they fail to use $deg_2$ as the book of Guillemin & Pollack suggest. Heres an approach that use the notion of $deg_2$:
Let $f:\mathbb{C} \to \mathbb{C}$ be defined as
$$
f(z)=z^7+\cos(|z^2|)(1-93z^4).
$$
Consider the homotopy $F(z,t)=tf(z)+(1-t)z^7$ between $f(z)$ and $z^7$. Now if $W=\left\{z: |z|\leq R \right\}$ (where $R$ is taken large enough such that $F(z,t)\neq0$ for all $(z,t) \in \partial W \times [0,1]$), then the maps 
$$
\frac{F(\cdot,t)}{|F(\cdot, t)|} : \partial W \to S^1
$$
are well define for all $t \in [0,1]$, ($S^1=\left\{z: |z|=1 \right\}$), more over since $F(z,1)=f(z)$ and $F(z,0)=z^7$ are homotpic we have that
$$
deg_2 \left(\frac{f}{|f|}\right)=deg_2 \left( \frac{z^7}{|z^7|}\right)
$$
but clearly $g(z)=z^7/|z^7|$ makes seven turns around any point $y \in S^1$, then #$\left(g^{-1}(y)\right) = 7$, and since $7 \equiv 1 \pmod 2$, we have that $deg_2(g)$ is nonzero, that is
$$
deg_2 \left(\frac{f}{|f|}\right)\neq 0
$$
this means that there exist $z\in W$ such that $f(z)=0$ as required.
