Rouché's theorem

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May 12th 2011, 12:13 PM
MathMan170

Rouché's theorem

I have a question that is as follows,

Let $p(z)=z^{2}-ce^{-z}$ where $0<c<1$

Use Rouches Theorem to show that if $c<{\frac{1}{e}}$ then there are two zeros of p(z) within the unit circle.

Now i chose $g(z)=z^{2}$ and $h(z)=-ce^{-z}$

and then evaluated these at z=1, as $\left | z \right |=1$ on the boundary of the circle to get,

$\left | h(1) \right |=\left | \frac{-c}{e} \right |=\frac{c}{e}$

$\left | g(1) \right |=\left | 1^{2} \right |=1$

i then said that by rouches theorem, $g(z)+h(z)$ (i.e p(z)) has the same number of zeros as g(z), i.e. 2, if

$\left | h(1) \right |<\left | g(1) \right |$

which leads to me concluding that this is when

$c<e$

Does anyone know what i am doing wrong?
May 12th 2011, 12:23 PM
Tinyboss

Looks right to me. Of course, you have also proved the problem as stated. ;-) But I think you're right, there's probably an error in the statement.
May 12th 2011, 12:31 PM
MathMan170

Yes, i guess it does satisfy it. But the next part of the question is that if c>1/e then we might think that there is only one solution in the cicle, then by setting c=1/2, show using the integral form of the arguement principle that there is only one zero in this case.