# Thread: Determine the number of zeros of a given function by using Rouche's Th

1. ## Determine the number of zeros of a given function by using Rouche's Th

please help me to determine the number of zeros of the following function in the disk D(0,2) by using Rouche's Th:

$\displaystyle f(z)=z^2e^z-z, z\in D(0,2)$

z is a complex number!~

2. Originally Posted by frankmelody
please help me to determine the number of zeros of the following function in the disk D(0,2) by using Rouche's Th:

$\displaystyle f(z)=z^2e^z-z, z\in D(0,2)$
Is there a problem with saying that $\displaystyle z^2e^z\geq{z}$ and $\displaystyle z^2e^z$ has two zeros on that range, thus the actual function has two? Or am I remembering the theorem wrong...

3. Originally Posted by Mathstud28
Is there a problem with saying that $\displaystyle z^2e^z\geq{z}$ and $\displaystyle z^2e^z$ has two zeros on that range, thus the actual function has two? Or am I remembering the theorem wrong...
what do you mean by $\displaystyle z^2e^z\geq{z}$? I think z is a complex number here......

4. Originally Posted by frankmelody
please help me to determine the number of zeros of the following function in the disk D(0,2) by using Rouche's Th:

$\displaystyle f(z)=z^2e^z-z, z\in D(0,2)$

z is a complex number!~
Let $\displaystyle g(z) = z^2e^z$ and $\displaystyle h(z) = -z$.
Then on $\displaystyle |z|=2$ we have $\displaystyle |g(z)| > |h(z)|$.

Therefore, $\displaystyle g(z)+h(z)$ has as many zeros as $\displaystyle g(z)$ in $\displaystyle D(0,2)$

5. Originally Posted by ThePerfectHacker
Let $\displaystyle g(z) = z^2e^z$ and $\displaystyle h(z) = -z$.
Then on $\displaystyle |z|=2$ we have $\displaystyle |g(z)| > |h(z)|$.

Therefore, $\displaystyle g(z)+h(z)$ has as many zeros as $\displaystyle g(z)$ in $\displaystyle D(0,2)$
I am afraid you are wrong, you said on "$\displaystyle |z|=2$ we have $\displaystyle |g(z)| > |h(z)|$", but if you let z=-2, obviously you are wrong.