Help me please with this one:
Locate the zeros of
$\displaystyle f(z)=e^{z}-4z^{2}+1 \;, \left \{ |z|<1 \right \}$
and show that they are simple
If it is compulsory to apply Rouche's theorem as title says, then, for all $\displaystyle z$ such that $\displaystyle |z|=1$:
$\displaystyle |e^z+1|\leq 2<|-4z^2|$
According to Rouche's theorem, the number $\displaystyle N$ of zeros of $\displaystyle f(z)$ in $\displaystyle |z|<1$ including multiplicities is equal to the number of zeros of $\displaystyle h(z)=-4z^2$ in $\displaystyle |z|<1$ . That is, $\displaystyle N=2$ .
Tonio's hint for the restriction of $\displaystyle f$ to $\displaystyle \mathbb{R}$ allows to locate those two zeros and prove that both are simple.
Fernando Revilla