What is the Limit of a Complex Integral as the Radius Approaches Zero?

[iluvphysics]

Homework Statement

Let U be open in C, f : U -> C continuous.
Prove that

<br /> \lim_{R\rightarrow 0} \int_0^{2\pi} f(Re^{it}) dt = 2\pi f(0)<br />

Homework Equations

<br /> \lim_{R\rightarrow 0} f(Re^{it}) dt = f(0)<br />

Also

<br /> \int_0^{2\pi} \lim_{R\rightarrow 0} f(Re^{it}) dt = \int_0^{2\pi} f(0) = 2\pi f(0)<br />

The Attempt at a Solution

The question then just resolves to passing the limit inside the integral.
I would think then we'd use either uniform convergence or DCT, since the integral of a complex function is simply the sum of the integrals of its real and imaginary parts.

Would this be the correct way of doing it, or is there some other way from complex analysis instead?

 

[snipez90]

I don't know any complex analysis, but for DCT, the fact that the integral of a complex function is the sum of the integrals of its real and imaginary parts is irrelevant right? The dominating function would just be the upper bound guaranteed by continuity of f right?

 

[ninty]

How would continuity imply boundedness in this case?
I would understand that if U was compact.