09/08/2007, 10:34 AM

Proposition. All fixed points of are hyperbolic, i.e. for each complex with .

Clearly . So we want to show that .

We can exclude the case as this implies and we know that 0 is not a fixed point of .

For the other case we set and get

and hence the equation system:

(1) and (2).

We square both equation and add them:

.

But beware, this is only a necessary condition on the fixed points. The fixed points lie discretely on the complex plane. Not every point satisfying this equation is a fixed point.

But from this condition we can look what happens for . For we get and we see that both values of do not satisfy equation (2). So there is no fixed point with .

Clearly . So we want to show that .

We can exclude the case as this implies and we know that 0 is not a fixed point of .

For the other case we set and get

and hence the equation system:

(1) and (2).

We square both equation and add them:

.

But beware, this is only a necessary condition on the fixed points. The fixed points lie discretely on the complex plane. Not every point satisfying this equation is a fixed point.

But from this condition we can look what happens for . For we get and we see that both values of do not satisfy equation (2). So there is no fixed point with .