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Can something explain to me why the absolute value of a complex exponential is 1? (Or at least that's what my textbook is telling me.)

For example:

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

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What is $ j $? $ $ –  SDevalapurkar Mar 22 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 at 3:28

4 Answers 4

up vote 4 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 $\sqrt{\cos^2(x)+\sin^2(x)}=1$

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ah. of course. Should have known that. Thanks! –  codedude Mar 22 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 at 3:31
Question - shouldn't that be sqrt(cos^2 + sin^2) = 1 ? –  codedude Mar 22 at 3:33
yeah I forgot to add that in. Simplifies to the same thing though. –  ruler501 Mar 22 at 3:34
Yeah, just checking. –  codedude Mar 22 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 $\left|e^{-2j}\right| = 1$.

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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.

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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 at 3:39

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

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Any particular reason to use $j$ and not $i$? –  Cole Johnson Mar 22 at 4:31

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