# Rouché's theorem

• February 2nd 2011, 02:20 AM
sinichko
Rouché's theorem
Help me please with this one:

Locate the zeros of

$f(z)=e^{z}-4z^{2}+1 \;, \left \{ |z|<1 \right \}$

and show that they are simple
• February 2nd 2011, 02:28 AM
tonio
Quote:

Originally Posted by sinichko
Help me please with this one:

Locate the zeros of

$f(z)=e^{z}-4z^{2}+1 \;, \left \{ |z|<1 \right \}$

and show that they are simple

Hint: check the values $f(-1)\,,\,\,f(0)\,,\,\,f(1)$ , for example.

Tonio
• February 2nd 2011, 02:57 AM
FernandoRevilla
Quote:

Originally Posted by sinichko
Locate the zeros of $f(z)=e^{z}-4z^{2}+1 \;, \left \{ |z|<1 \right \}$ and show that they are simple

If it is compulsory to apply Rouche's theorem as title says, then, for all $z$ such that $|z|=1$:

$|e^z+1|\leq 2<|-4z^2|$

According to Rouche's theorem, the number $N$ of zeros of $f(z)$ in $|z|<1$ including multiplicities is equal to the number of zeros of $h(z)=-4z^2$ in $|z|<1$ . That is, $N=2$ .

Tonio's hint for the restriction of $f$ to $\mathbb{R}$ allows to locate those two zeros and prove that both are simple.

Fernando Revilla