Does $i^i$ and $i^{1\over e}$ have more than one root in $[0, 2 \pi]$ How to find all roots if power contains imaginary or irrational power of complex number? How do I find all roots of the following complex numbers?
$$(1 + i)^i, (1 + i)^e, (1 + i)^{ i\over e}$$
EDIT:: Find the value of $i^i$ and $i^{1\over e}$ in $[0, 4\pi]$?

Let $e^z=N≠0$ ,So, $N=e^z=e^z\cdot e^{2n\pi i}$ as $e^{2n\pi i}=1$ for any integer $n$.
$N=e^{z+2n\pi i}\implies Log N=z+2n\pi i$ 
If $n=0$ ,we get the principal value $z=\log N$, so $LogN=\log N+2n\pi i$
Let $A=(a+ib)^{x+iy}$
$Log A=(x+iy)Log(a+ib)$
Let $a=r\cos \theta$ and $b=r\sin \theta\implies a+ib=r(\cos \theta+i\sin \theta)=re^{i\theta}$
So, $r=\sqrt{a^2+b^2}$ and $\theta=\tan ^{-1}\frac{b}{a}$
then $Log(a+ib)=\frac{1}{2}\log (a^2+b^2)+i(2k\pi+\tan ^{-1}\frac{b}{a})$ where $k$ is any integer.
So, $Log A=(x+iy)Log(a+ib)=(x+iy)(\frac{1}{2}\log (a^2+b^2)+i(2k\pi+\tan ^{-1}\frac{b}{a}))$
So, $Log A=\frac{1}{2}x\log (a^2+b^2)-y(2k\pi+\tan ^{-1}\frac{b}{a})+i(\frac{1}{2}y\log (a^2+b^2)+x((2k\pi+\tan ^{-1}\frac{b}{a}))$
So, $$A=(a+ib)^{x+iy}=e^{\frac{1}{2}x\log (a^2+b^2)-y(2k\pi+\tan ^{-1}\frac{b}{a})}\cdot e^{i({\frac{1}{2}y\log (a^2+b^2)+x(2k\pi+\tan ^{-1}\frac{b}{a}})}$$
$$=e^{\frac{1}{2}x\log (a^2+b^2)-y(2k\pi+\tan ^{-1}\frac{b}{a})} cis(({\frac{1}{2}y\log (a^2+b^2)+x(2k\pi+\tan ^{-1}\frac{b}{a}})) $$ where $cis(P)=\cos P+i\sin P$
(1)For $(1+i)^i$ $a=b=y=1$ and $x=0$
So, $r^2=a^2+b^2=2$ and $\sin \theta=\cos \theta=\frac{1}{\sqrt2}\implies \theta=\frac{\pi}{4}$
$(1+i)^i=e^{-(2k\pi+\frac{\pi}{4})}(\cos (\frac{1}{2}\log 2)+i\sin (\frac{1}{2}\log 2))$ where $k$ is any integer.
(2)For $(1+i)^e, a=b=1$ and $x=e,y=0$
So, $r^2=a^2+b^2=2$ and $\sin \theta=\cos \theta=\frac{1}{\sqrt2}\implies \theta=\frac{\pi}{4}$
$(1+i)^e=e^{\frac{1}{2}e\log 2}\cdot cis(e(2k\pi+\frac{\pi}{4}))$ where $k$ is any integer.
$(1+i)^e=2^{(\frac{e}{2})}\cdot cis(e(2k\pi+\frac{\pi}{4}))$
(3) For $(1+i)^{\frac{i}{e}}, a=b=1$ and $x=0,y=\frac{1}{e}$
$(1+i)^{\frac{i}{e}}=e^{-\frac{1}{e}(2k\pi+\frac{\pi}{4})}\cdot cis(\frac{1}{2e}\log 2)$ where $k$ is any integer.
(4) For $i^i,a=0,b=1,x=0,y=1\implies r^2=1$ and $\sin \theta=1$ and $\cos \theta=0\implies \theta=\frac{\pi}{2}$
So, the general value of $i^i=e^{-(2k\pi+\frac{\pi}{2})}$ where $k$ is any integer.
Now by the given condition, $0 ≤e^{-(2k\pi+\frac{\pi}{2})} ≤4\pi$
$0 ≤e^{-(2k\pi+\frac{\pi}{2})}$ is true for all real finite $k$.
So, we need check $ -(2k\pi+\frac{\pi}{2})≤\ln _e{4\pi}$ .
or $2k\pi+\frac{\pi}{2}≥-\ln _e{4\pi}$
Without using calculator, we can observe that this inequality is satisfied by all integer $k≥0$
(5) For $i^{\frac{1}{e}}, a=0,b=1,x=\frac{1}{e},y=0$
$\implies r^2=1$ and $\sin \theta=1$ and $\cos \theta=0\implies \theta=\frac{\pi}{2}$
$i^{\frac{1}{e}}=e^{\frac{1}{2e}\log 2}\cdot cis(\frac{1}{e}(2k\pi+\frac{\pi}{2}))$
can be further simplified to $2^{\frac{1}{2e}}\cdot cis(\frac{1}{e}(2k\pi+\frac{\pi}{2}))$ where $k$ is any integer.
It does have an imaginary part, so can not have any pure real value in any specified range.
