Now let us take it a bit further for a general real base \( b>1 \).
Proposition. The only real base \( b>1 \) for which the complex function \( b^z \) has a fixed point \( a \) with \( |\exp_b'(a)|=1 \) is \( b=e^{1/e} \) and in this case the only such fixed point is \( e \).
Simalarly to the previous post we assume a fixed point being given by
\( a=re^{i\alpha} \) then for parabolicity we have to prove
\( |\exp_b'(a)|=\ln(b)|\exp_b(a)|=\ln(b)|a|=\ln(b)r=:s \)
and we have the equation system:
\( \ln(b)r\cos(\alpha)=\ln( r) \) (1) and \( \ln(b)r\sin(\alpha)=\alpha \) (2)
Both equation squared and added yields
\( (\ln(b)r)^2=\ln ( r)^2+\alpha^2 \) and hence
\( \alpha=\pm\sqrt{(\ln(b)r)^2-\ln( r)^2} \)
For the case \( s=1 \): \( r=1/\ln(b) \) and \( \ln( r)=-\ln(\ln(b)) \) and so
\( \alpha=\pm\sqrt{1-(\ln(\ln(b)))^2} \).
Now we put this into equation (1) with \( x=-\ln(\ln(b)) \):
\( \cos(\pm\sqrt{1-x^2})=x \)
We substitute \( \sqrt{1-x^2}=:y \), \( x=\pm\sqrt{1-y^2} \)
\( \cos(\pm y)=\pm\sqrt{1-y^2} \). But the functions on both sides are well known. The right side is a circle with radius 1 and the left side is always above or below the right side in the region \( y=-1..1 \). So equality happens exactly at \( y=0 \).
This implies \( \ln( r)=x=\pm 1 \) however only \( \ln( r)=1 \) satisfies equation (1). Then \( b=\exp( \exp( -x))=e^{1/e} \) is the only real base for which \( \exp_b \) has a parabolic fixed point and this fixed point is given by \( \alpha=0 \) and \( \ln( r)=1 \) and is hence the real number \( e \).
Proposition. The only real base \( b>1 \) for which the complex function \( b^z \) has a fixed point \( a \) with \( |\exp_b'(a)|=1 \) is \( b=e^{1/e} \) and in this case the only such fixed point is \( e \).
Simalarly to the previous post we assume a fixed point being given by
\( a=re^{i\alpha} \) then for parabolicity we have to prove
\( |\exp_b'(a)|=\ln(b)|\exp_b(a)|=\ln(b)|a|=\ln(b)r=:s \)
and we have the equation system:
\( \ln(b)r\cos(\alpha)=\ln( r) \) (1) and \( \ln(b)r\sin(\alpha)=\alpha \) (2)
Both equation squared and added yields
\( (\ln(b)r)^2=\ln ( r)^2+\alpha^2 \) and hence
\( \alpha=\pm\sqrt{(\ln(b)r)^2-\ln( r)^2} \)
For the case \( s=1 \): \( r=1/\ln(b) \) and \( \ln( r)=-\ln(\ln(b)) \) and so
\( \alpha=\pm\sqrt{1-(\ln(\ln(b)))^2} \).
Now we put this into equation (1) with \( x=-\ln(\ln(b)) \):
\( \cos(\pm\sqrt{1-x^2})=x \)
We substitute \( \sqrt{1-x^2}=:y \), \( x=\pm\sqrt{1-y^2} \)
\( \cos(\pm y)=\pm\sqrt{1-y^2} \). But the functions on both sides are well known. The right side is a circle with radius 1 and the left side is always above or below the right side in the region \( y=-1..1 \). So equality happens exactly at \( y=0 \).
This implies \( \ln( r)=x=\pm 1 \) however only \( \ln( r)=1 \) satisfies equation (1). Then \( b=\exp( \exp( -x))=e^{1/e} \) is the only real base for which \( \exp_b \) has a parabolic fixed point and this fixed point is given by \( \alpha=0 \) and \( \ln( r)=1 \) and is hence the real number \( e \).
