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 aresimple

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- Feb 2nd 2011, 02:20 AMsinichkoRouché's theorem
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* - Feb 2nd 2011, 02:28 AMtonio
- Feb 2nd 2011, 02:57 AMFernandoRevilla

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