TITLE: Number of solutions to $z=\frac{1}{4}e^z$ inside the unit circle
QUESTION [5 upvotes]: I'm trying to solve this using Rouche's theorem and I think I did. I'm completely winging this based on Wiki's article about it so please tell me if I'm off the ball.
So the problem is actually finding the number of zeroes of $z-\frac{1}{4}e^z$ inside the unit circle.
So I choose my $f$ function to be $z$ and my $g$ function to be $\frac{1}{4}e^z$. No I have to show that $|f(z)|>|g(z)|$ on the unit circle.
$$|z|=1\Rightarrow z =\cos\phi+i\sin\phi\Rightarrow e^z=e^{\cos\phi}+e^{i\sin\phi}=e^{\cos\phi}+\cos\sin\phi+i\sin\sin\phi$$
Now the imaginary part of $-\frac{1}{4}e^z$ is in $[-\frac{1}{4},\frac{1}{4}]$ because $\sin$ is in $[-1,1]$, and the real part is in $[-\frac{e+1}{4},\frac{e+1}{4}]\approx[-0.93, 0.93]$. The point with the greatest absolute value in that rectangle is at $(0.93, 1/4)$ and it's absolute value is less than $1$.
So I conclude that $|f(z)|$ is indeed greater than $|g(z)|$ on the unit circle, and since $f$ has one zero inside of it, $f+g$ which is the original function, also has one zero inside of the circle.

REPLY [4 votes]: I think that you are overcomplicating it a bit. Also, the function having a certain amount of zeros is not the same thing as the equation having the same amount of solutions. You'd have to check that the order of each zero is one.
Suppose $|z| = 1$. We have ${\rm Re}(z) \leq 1$ and hence: $$\left|\frac{1}{4}e^z\right| = \frac{1}{4}|e^z| = \frac{1}{4}e^{{\rm Re}(z)} \leq \frac{1}{4}e^{1}  = \frac{e}{4}< 1 = |z|,$$ so $z - \frac{1}{4}e^z$ has the same quantity of zeros that $z$ in $B(0,1)$: one zero. So that equation has one solution.