One More Complex Analysis Problem

**taypez** · December 18th 2006, 10:45 AM

Here is another Complex problem I can't figure out

If f is analytic for │z│≤1 show that

∫D f(x+iy) dx dy = f(0)pi

D = {(x,y) : x² + y² ≤ 1}

Thanks for the help.

Taypez

**CaptainBlack** · December 19th 2006, 12:54 AM

Originally Posted by taypez

Here is another Complex problem I can't figure out

If f is analytic for │z│≤1 show that

∫D f(x+iy) dx dy = f(0)pi

D = {(x,y) : x² + y² ≤ 1}

Thanks for the help.

Taypez

Let:

$f(z)=\sum_{n=0}^{\infty}a_n\,z^n,\ \ z \in D$

then:

$\int_D f(x+iy) dxdy=\sum_{n=0}^{\infty}a_n\,\int_D (x+iy)^n dxdy$

(the change in order of integration is permitted here).

So we now need only consider the following integrals:

$a_n \,\int_D (x+iy)^n\, dxdy$ .

Now change to polars:

$a_n \,\int_D (x+iy)^n\, dxdy=a_n \,\int_D r^n e^{i\,n \theta}\, r\,d\theta dr=a_n \,\int_0^{1}r^{n+1}\int_0^{2 \pi} e^{i\,n \theta}\, d\theta dr$ .

This last term is $a_0\, \pi$ when $n=0$ and $0$ otherwise, which is sufficient to prove the required result:

$\int_D f(x+iy) dxdy=\pi\,a_0=\pi \, f(0)$ .

RonL

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