# Inverse function and poles

1. Jul 5, 2010

### zetafunction

let be a function $$y=f(x)$$ with poles $$f(a_{i} ) = \infty$$ for some real 'a'

my question is if we define the inverse function g(x) so $$g(f(x))=x$$ ,then is this true

$$g(\infty)=a_{i}$$ my question is that it seems that g(x) would have several asymptotes as x-->oo how it can be ??

2. Jul 5, 2010

### Dickfore

The function

$$f(x) = \frac{A}{x - a} + \frac{B}{x - b}$$

has two simple poles at $x = a, b$. However, if you try to solve for its inverse, you get:

$$x = \frac{A}{y - a} + \frac{B}{y - b}$$

$$x (y - a) (y - b) = A (y - b) + B (y -a)$$

$$x y^{2} - (a + b) x y + a b x = (A + B) y - A b - B a$$

$$x y^{2} - [(a + b) x + (A + B)] y + (a b x + A b + B a) = 0$$

This is a quadratic equation that has two roots in the set of complex numbers, corresponding to two branches of the inverse function. For the behavior at $x = \infty$, you need to make the substitution $x \rightarrow 1/x$

$$\frac{1}{x} y^{2} - [\frac{a + b}{x} + (A + B)] y + (\frac{a b}{x} + A b + B a) = 0$$

Multiply out with $x$

$$y^{2} - [(a + b) + (A + B) x] y + [a b + (A b + B a) x] = 0$$

and take $x = 0$, and you get:

$$y^{2} - (a + b) y + a b = 0 \Rightarrow (y - a)(y - b) = 0$$

As you can see, there are two solutions even at infinity.