consider the function

 

[math]\frac{1}{\epsilon^2 + z^2}[/math]

 

So we know that there are two poles, one at [math]z = i \epsilon[/math], one at [math]z = - i \epsilon[/math]. So when this function never hits 0 on the real line, how do the singularities affect its behavior on the line?

 

Okay, so poles are a subclass of singularities. I think that [math]z = i \epsilon[/math] and [math]z = - i \epsilon[/math] are poles - I may be wrong here. The question is - how do complex singularities (complex poles in this case) affect a function's behavior when the function is plotted on the real line?

No, you are correct.

[math]\frac{1}{\epsilon^2+ z^2}= \frac{1}{(z- i\epsilon)(z+ i\epsilon)}[/math]

and so has a pole of order 1 at [math]z= i\epsilon[/math] and [math]z= -i\epsilon[/math]. As to how that affects its behavior on the real line, here is a simple example: If you were to construct its Taylor's series, about z= 0, say, you would find that the radius of convergence was [math]\epsilon[/math]. That's because, in the complex plane, the "radius of convergence" really is a "radius"! In the complex plane, the series converges as long as it doesn't hit a "problem"- which, in the case of [math]\frac{1}{\epsilon^2+ z^2}[/math] is at [math]i\epsilon[/math] and [math]-i\epsilon[/math].

 

Here's a nice application: Suppose you were to construct the Taylor's series for [math]\frac{1}{\epsilon^2+ x^2}[/math] about the point [math]x= x_0[/math]. What would its radius of convergence be? You don't need to actually calculate the series to answer that- it is the distance from [math]x= x_0[/math] to [math]i\epsilon[/math] and [math]-i\epsilon[/math], [math]\sqrt{x_0^2+ \epsilon^2}[/math].