# complex analysis: integrals

• Dec 3rd 2007, 09:03 PM
hanahou
complex analysis: integrals
http://i3.photobucket.com/albums/y57...ty/prob3-1.jpg

how can i do this, applying Rouche Theorem and Cauchy Argument Principle?
• Dec 4th 2007, 07:38 AM
ThePerfectHacker
Use Cauchy's Argument Principle.

1)$\displaystyle \frac{1}{2\pi i}\oint_{\Gamma}\frac{f'(z)}{f(z)}dz = \mathbb{Z} - \mathbb{P}$ where $\displaystyle \mathbb{P}$ is the number of poles and $\displaystyle \mathbb{Z}$ is the number of zeros (both counting multiplicity). Thus, this equation tells us that $\displaystyle \mathbb{Z} - \mathbb{P} = 2$ but $\displaystyle \mathbb{P} = 0$ because $\displaystyle f(z)$ ia analytic on $\displaystyle \Omega$. Thus, the function $\displaystyle f(z)$ has two zeros $\displaystyle z_1$ and $\displaystyle z_2$ within the unit disk (and I let $\displaystyle z_1 = z_2$ so if it is the same zero of multiplicity two).

2)$\displaystyle \frac{1}{2\pi i}\oint_{\Gamma}z\frac{f'(z)}{f(z)}dz = z_1 + z_2$. Thus, $\displaystyle z_1+z_2 = 0$.

3)$\displaystyle \frac{1}{2\pi i}\oint_{\Gamma}z^2 \frac{f'(z)}{f(z)}dz = z_1^2+z_2^2 = \frac{1}{2}$.

Now solve for $\displaystyle z_1 \mbox{ and }z_2$.