# Complex Analysis-Analytic Functions

• Jun 19th 2010, 06:20 AM
WannaBe
Complex Analysis-Analytic Functions
1. Prove that there is no analytic function inside
Code:

`z:0<|z|<1`
such as every z in this region satysfies:
Code:

`f^2(z)=z`
.

2. Let f be analytic at the annulus:
Code:

` D=z:0<|z-a|<r`
.
Prove that if f has an antideriative function then Res(f,a)=0 .

Hope you'll be able to help me

Thanks a lot!
• Jun 19th 2010, 06:31 AM
Ackbeet
Regarding #2, what is a "former" function?
• Jun 19th 2010, 08:51 AM
WannaBe
iI meant antideriative...sry
• Jun 19th 2010, 08:53 AM
Bruno J.
For #1, suppose that such a function exists, and show that the Cauchy-Riemann equations are not satisfied at z=0.
Or you can differentiate implicitly with respect to \$\displaystyle z\$ and show that \$\displaystyle f'(z)\$ cannot be analytic at \$\displaystyle 0\$.
• Jun 19th 2010, 08:59 AM
Bruno J.
For #2, suppose that an antiderivative exists. Show that this implies that the integral along a simple closed contour inside the annulus vanishes... so what if the interior of the contour contains the point \$\displaystyle z=0\$?
• Jun 19th 2010, 10:12 AM
WannaBe
We can't use Cauchy-Riemann,because f isn't analytic at z=0...
How can we solve this problem?

2 is completely understandable...

Thanks @!
• Jun 19th 2010, 10:42 AM
Bruno J.
Quote:

Originally Posted by WannaBe
We can't use Cauchy-Riemann,because f isn't analytic at z=0...
How can we solve this problem?

2 is completely understandable...

Thanks @!

I'm sorry, I hadn't seen the point 0 was excluded. I hadn't had my coffee just yet! :)

One way which I can think of is this : a function analytic inside a domain maps a closed contour to a closed contour. So assume there exists such a function. Now take a simple closed contour around the origin, and show that \$\displaystyle f\$ does not map it to a closed contour (use the argument principle, and the fact that \$\displaystyle 2f(z)f'(z)=1\$ inside the punctured disc).
• Jun 19th 2010, 11:39 AM
WannaBe
Thanks...