
Rouché's theorem
I have a question that is as follows,
Let $\displaystyle p(z)=z^{2}ce^{z}$ where $\displaystyle 0<c<1$
Use Rouches Theorem to show that if $\displaystyle c<{\frac{1}{e}}$ then there are two zeros of p(z) within the unit circle.
Now i chose $\displaystyle g(z)=z^{2}$ and $\displaystyle h(z)=ce^{z}$
and then evaluated these at z=1, as $\displaystyle \left  z \right =1$ on the boundary of the circle to get,
$\displaystyle \left  h(1) \right =\left  \frac{c}{e} \right =\frac{c}{e}$
$\displaystyle \left  g(1) \right =\left  1^{2} \right =1$
i then said that by rouches theorem, $\displaystyle g(z)+h(z)$ (i.e p(z)) has the same number of zeros as g(z), i.e. 2, if
$\displaystyle \left  h(1) \right <\left  g(1) \right $
which leads to me concluding that this is when
$\displaystyle c<e$
Does anyone know what i am doing wrong?

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.

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.