This question was given in an exam in complex analysis: Let $f \left( z \right) = 3z^{100} - e^z$. Find all of $f$'s roots in $D \left ( 0,1 \right)$ and show that they are simple roots.
I've seen these types of questions with polynomials and the usual technique is to use Rouche's Theorem but it seems that the conditions are not satisfied here. Neither function $3z^{100}$ nor $e^z$ is strictly greater than the other in the unit disc.
EDIT: I've misinterpreted the conditions for Rouche's theorem. As mentioned in the comments, the theorem only requires that one function be strictly greater than the other on the boundary of the compact domain (in this case, the unit circle) and indeed that condition is met.
For $\left|z \right|=1$ we have that $\left| 3z^{100} \right| = 3$ and $\left|e^z \right| = \left| e^{ \mathfrak{R}z } \right| \leq e<3$. Thus we have that $\left| 3z^{100} \right|>\left| e^z \right|$ on the unit circle.
Therefore, using Roche's theorem we obtain that $3z^{100}$ and $3z^{100} - e^z$ have the same number of zeros in the unit disc. $3z^{100}$ clearly has a single zero $z=0$ there, of multiplicity 100 and therefore $3z^{100} - e^z$ also has 100 zeroes there.
To show that these zeroes are simple, we can simply differentiate. Let $z_0$ be a zero of $f$ in the unit disc. Observe $f^\prime \left( z \right) = 300z^{99} - e^z$. Assume towards contradiction that $z_0$ is a zero of $f^\prime$. Then $ 300{z_0}^{99} - e^{z_0}=0=3{z_0}^{100}- e^{z_0}$. This yields that $z_0=0$ or $z_0=100$ (not in the unit circle thus we can disregard it), thus $z_0=0$ which is clearly a contradiction since $0$ is not a zero of $f$.