Absolute value of complex exponential

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 $\sqrt{\cos^2(x)+\sin^2(x)}=1$

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, $\lvert e^{i\,\text{Im}}\rvert=1.$

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 $\left|e^{-2j}\right| = 1$.