Use Rouches Theorem to show that if then there are two zeros of p(z) within the unit circle.
Now i chose and
and then evaluated these at z=1, as on the boundary of the circle to get,
i then said that by rouches theorem, (i.e p(z)) has the same number of zeros as g(z), i.e. 2, if
which leads to me concluding that this is when
Does anyone know what i am doing wrong?
May 12th 2011, 12:23 PM
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
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.