Show that the equation $$e^{-z}-2z^2=1$$ has exactly two solutions in the disk $|z|<1$, and that they are both real.
I'm trying to use Rouché's theorem to find the zeros for $f(z)=e^{-z}-2z^2-1=0$ but I can't seem to find a way to divide $f(z)=g(z)+h(z)$ such that $|g(z)|<|h(z)|$ for all $|z|<1$. It doesn't matter how I choose my functions, at $z=0$ I always get $|g(z)|=|h(z)|$.
Is Rouché's the wrong way to tackle the problem? If so, what method could you use instead?
Regarding the second part of the problem. Any push in the right direction would be appreciated since I have no idea how to show why the solutions have to be real.
The requirement $|g(z)|<|h(z)|$ only applies to the boundary, i.e. $C=\{z:|z|=1\}$.
On the set $C$ we have the following estimates,
$$|-2z^2| \geq 2$$
$$|e^{-z}-1|\leq e-1$$
So $f(z)$ has as many zeros (counting multiplicity) as $-2z^2$ in the open ball of radius $1$. Since $-2z^2$ has two zeros, $f$ has two zeros.
There is a clunk way to show that the roots in the ball must be real. We can show that the real function $$f(x) = e^{-x}-2x^2-1=0$$
has two roots in the interval $]-1,1[$ using the intermediate value theorem. We have the following,
$$f(-1) = e-3<0 \\ f(1) = e^{-1}-3<0\\ f(0) = 0\\ f(-\frac{1}{2}) = \sqrt{e}-\frac{1}{2}-1>0$$
By the IVT the function $f$ has a root in the interval $]-1,-\frac{1}{2}[$ and we also found a root at $0$. So $f(z)$ has two real roots in the open ball and these are the only roots by Rouche's theorem.