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If you guys are not familiar with the $\operatorname{cis}$ notation

$$ \operatorname{cis}(x) = (\cos(x) + i\sin(x)) $$

I've tried doing this:

By comparing arguments

$$ x^2 - x = 0\Longrightarrow x(x-1) = 0\Longrightarrow x = 0 ,~ x = 1 $$ But clearly from the graph there are more intersections and searching the internet I learnt you can do this

$$ x^2 = x + 2k\pi $$

But why can we plus $2k\pi$ and does the $2k\pi$ have to be on the right side of the equation why can't it be on the left hand side?

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  • $\begingroup$ Rather expand and compare real and imaginary parts $\endgroup$ – Archis Welankar Jan 12 '16 at 2:55
  • $\begingroup$ The key idea is that $\operatorname{cis}x=\operatorname{cis}y$ if and only if $x$ and $y$ differ by an integer multiple of $2\pi$. Try proving that. (And yes, $2k\pi$ can be on either side of your equation.) $\endgroup$ – Rahul Jan 12 '16 at 3:08
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$$\text{cis}\left(x^2\right)=\text{cis}(x)\Longleftrightarrow$$ $$\cos\left(x^2\right)+\sin\left(x^2\right)i=\cos(x)+\sin(x)i\Longleftrightarrow$$ $$e^{ix^2}=e^{ix}\Longleftrightarrow$$ $$\ln\left(e^{ix^2}\right)=\ln\left(e^{ix}\right)\Longleftrightarrow$$ $$ix^2=ix\Longleftrightarrow$$ $$x^2=x\Longleftrightarrow$$


Notice, if $a,b\in\mathbb{R}$ and $k\in\mathbb{Z}$

$a+bi=\sqrt{a^2+b^2}e^{\arg(a+bi)i}=\sqrt{a^2+b^2}e^{\left(2\pi k+\arg(a+bi)\right)i}$


$$x^2-x+2\pi k=0\Longleftrightarrow$$


Use the abc-formula:


$$x=\frac{1}{2}\pm\frac{1}{2}\sqrt{1-8\pi k}\space\space\space\space\space\space\space\space\space\space\space\space\text{with}\space k\in\mathbb{Z}$$

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