Find the number of zeros in $D(0;1)$ of $\cos \pi z -100z^{100}$
The goal is to apply Rouché's Theorem to either $\cos \pi z$ or $-100z^{100}$. Here is what I did:
Let $f(z)=-100z^{100}$ and $g(z)=\cos \pi z$. Now $$\begin{align*} |\cos\pi z|&=\left \vert \sum_{n=0}^{\infty}(-1)^n\frac{\pi z^{2n}}{(2n)!} \right \vert\\ &\leq \sum_{n=0}^{\infty}|(-1)^n|\frac{|\pi z|^{2n}}{(2n)!}\\ &= 1+\frac{\pi^2}{2!}+\frac{\pi^4}{4!}+\dots\\ &<1+\frac{4^2}{2!}+\frac{4^4}{4!}+\dots\\ &=\sum_{n=0}^{\infty}\frac{4^{2n}}{(2n)!}\\ &=\cosh 4 \end{align*}$$ And on the other hand, $|f(z)|=|-100^{100}|=100$. Since $\cosh 4<100$, $f(z)$ has the same number of zeros that $f(z)+g(z)$ in $D(0;1)$, that is, $100$.
I don't know if the bound for $g$ is right, because Wolfram Alpha gives that the equation has two solutions. Thanks in advance.
Your solution looks fine. A slight nitpick is that in doing your estimates, you should mention you are considering $z$ on the boundary of the disc, i.e. $|z|=1$.
Now, there are two real solutions to your equation, which is what you probably found on WolframAlpha. If you ask it to search for complex roots of your equation, you will find many more solutions.