Why do $e^i$ and $i^e$ both have absolute value 1?

Question

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?

Answer 1

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.

Answer 2

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 $1$.

Answer 3

From definition, $|e^z| = e^{\text{Re} \ z}$ $$ |i^e| = |e^{\frac{i\pi e}{2}}| = e^{\text{Re} \ \left(\frac{i\pi e}{2}\right) } = e^0 = 1$$ $$|e^i|= e^{\text{Re} \ i}= e^0= 1$$

Answer 4

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 $|i^e| = |i|^e = 1^e=1$.

======

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

$z^w = [e^{\ln r + i\theta}]^{a+bi} = e^{(\ln r + i\theta)(a+bi)} =$

$e^{(\ln ra-\theta b)+(\ln rb+\theta a)i} =$

$e^{\ln ra-\theta b} e^{i(\ln rb+\theta a)} =$

$\frac {r^a}{e^{\theta b}}(\cos(\ln rb+\theta a)+i\sin (\ln rb+\theta a))$.