15
$\begingroup$

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:

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

$\endgroup$
  • 2
    $\begingroup$ What is $ j $? $ $ $\endgroup$ – user122283 Mar 22 '14 at 3:27
  • $\begingroup$ 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. $\endgroup$ – codedude Mar 22 '14 at 3:28
  • $\begingroup$ absolute value of any complex number is always real. $\endgroup$ – Bill Moore Nov 19 '18 at 5:30
20
$\begingroup$

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

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

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

$\endgroup$
3
$\begingroup$

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

$\endgroup$
  • $\begingroup$ Is the plus sign correct? $\endgroup$ – Ramen Sep 17 '17 at 16:33
  • $\begingroup$ @Ramen Thank you. $\LaTeX$ is prone to distract from the actual math... It is corrected now. $\endgroup$ – Antoni Parellada Sep 17 '17 at 20:29
2
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – ruler501 Mar 22 '14 at 3:39
2
$\begingroup$

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

$\endgroup$
  • $\begingroup$ Any particular reason to use $j$ and not $i$? $\endgroup$ – Cole Johnson Mar 22 '14 at 4:31
  • 3
    $\begingroup$ for anyone else wondering, it's probably an engineering thing. In engineering i can be current, whereas j is unused. $\endgroup$ – 1mike12 Nov 25 '14 at 6:15
0
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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