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