09/08/2007, 10:34 AM
Proposition. All fixed points of \( e^x \) are hyperbolic, i.e. \( |\exp'(a)|\neq 0,1 \) for each complex \( a \) with \( e^a=a \).
Clearly \( \exp'(a)=\exp(a)=a \). So we want to show that \( |a|\neq 0,1 \).
We can exclude the case \( |a|=0 \) as this implies \( a=0 \) and we know that 0 is not a fixed point of \( \exp(x) \).
For the other case we set \( a=re^{i\alpha}=r\cos(\alpha)+ir\sin(\alpha) \) and get
\( re^{i\alpha}=e^{r\cos(\alpha)+ir\sin(\alpha)}=e^{r\cos(\alpha)}e^{ir\sin( \alpha )} \) and hence the equation system:
\( \ln ( r )=r\cos(\alpha) \) (1) and \( \alpha=r\sin(\alpha) \) (2).
We square both equation and add them:
\( \ln ( r)^2+\alpha^2=r^2 \cos(\alpha)^2+ r^2\sin(\alpha)^2=r^2 \)
\( \alpha=\pm\sqrt{r^2-\ln ( r)^2} \).
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 \( |a|=r=1 \). For \( r=1 \) we get \( \alpha=\pm 1 \) and we see that both values of \( \alpha \) do not satisfy equation (2). So there is no fixed point with \( |a|=1 \).
Clearly \( \exp'(a)=\exp(a)=a \). So we want to show that \( |a|\neq 0,1 \).
We can exclude the case \( |a|=0 \) as this implies \( a=0 \) and we know that 0 is not a fixed point of \( \exp(x) \).
For the other case we set \( a=re^{i\alpha}=r\cos(\alpha)+ir\sin(\alpha) \) and get
\( re^{i\alpha}=e^{r\cos(\alpha)+ir\sin(\alpha)}=e^{r\cos(\alpha)}e^{ir\sin( \alpha )} \) and hence the equation system:
\( \ln ( r )=r\cos(\alpha) \) (1) and \( \alpha=r\sin(\alpha) \) (2).
We square both equation and add them:
\( \ln ( r)^2+\alpha^2=r^2 \cos(\alpha)^2+ r^2\sin(\alpha)^2=r^2 \)
\( \alpha=\pm\sqrt{r^2-\ln ( r)^2} \).
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 \( |a|=r=1 \). For \( r=1 \) we get \( \alpha=\pm 1 \) and we see that both values of \( \alpha \) do not satisfy equation (2). So there is no fixed point with \( |a|=1 \).
