How did they get $e^{-2n\pi}$ ?
I found this question in the college course material. The answer was provided and it doesn't matches mine.
Express $\pi^{i}$ in $A+iB$ form.
What I did:
Let $\pi^{i}$ =$m$
taking logarithm both sides
$ilog{\pi}=log(m)$
$e^{(i log{\pi})}=m$
From the demoivre's theorem we have :
$cos(log{\pi})+isin(log{\pi})=m$
Which is in the form of $A+iB$ but the answer in the solution manual is:
$e^{-2n\pi}(cos(log{\pi})+isin(log{\pi}))$
Point out my fault, Kindly!
Thanks in advance!
The complex exponentiation is defined as $$z^w=e^{w\log z}$$ where $\log z=\ln |z|+i\arg z.$ So,
$$\pi^i=e^{i\log \pi}=e^{i(\ln \pi+i\arg \pi)}.$$ Now, $\arg \pi=2n\pi,$ depending on the branch you choose for the complex logarith. Thus
$$\pi^i=e^{i(\ln \pi+2n\pi i)}=e^{-2n\pi}e^{i\ln\pi}=e^{-2n\pi}(\cos(\ln \pi)+i\sin(\ln \pi)).$$
Obviously if you work with the principal value of the complex logarith (that is, the argument is in $(-\pi,\pi]$ then $n=0$ and you have $$\pi^i=\cos(\ln \pi)+i\sin(\ln \pi).$$
The problem with complex exponential is that it is not injective as in the real case (note that $e^{2n\pi i}=1$). For this reason the logarithm is a multivalued function (that is, you have to choose a branch to work with).