Absolute value of complex exponential Can somebody explain to me why the absolute value of a complex exponential is 1? (Or at least that's what my textbook says.) 
For example:
$$|e^{-2i}|=1, i=\sqrt {-1}$$
 A: If it is purely complex then you have $e^{xi}=\cos(x)+i\sin(x)$ the absolute value($|a+bi|=\sqrt{a^2+b^2}$) is then equal to $\sqrt{\cos^2(x)+\sin^2(x)}=1$
A: By Euler's formula, $e^{j\theta}=\cos(\theta)+j\sin(\theta)$, which is a point on the unit circle at an angle of $\theta$. Let $\theta = \frac{-2j}{j} = -2$, so $e^{-2j}$ is one of the points on the unit circle, which of course is one unit from the origin, so $\left|e^{-2j}\right| = 1$.
A: Hint: $e^{-2j} = \cos(-2) + j \sin(-2)$ ...
A: When we extend exponential function $f(x)=e^x$ to complex numbers
so that the extension is differentiable, it is the only way to define
$$
f(x+iy)=e^x(\cos y+i\sin y)
$$
I hop that it help your question.
One should know that why the Euler formula comes.
A: Definition of absolute value:
$$\left|a+i\ b\right|=\sqrt{a^2+b^2}$$
Euler's Formula:
$$e^{i\theta}=cos\ \theta + i\sin\ \theta$$
Trig Identity:
$$cos^{2} \theta + sin^{2} \theta = 1$$
Steps:
$$ \left|e^{-i2}\right| $$
$$\theta=-2$$
$$\left|e^{i\theta}\right|=\sqrt{\cos^2\ \theta + \sin^2\ \theta}$$
$$\left|e^{i\theta}\right|=\sqrt1$$
$$\left|e^{i\theta}\right|=1$$
$$ \left|e^{-i2}\right| = 1 $$
A: After +1 accepted answer, just an extension on the same...
$$\begin{align}
\left\vert e^{\text{Re}\,+\,i\text{ Im}}\right\vert 
& = \lvert e^\text{Re}\cdot e^{i\text{ Im}} \rvert\\[2ex]
&=\lvert e^{\text{Re}}\rvert\,\cdot \lvert e^{\,i\text{ Im}}\rvert\\[2ex]
&=  e^{\text{Re}}
\end{align}$$
because $ e^{ix} \in S^1,$ and hence, $\lvert e^{i\,\text{Im}}\rvert=1.$ 
A: I would like to provide an intuitive understanding, in addition to the previous excellent answers.
Recall that a complex number in Euler’s form can be expressed as $r e^{i \theta}$, in this case, the modulus is 1 and the argument is -2. Graphically, we can visualize complex numbers of modulus 1 as points on a unit circle.
Algebraically, we say that the unitary group of degree 1 is isomorphic to the unit circle, namely $U(1) \cong S^1$
