Sign up ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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:

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

share|cite|improve this question
What is $ j $? $ $ –  Sanath Devalapurkar 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

4 Answers 4

up vote 6 down vote accepted

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

share|cite|improve this answer
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 $\left|e^{-2j}\right| = 1$.

share|cite|improve this answer

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.

share|cite|improve this answer
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: $e^{-2j} = \cos(-2) + j \sin(-2)$ ...

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.