Why do $e^i$ and $i^e$ both have absolute value 1? Why do $e^i$ and $i^e$ both have absolute value 1? I don't know how to view $i^e$ as a complex number. What is its real part and imaginary part?
 A: We define $z^w$ as $z^w=e^{w\log(z)}$ where $\log(z)$ is the multivalued function
$$\log(z)=\text{Log}(|z|)+i\arg(z)$$
where $\arg(z)$ is the multivalued argument of $z$.
Here, we have $i^e=e^{e\log(i)}$.  Inasmuch as $\text{Log}(|i|)=\text{Log}(1)=0$ and $\arg(i) =\frac\pi2+2n\pi$, $n\in \mathbb{Z}$, we have
$$\begin{align}
i^e&=e^{e\log(i)}\\\\
&=e^{e\left(\text{Log}(|i|)+i\arg(i)\right)}\\\\
&=e^{e\left(0+i\left(\frac\pi2+2n\pi\right)\right)}\\\\&=e^{ie(\pi/2+2n\pi)}
\end{align}$$
which clearly has unit magnitude.
A: Complex exponentiation is in general multivalued. Since $e^z=i$ iff $z=\pi i(1+4n) /2$ for some $n\in\Bbb Z$, the branches of $\ln z$ give $i^e=\exp(e\pi i(1+4n) /2)$. All such values are of modulus $1$.
A: From definition, $|e^z| = e^{\text{Re} \ z}$
$$ |i^e| = |e^{\frac{i\pi e}{2}}| = e^{\text{Re} \ \left(\frac{i\pi e}{2}\right) } = e^0 = 1$$
$$|e^i|= e^{\text{Re} \ i}= e^0= 1$$
A: This is almost more word play than mathematics but....
Every non-zero complex number, $z$, may be written in polar form as $z = re^{i\theta}$ were $r = |z|$ and $\theta = \arg z$.  If $Re(z) = a$ and $Im(z) = b$ then $r = \sqrt{a^2 + b^2} = \sqrt{z\overline z}$.  And $\arg = \arctan \frac ba$.
Thus $re^{i\theta} = r(\cos \theta + i \sin \theta) = r\cos \theta + ir\sin \theta = a + bi= z$ assuming that $a=r\cos \theta; b= r\sin \theta$ or alternatively (depending on which way you look at it) that $r = \sqrt{a^2 + b^2}$ and $\theta$ is so that $\cos \theta = \frac {a}{\sqrt {a^2 + b^2}}$ and $\sin \theta = \frac {b}{\sqrt{a^2 + b^2}}$ or when $\tan \theta = \frac ba$.
So.... $$e^i = \cos 1 + i \sin 1 \approx 0.54030230586813971740093660744298 + i 0.8414709848078965066525023216303$$.  And $|e^i| =1$ because that is simply how polar notation works.  The $e^{i\theta}$ are all on the unit circle. And all complex numbers are represented by a point on the unit circle, times a scalar $r$ to represent how large it is.  In this case $z = 1\cdot e^i$ were $r = 1$ means it has a magnitude of $1$.
....
Using polar notation we can make sense of an expression such as $z^w$.
$i = 0 + i*1 = \cos \frac \pi 2 + i\sin \frac \pi 2$ so $i = e^{i\frac \pi 2}$.
So $i^{e} = [e^{i\frac \pi 2}]^e = e^{i(\frac {\pi e}2)} = \cos \frac {\pi e}2+i \sin \frac {\pi e}2\approx-0.42821977341382775376026209125985 -0.90367462377639553660085345933434i$.
And why does $|i^e| = 1$?  Well, simply because $|i|=1$ so $|i^e| = |i|^e = 1^e=1$.
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More complex:  If $z = re^{i\theta}$ and $w = a+bi$ then
$z = r e^{i\theta} = e^{\ln r}e^{i\theta} = e^{\ln r + i\theta}$ and so
$z^w = [e^{\ln r + i\theta}]^{a+bi} = e^{(\ln r + i\theta)(a+bi)} =$
$e^{(\ln ra-\theta b)+(\ln rb+\theta a)i} =$
$e^{\ln ra-\theta b} e^{i(\ln rb+\theta a)} =$
$\frac {r^a}{e^{\theta b}}(\cos(\ln rb+\theta a)+i\sin (\ln rb+\theta a))$.
