Contour integral in complex plane (tricky)

215 Views Asked by At

Let U be a simply connected domain with a simple closed boundary curve C oriented anticlockwise, and define for all w ∈ C \ C $$ g(w)=\oint_C \frac{e^zdz}{(z-w)^2}$$ Find a formula for g(w) which does not involve integration.

I haven't seen something like this before...We have f(z) and g(w) together. Is the procedure for Cauchy's integration formula same as ordinary ones? Or is there something special here?

1

There are 1 best solutions below

0
On BEST ANSWER

The residue theorem will give the result immediately. But if you don't know the theorem, there's an informal way to see that: First we may expand $e^z=\sum_n z^n/n!$. Then the integration is just $e^{z+w}/z^2$, expand and notice that $\oint_C z^n$ vanishes when $n\not = -1$. The only term remain is just $$\oint e^w \frac{z}{1! z^2}dz = 2\pi i e^w$$ (PS: this way is informal since it does not prove that the integration and summation can be interchanged, but this can be proved because $e^z$ converges uniformly in a bounded area.)