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Question Number 19434 by Tinkutara last updated on 11/Aug/17

$$\mathrm{If}{\left(\frac{1+i\sqrt{3}}{1-i\sqrt{3}}\right)}^n\mathrm{is}\mathrm{an}\mathrm{integer},\mathrm{then}n\mathrm{is}$$

Answered by ajfour last updated on 11/Aug/17

$${\left(\frac{1+i\sqrt{3}}{1-i\sqrt{3}}\right)}^{\mathrm{n}}={\left(\frac{{\mathrm{e}}^{i\pi /6}}{{\mathrm{e}}^{-i\pi /6}}\right)}^{\mathrm{n}}=\left[\mathrm{x}\right]$$$$\Rightarrow {\mathrm{e}}^{\mathrm{in}\pi /3}=\pm 1={\mathrm{e}}^{\mathrm{ik}\pi }$$$$\Rightarrow \mathbf{n}=3\mathbf{k}\mathrm{where}\mathrm{k}\in Z.$$

Commented by Tinkutara last updated on 11/Aug/17

$$\mathrm{Thank}\mathrm{you}\mathrm{very}\mathrm{much}\mathrm{Sir}!$$

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