# Finding the minimal $n$ so the equations are true

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?

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

hint:

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

Hint:

Squaring, you find

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

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

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

• What do you mean by "Squaring"? I could not understand how you got the first equation. – vesii Nov 20 '18 at 17:42
• @vesii: come on, you know how to square a complex number, don't you ? – Yves Daoust Nov 20 '18 at 17:43
• @vesii: the formula doesn't need a translation. – 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 $$n=8$$.