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:
$$|e^{-2i}|=1, i=\sqrt {-1}$$
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2$\begingroup$ What is $ j $? $ $ $\endgroup$– user122283Mar 22, 2014 at 3:27
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$\begingroup$ 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. $\endgroup$– codedudeMar 22, 2014 at 3:28
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$\begingroup$ absolute value of any complex number is always real. $\endgroup$– Bill MooreNov 19, 2018 at 5:30
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7 Answers
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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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$\begingroup$ ah. of course. Should have known that. Thanks! $\endgroup$– codedudeMar 22, 2014 at 3:31
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2$\begingroup$ It's an identity most people don't remember(I know it took a while for me to start being able to use it). $\endgroup$– ruler501Mar 22, 2014 at 3:31
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$\begingroup$ Question - shouldn't that be sqrt(cos^2 + sin^2) = 1 ? $\endgroup$– codedudeMar 22, 2014 at 3:33
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$\begingroup$ yeah I forgot to add that in. Simplifies to the same thing though. $\endgroup$– ruler501Mar 22, 2014 at 3:34
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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, $\lvert e^{i\,\text{Im}}\rvert=1.$
answered Jun 8, 2017 at 12:44
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$\begingroup$ @Ramen Thank you. $\LaTeX$ is prone to distract from the actual math... It is corrected now. $\endgroup$ Sep 17, 2017 at 20:29
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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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$\begingroup$ 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. $\endgroup$– ruler501Mar 22, 2014 at 3:39
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Hint: $e^{-2j} = \cos(-2) + j \sin(-2)$ ...
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$\begingroup$ Any particular reason to use $j$ and not $i$? $\endgroup$ Mar 22, 2014 at 4:31
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6$\begingroup$ for anyone else wondering, it's probably an engineering thing. In engineering i can be current, whereas j is unused. $\endgroup$– 1mike12Nov 25, 2014 at 6:15
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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:
$$cos^{2} \theta + sin^{2} \theta = 1$$
Steps: $$ \left|e^{-i2}\right| $$ $$\theta=-2$$ $$\left|e^{i\theta}\right|=\sqrt{\cos^2\ \theta + \sin^2\ \theta}$$ $$\left|e^{i\theta}\right|=\sqrt1$$ $$\left|e^{i\theta}\right|=1$$ $$ \left|e^{-i2}\right| = 1 $$
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I would like to provide an intuitive understanding, in addition to the previous excellent answers.
Recall that a complex number in Euler’s form can be expressed as $r e^{i \theta}$, in this case, the modulus is 1 and the argument is -2. Graphically, we can visualize complex numbers of modulus 1 as points on a unit circle.
Algebraically, we say that the unitary group of degree 1 is isomorphic to the unit circle, namely $U(1) \cong S^1$
