Is there a simplification for this relation?
$$ \frac{x}{\left| x\right| } $$
where $x=a+i b$, $a$ and $b$ are reals.
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1Assuming $x\neq 0$ (so that the division is valid) you can write this as $\dfrac{a+ib}{\sqrt{a^2+b^2}}$... though I'm not sure if you consider this simpler. – peek-a-boo Aug 06 '20 at 11:17
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By polar form $x=|x|e^{i\theta}$ we obtain
$$\frac{x}{\left| x\right| }=e^{i\theta}$$
with $\theta = \operatorname {atan2} \left(b, a \right)$ (see atan2).
user
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1$\operatorname {atan2} \frac b a$ makes no sense. If you are referring to 2-argument arctangent then it should be $\operatorname {atan2}(b, a)$. – Martin R Aug 06 '20 at 11:40
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Assuming $x \neq 0$, you can reduce it to:-
$e^{i\theta}$, where
$\theta = \tan^{-1}(\frac b a)$ if $a > 0$
$\theta = \tan^{-1}(\frac b a) + \pi$ if $a<0$
$\theta = \frac \pi 2$ if $a=0$ and $b>0$
$\theta = -\frac \pi 2$ if $a=0$ and $b<0$
Anuneet Anand
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We have $\theta = \frac \pi 2$ if $a=0$ and $b>0$, $\theta =- \frac \pi 2$ if $a=0$ and $b<0$. – user Aug 06 '20 at 18:21
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any complex number say '$z$' can be written in the form of : $z = a + ib$
so we define a function $f(z) = \frac{z}{|z|} \,\, (z \not= 0)\,\,$ then we can show that : $$f(a+ib) = \frac{a}{\sqrt{a^2 +b^2}} + i\frac{b}{\sqrt{a^2 +b^2}} = cos(\theta) + sin(\theta) = \cos(\tan^{-1}(\frac{b}{a})) + i\sin(tan^{-1}(\frac{b}{a})) $$
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