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.

Just use Cauchy's argument principle Wikipedia, Wolfram Mathworld:
For a function $f$ meromorphic on a set enclosed by contour $C$:
$$\oint_{C}\frac{f'(z)}{f(z)}dz = 2\pi i(N-P)$$
Where $N$ are number of zeros and  $P$ are number of poles of  $f(z)$ inside contour $C$ (on said set).
I could not find right symbol for direction, but should of course be correctly oriented Jordan curve and all that. The unit circle counterclockwise will probably do.
Wolfram Alpha command for integral.
