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 $[0, 4\pi]$?
 A: So we want a general rule for evaluating formal expressions $z^w$, where
$z=r\, e^{i\phi}\ne 0$ and $w=u+iv$ are complex numbers.
Depending on circumstances one  wants the full infinite set of values for $z^w$, which arises from the multivaluedness of the logarithm function in the complex domain, or one wants only the so called principal value of $z^w$. Denote by $\dot{\mathbb C}$ the punctured compex plane and by ${\mathbb C}^-$ the complex plane slit along the negative real axis. Then $\arg:\ \dot{\mathbb C}\to{\mathbb R}/(2\pi{\mathbb Z})$ is the multivalued argument function and ${\rm Arg}:\ {\mathbb C}^-\to\ ]{-\pi},\pi[\ $ the principal value of the argument. Correspondingly one has two forms of the logarithm:
$$\log z:=\log|z| + i\arg(z)\ ,\qquad{\rm Log}z:=\log|z| + i{\rm Arg}(z)\ .$$
In the real domain the "general power function" $a^b$ is defined for $a>0$ by $a^b:=\exp(b\,\log a)$. Transferring this principle into the complex world one arrives at
$$z^w:=\exp(w\ \log z)=\exp\bigl((u+iv)(\log r+i\phi +2k\pi i)\bigr)\qquad (r>0)\ ,$$
resp. at
$$z^w:=\exp(w\ {\rm Log} z)=\exp\bigl((u+iv)(\log r + i\phi)\bigr)\qquad(r>0, \ |\phi|<\pi)\ .$$
It seems that the OP is interested in the first of the two interpretations. Therefore let's look at the example $i^i$. As $i=e^{i\pi/2}$ one has $r=1$, $\phi={\pi\over2}$. Therefore we get
$$i^i=\exp\bigl( (0+i)(\log 1+i{\pi\over2}+2k\pi i)\bigr)=\exp\bigl(-{\pi\over2}-2k\pi \bigr)\ ,$$
where this is to be interpreted as the  resulting set when $k$ runs through ${\mathbb Z}$. We see that this set $\{c_k\}_{k\in{\mathbb Z}}$ consists of positive real numbers. For $k\leq-1$ the $c_k$ lies outside the interval $[0,4\pi]$; but for all $k\geq0$ one has $c_k\in[0,4\pi]$, so that, yes: there are infinitely many values of $i^i$ in the given interval.
A: 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.
A: A number $x$ of $\mathbb{C} $ is a primitive $n^{th}$ root if $x^n - 1 = 0$, but $x^j -1 \not= 0$ for any $0 \leq j < n$.
If we for each natural number $n$ finds the primitive roots of $x^n -1 = 0$ and collects them for all such $n$, 
then we find all roots once. In number, the primitive $n^{th}$ roots are $\varphi (n)$ where the function is Eulers phi function,  counting  how many  numbers  in  the interval $[0, n)$ being relative prime to $n$, where  $j$ is relative prime to $n$ if $gcd (n,j)=1$ (greatest common divisor).For such $j$ we have a  root 
\begin{equation}\nonumber 
e^{i\frac{ 2 \pi  }{n }j}.
\end{equation}
If $z$ is a complex number, $z$ can be written $z = |z| e^{2 \pi i( \gamma+ m )} = (|z|^{1/n} e^{2 \pi i (\gamma+  m )  /n})^n$. Therefore an equation $x^n - z = 0$ has an expression 
\begin{equation}\nonumber 
\Big(\frac{x}{|z|^{1/n} e^{2 \pi i (\gamma + m)/n}} \Big)^n - 1 = 0
\end{equation}
From this, the  $n^{\text{th }}$ roots are 
\begin{equation}\nonumber 
x =|z|^{1/n}e^{i \frac{2 \pi }{n }\big(\gamma    + j + m\big) } 
\end{equation} 
If $|z|\not =1$,  roots belonging to different $n$ will be different. If $|z|=1$, $m+j$ must be prime to $n$. In that case, we get the contribution from $n$ to be the roots 
\begin{equation}\nonumber 
x =|z|^{1/n}e^{i \frac{2 \pi }{n }\big(\gamma  + j + m\big) }  \qquad \text{where } \gcd(n, j+m)=1.
\end{equation} 
or, what is the same,
\begin{equation}\nonumber 
x =|z|^{1/n}e^{i \frac{2 \pi }{n }\big(\gamma  +k\big) }  \qquad \text{where } \gcd(n, k)=1, \qquad 1 \leq k < n. 
\end{equation} 
For example, when  $z$ equals  $i^i$, we have a real number, so take $\gamma = 0$ and the real part is $\exp (- \pi /2)$. This should give 
the $n^{th}$ roots 
\begin{equation}\nonumber 
  x = e^{-\frac{\pi }{2 n } +i \frac{2 \pi }{n } k} . \qquad \text{where }  1 \leq k < n. 
\end{equation}
When $z$ equals $i^{1/e}$, rewritten ($(e^{i \pi /2})^{1/e})=e^{i \pi /(2 e)}$, $|z|=1$ and $\gamma $ from above is $\pi /(2 e)+ 2 \pi m $.
Then the  roots which are $n^{th}$ roots but not $k^{th}$ roots for $k < n$ are 
\begin{equation}\nonumber 
x = e^{i \frac{2 \pi }{n }\big(\frac{1}{4 e  } + k\big) } \qquad \text{where } \gcd(n,k) = 1. 
\end{equation}
Collecting together for all natural numbers $n$ should give all possible roots in  the respective  cases.
A: Since you probably mean the multi-valued complex exponential, you can go straight to the definition:
$$ x^y := \exp(y \log x) $$
The multi-valued complex logarithm can be computed as
$$ \log x = \ln |x| + \mathbf{i} \arg(x) $$
where $\arg(x)$ is the polar angle of $x$. This, of course, takes values in the set
$$ \arg(x) = \mathop{\text{Arg}}(x) + 2 \pi n $$
where $n$ is an integer, and $\mathop{\text{Arg}}$ is the "principal value". For your particular question, all you need to know about the idea of a principal value is that $\mathop{\text{Arg}}(x)$ is a particular choice of polar angle for $x$; here it doesn't matter which one you choose.
To demonstrate that this reproduces what you already know, consider the case of $(-8)^{1/3}$. In this case, we have:


*

*$\mathop{\text{Arg}}(x) = \pi$

*$\arg(x) = \pi + 2 \pi n$

*$\log(x) = \ln(8) + \mathbf{i} \pi (1 + 2n)$

*$\exp(y \log x) = \exp((1/3) \ln(8) + \mathbf{i} \pi (1 + 2n)/3) = \exp(\ln(2)) \exp(\mathbf{i} \pi (1 + 2n)/3)$

*$x^y = 2 (\cos(\frac{1+2n}{3} \pi) + \mathbf{i} \sin(\frac{1 + 2n}{3} \pi))$


and so the values for $(-8)^{1/3}$ are all of the numbers $2 \exp(\mathbf{i}  \pi \frac{1 + 2n}{3})$ where $n$ is an integer. $n=1$ gives you the familiar value $-2$. $n=0$ and $n=2$ give the other two complex cube roots of $-8$. $n=4$ also gives $-2$, because $3 \pi \mathbf{i}$ and $\pi \mathbf{i}$ are the same angle. In fact, as $n$ ranges over all integers, only three distinct values are produced. This agrees with the familiar fact that nonzero complex numbers have three cube roots.
A: I think you mean, "all possible meaningful values"  You know that
$$1 + i = \sqrt{2}e^{i\pi/4 + 2n\pi} = 
e^{i\pi(2n + 1/4) +\log(2)/2}.$$
If you exponentiate by $i$, you get
$$(1 + i)^i = e^{-\pi(2n + 1/4) + i\log(2)/2}  $$
You can deal similarly with the other cases.   
Exponentiation to a complex power requires restricting of domain; this all boils down to the problem of dealing with the complex logarithm.
