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.
 A: Hint: Use $f(z)=\exp(z)-1-3z$, $g(z)=-3z$. then $f(z)-g(z)=\exp(z)-1$, and on $|z|=1$, $|\exp(z)-1|\leq \exp(|z|)-1=e-1<3$ (use that $\exp(z)-1=\sum_{n\geq 1} z^n/n!$).
A: For $z\ne 0$, let $\gamma$ be the line segment connecting $0$ and $z$. Then,
$$|e^z -1| = \left| \int _\gamma e^u du \right| \le \int _\gamma \sup_{u\in \gamma}|e^u| |du| \le e|z| $$ 
A: 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.
A: There is only the one solution.
An elementary proof goes as follows:
Let $z=x+iy$ with $x,y$ real and $x^2+y^2 \leq 1$.  Then 
$$
e^x\cos y = 3x+1\\
e^x\sin y = 3y
$$
The first of these can be solved for $y$, giving (using the fact that  $|y|$ cannot be greater than $1$ to eliminate all the other values of arc cosine)
$$
y = \pm \cos^{-1} (e^{-x}(1+3x))
$$
Using the $-$ sign, the second equation becomes 
$$
e^x \sqrt {1-e^{-2x}(1+3x)^2} = 3 \cos^{-1}(e^{-x}(1+3x))
$$
The argument of the arc cosine is greater than $1$ for all $0<x\leq 1$ so for this equation to hold, $x$ must not be positive. However, for $-1 \leq x\leq 0$, 
 $
e^x \sqrt {1-e^{-2x}(1+3x)^2} $
 is concave upward and its derivative at $x=0$ is less negative than the derivative of $
 3 \cos^{-1}(e^{-x}(1+3x)) $.  Thus for $x\leq 0$ these are only equal at $x=0$.
Using the $+$ sign, of course is even more hopeless for finding another solution, since for $-1\leq x <0$ the LHS is negative and the RHS is positive.
So the only solution is $x=0$.  And then $\sin y = 3y$ forces $y=0$.
