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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$\begingroup$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 $\sqrt{\cos^2(x)+\sin^2(x)}=1$
$\endgroup$ 8 $\begingroup$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, $\lvert e^{i\,\text{Im}}\rvert=1.$
$\endgroup$ 2 $\begingroup$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 $\left|e^{-2j}\right| = 1$.
$\endgroup$ $\begingroup$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.
$\endgroup$ 1 $\begingroup$Hint: $e^{-2j} = \cos(-2) + j \sin(-2)$ ...
$\endgroup$ 2 $\begingroup$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:
$$cos^{2} \theta + sin^{2} \theta = 1$$
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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