# Absolute value of complex exponential

by codedude   Last Updated November 09, 2018 01:20 AM

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:

$$|e^{-2i}|=1, i=\sqrt {-1}$$

Tags :

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

ruler501
March 22, 2014 03:28 AM

Hint: $e^{-2j} = \cos(-2) + j \sin(-2)$ ...

DeepSea
March 22, 2014 03:29 AM

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

user3213784
March 22, 2014 03:32 AM

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.

Chung. J
March 22, 2014 03:33 AM

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

June 08, 2017 12:44 PM

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:

$$cos^{2} \theta + sin^{2} = 1$$

Steps: $$\left|e^{-i2}\right|$$ $$\theta=-2$$ $$\left|e^{i\theta}\right|=\sqrt{\cos^2\ \theta + \sin^2\ \theta}$$ $$\left|e^{i\theta}\right|=\sqrt1$$ $$\left|e^{i\theta}\right|=1$$ $$\left|e^{-i2}\right| = 1$$

Bill Moore
November 09, 2018 01:09 AM