Limits and continuity for complex functions

[jaejoon89]

Homework Statement

Given

f(z) = (1/(z-a))(1/z^2 - 1/a^2)

a is a fixed complex value


If you define a function over the complex numbers by mapping z to f(z) when z is not equal to a, how should this function be defined at a s.t. it's continuous at point a? Explain.

Homework Equations

A function will be continuous at a if

lim(z->a) f(z) = f(a)

The Attempt at a Solution

f(z) = -(z+a) / (z^2 a^2)
lim(z->a) = -2/a^2 = f(a)

I'm really not sure how to explain it or "justify it" as I'm supposed to beyond the 2 lines written above.

I'm really not sure what else is needed... If a is an interior point of the domain (?) then can't continuity be shown using the delta/epsilon definition?

 

[snipez90]

Well if you took the limit as z -> a, then that expression is just the derivative of 1/z^2 at a, and differentiability implies continuity so...

 

[jaejoon89]

Thanks, but I think I'm supposed to justify it more along the lines of the delta/epsilon definition, or some other way without talking about differentiation. Which is back to my original question: how do you do that here?

I know that for any real number ε > 0 there exists a real number δ > 0 s.t. | f(z) − a | < ε for all complex numbers that satisfy | z − a | < δ. But how do I show this for this function?

 

[snipez90]

Well either way, it seems like you need to know about differentiability to even figure out how to define f at a so that f is continuous or else you can't really prove continuity but then this all seems kind of pointless since differentiability implies continuity is the most basic result once differentiability has been defined.

But whatever, so basically f(a) = -2/a^3 right. So then you start with

\left|\frac{\frac{1}{z^2} - \frac{1}{a^2}}{z-a} + \frac{2}{a^3}\right| &lt; \varepsilon

and you should end up with

\left|\frac{(z-a)(2z + a)}{a^3 z^2}\right| &lt; \varepsilon.

Fill in the steps in between and/or check that what I wrote down is actually correct. Can you do the rest from here?