Thread: Removable Singularities Proof? (complex analysis)

assumptions:
1)a function has an isolated singularity at a point k (k is in the complex plane)
2)the function is continuous at the point k


How does one prove that the function has a removable singularity at k?


so far i have all the definitions...
the function is analytic in a punctured neighbourhood around k (isolated sing), and limf(z)=f(k) as z tends to k

Is that a proposed problem? It is a particular case of the Riemann Theorem (long proof): If is analytic and , then has a removable singularity at .

OK, thanks. It's a homework problem, and we haven't covered the Riemann Theorem in lectures.. and looking ahead at past year's notes, it's not covered either. I'm not sure if we can use it, or prove otherwise. thank you nevertheless

ah, yes, i found riemann's theorem online.. makes sense, though it is quite lengthy. but it supposes that lim(z-z_0)f(z)=0 as z tends to z_0

Yes, this is essentially Riemann's theorem on removable singularities. I don't know what proof you saw online, but it shouldn't be particularly long or complicated. I will sketch a proof for you.

Essentially, if the function is bounded on a punctured disk and is holomorphic everywhere in that region, the singularity is removable. The intuition behind this is pretty simple; think back to some basic calculus course where you had a function like . The function looks like it should blow up at 1, but of course this is just in disguise, with there being a removable singularity at 1.

Suppose then that f is holomorphic on an open set containing and its closure. Denote the boundary of by . The idea is to basically prove that the Cauchy integral formula holds, and we will then get the desired extension.

What we do is set up a double keyhole contour around any point and . You can then find that

for all , and since this function agrees everywhere with the original function and is holomorphic function, it must be our desired extension. Thus the singularity is removable.