Convergent series of complex numbers

By David Loeffler (P865) on Wednesday, June 21, 2000 - 09:25 pm :

I am rather stuck on what looks like it ought to be a fairly easy problem: to show that if {z_n} is a series of complex numbers converging to a, then {(z_n)^1/2} converges to (a)^1/2. Essentially this seems to boil down to showing that the square root function is continuous; this is obvious because it is differentiable (except at 0), but I still can't actually find any way of constructing suitable N's for given e's.
David Loeffler

By Sean Hartnoll (Sah40) on Thursday, June 22, 2000 - 12:32 am :

Suppose |z_n-a| < e for n > N.

Then |z_n^1/2-a^1/2|=|z_n-a|/|z_n^1/2+a^1/2| < e/|z_n^1/2+a^{1 /2}|

But also |z_n| > |a|-e, so

< e/K where K is a constant =|(|a|-e)^1/2+a^1/2|

This essentially proves it, if you can't see the remaining steps let me know (or if you don't follow something). There are probably many other ways of doing this, in fact I think I have been a bit sloppy, but I think the idea is correct.

Sean

By David Loeffler (P865) on Thursday, June 22, 2000 - 08:24 pm :

Thanks. I hadn't thought of that lower bound for the denominator. I think I can see the remaining steps.

David.