# Rouche's theorem

1. Mar 20, 2012

### soofjan

I have been trying to solve the following question for a while now. I need to find the number of roots of:
f(z) = 6z^2 - 6 + Log(1+z)
in the area
D = { z is complex : |z-1| < 1 }

I assume this is solved by Rouche's theorem that requires me to find 2 analytic functions, h(z) and g(z) in D, where | g(z) | < | h(z) | on the verge of D, and f(z) = g(z) + h(z).

So I tried all possibilites, but none of them seem to work. Am I missing something?
Thanks.

2. Mar 21, 2012

### jackmell

How about letting:

$$g=6z^2-6$$
$$f=6z^2-6+Log[1+z]$$

Then is $|f-g|\leq |g|$ for $z=1+e^{it}$
If so then $f$ has one zero in that domain.

Also, you can confirm this by plotting the argument change around the circle. This is the Mathematica code to see this:

Code (Text):

Plot[Arg[6 z^2 - 6 + Log[1 + z]] /.
z -> 1 + Exp[I t], {t, -\[Pi], \[Pi]}]

Last edited: Mar 21, 2012
3. Mar 21, 2012

### soofjan

If you draw the circle, you can see that it is possible for the distance from (0,0) to a z on the circle to be 1 (when the argument is 60 degrees), and then | 6z^2 - 6 | = 0. In that case, g(z) won't be bigger than f(z) - g(z) for every z, as Rouche requires.

This is what buffles me about this question. Maybe I'm missing something in the pure definition and I'll be happy if someone could point out my mistake. Thanks again.

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