As part of a proof I came across with the following equations:

$$ \begin{cases} \left(\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}i\right)^{n}=1\\ \left(\frac{\sqrt{2}}{2}-\frac{\sqrt{2}}{2}i\right)^{n}=1 \end{cases}$$

I was solving a matrix and got those equations. I think that finding the formula of $n\in \mathbb{N}$ is quite difficult, although its possible (I saw the formula on Wolfram but have not Idea how to prove it). If so, how can I find the minimal $n$ so the equations are true?

  • $\begingroup$ Im curious what is the motivation $\endgroup$ – sss89 Nov 20 '18 at 17:20


those are both rotations, with well known angles. Finding their cycle duration is finding the answer



Squaring, you find

$$\left(\frac{\sqrt{2}}{2}\pm\frac{\sqrt{2}}{2}i\right)^{2}=0\pm i.$$

Now solve $(\pm i)^m=1$.

$$\left(\frac{\sqrt{2}}{2}\pm\frac{\sqrt{2}}{2}i\right)^2=\pm i,(\pm i)^2=-1,(-1)^2=1.$$

  • $\begingroup$ What do you mean by "Squaring"? I could not understand how you got the first equation. $\endgroup$ – vesii Nov 20 '18 at 17:42
  • $\begingroup$ @vesii: come on, you know how to square a complex number, don't you ? $\endgroup$ – Yves Daoust Nov 20 '18 at 17:43
  • $\begingroup$ @vesii: the formula doesn't need a translation. $\endgroup$ – Yves Daoust Nov 20 '18 at 17:57

Basic remark : the second equation is equivalent to the first one, by conjugation...

Thus we are left with the first equation, which can be written under the form : $$(\cos(\pi/4) + i \sin(\pi/4))^n=(e^{i\pi/4})^n=1$$ (by De Moivre formula) i.e.,

$$e^{i n \pi/4}=1$$

which clearly occurs iff $n$ is multiple of $8$.

Thus the minimal positive value of $n$ for which the equation is true is $n=8$.


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