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Question Number 127567 by mathocean1 last updated on 30/Dec/20

$$a=\frac{\sqrt{6}}{2}+\frac{\sqrt{2}}{2}i;b=1+i.$$$$Determinatepossiblevalues$$$$ofn\in \mathbb{N}suchsuchthat{a}^n\in \mathbb{R}and$$$$b^n\in i\mathbb{R}\left(pureimaginary\right).$$$$$$$$$$$$$$

Commented by malwan last updated on 01/Jan/21

$$a=\sqrt{3}(\frac{\sqrt{3}}{2}+\frac{i}{2})=[\sqrt{3},\frac{\pi }{6}]$$$$a^n=[{\left(\sqrt{3}\right)}^n,\frac{n\pi }{6}]\in \mathbb{R}$$$$\Rightarrow \frac{n\pi }{6}=m\pi ;m\in \mathbb{N}$$$$\Rightarrow n=6m;m\in \mathbb{N}$$$$$$$$b=1+i=[\sqrt{2},\frac{\pi }{4}]$$$$\Rightarrow b^n=[{\left(\sqrt{2}\right)}^n,\frac{n\pi }{4}]\in i\mathbb{R}$$$$\Rightarrow \frac{n\pi }{4}=(2m+1)\frac{\pi }{2}$$$$\Rightarrow n=4m+2;m\in \mathbb{N}$$

Answered by Ar Brandon last updated on 30/Dec/20

$$\mathrm{a}=\sqrt{2}{\mathrm{e}}^{\frac{\pi }{6}\mathrm{i}}\Rightarrow {\mathrm{a}}^{\mathrm{n}}=\sqrt{2^{\mathrm{n}}}{\mathrm{e}}^{\frac{\pi \mathrm{n}}{6}\mathrm{i}}$$$${\mathrm{a}}^{\mathrm{n}}\in \mathbb{R}\Rightarrow \frac{\pi \mathrm{n}}{6}=\mathrm{k}\pi ,\mathrm{k}\in \mathbb{Z}\Rightarrow \mathrm{n}=6\mathrm{k}$$$$\mathrm{b}=\sqrt{2}{\mathrm{e}}^{\frac{\pi }{4}\mathrm{i}}\Rightarrow {\mathrm{b}}^{\mathrm{n}}=\sqrt{2^{\mathrm{n}}}{\mathrm{e}}^{\frac{\pi \mathrm{n}}{4}\mathrm{i}}$$$${\mathrm{b}}^{\mathrm{n}}\in \mathrm{i}\mathbb{R}\Rightarrow \frac{\pi \mathrm{n}}{4}=\frac{(2\mathrm{k}+1)\pi }{2},\mathrm{k}\in \mathbb{Z}\Rightarrow \mathrm{n}=4\mathrm{k}+2$$

Answered by mr W last updated on 31/Dec/20

$$a=\sqrt{2}(\frac{\sqrt{3}}{2}+\frac{1}{2}i)=\sqrt{2}(\cos \frac{\pi }{6}+i\sin \frac{\pi }{6})$$$$a^n=\sqrt{2^n}(\cos \frac{n\pi }{6}+i\sin \frac{n\pi }{6})$$$$suchthat{a}^n\in R,$$$$\sin \frac{n\pi }{6}=0$$$$\Rightarrow \frac{n\pi }{6}=k\pi ,k\in \mathbb{Z}$$$$\Rightarrow n=6k...\left(1\right)$$$$$$$$b=1+i=\sqrt{2}(\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}i)=\sqrt{2}(\cos \frac{\pi }{4}+i\sin \frac{\pi }{4})$$$$b^n=\sqrt{2^n}(\cos \frac{n\pi }{4}+i\sin \frac{n\pi }{4})$$$$suchthat{b}^n\in iR,$$$$\cos \frac{n\pi }{4}=0,\sin \frac{n\pi }{4}\ne 0automatically$$$$\Rightarrow \frac{n\pi }{4}=\frac{(2k+1)}{2}\pi ,k\in \mathbb{Z}$$$$\Rightarrow n=2(2k+1)...\left(2\right)$$$$$$$$forboth\left(1\right)and\left(2\right):$$$$\Rightarrow n=6(2k+1)withk\in \mathbb{Z}$$

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