Number of Solutions to $e^{z}-3z-1=0$ in the Unit Disk I am working through some of the past qualifying exams in complex analysis and I am a bit stuck on the question I posed in the title. My immediately thought is use Rouche's Theorem. For instance, I tried letting $f(z)=e^{z}$ and $g(z)=3z+1$ in hopes of getting $|f(z)|\leq |g(z)|$ on $|z|=1$. But this is false since on $|z|=1$.
$$
|f(z)|\leq\sup_{x\in[-1,1]}e^{x}\leq e\not< 2\leq |3z+1|=|g(z)|.
$$
Clearly, there is at least one solution since $z=0$ works. I am thankful for any ideas as to how to proceed.

For $z\ne 0$, let $\gamma$ be the line segment connecting $0$ and $z$. Then,
$$|e^z -1| = \left| \int _\gamma e^u du \right| \le \int _\gamma \sup_{u\in \gamma}|e^u| |du| \le e|z| $$ 
