# 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:

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

• What is $j$?  – user122283 Mar 22 '14 at 3:27
• Oh sorry, it's the electrical engineering way of saying imaginary i. It's a habit I've gotten used to over the past 2 years. – codedude Mar 22 '14 at 3:28
• absolute value of any complex number is always real. – Bill Moore Nov 19 '18 at 5:30

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

• ah. of course. Should have known that. Thanks! – codedude Mar 22 '14 at 3:31
• It's an identity most people don't remember(I know it took a while for me to start being able to use it). – ruler501 Mar 22 '14 at 3:31
• Question - shouldn't that be sqrt(cos^2 + sin^2) = 1 ? – codedude Mar 22 '14 at 3:33
• yeah I forgot to add that in. Simplifies to the same thing though. – ruler501 Mar 22 '14 at 3:34
• Yeah, just checking. – codedude Mar 22 '14 at 3:34

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

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

• Is the plus sign correct? – Ramen Sep 17 '17 at 16:33
• @Ramen Thank you. $\LaTeX$ is prone to distract from the actual math... It is corrected now. – Antoni Parellada Sep 17 '17 at 20:29

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.

• I've always looked at the formula as the obvious result of the taylor series(if you plug in $iy$ it simplifies to $\cos y + i \sin y$) and then the $x$ is obvious from rules for exponentiation. – ruler501 Mar 22 '14 at 3:39

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

• Any particular reason to use $j$ and not $i$? – Cole Johnson Mar 22 '14 at 4:31
• for anyone else wondering, it's probably an engineering thing. In engineering i can be current, whereas j is unused. – 1mike12 Nov 25 '14 at 6:15

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