# Thread: One More Complex Analysis Problem

1. ## One More Complex Analysis Problem

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) : x&#178; + y&#178; ≤ 1}

Thanks for the help.

Taypez

2. 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) : x&#178; + y&#178; ≤ 1}

Thanks for the help.

Taypez
Let:

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

then:

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

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

So we now need only consider the following integrals:

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

Now change to polars:

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

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

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

RonL