Question Number 127567 by mathocean1 last updated on 30/Dec/20
$${a}=\frac{\sqrt{\mathrm{6}}}{\mathrm{2}}+\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}{i}\:;\:{b}=\mathrm{1}+{i}. \\ $$$${Determinate}\:{possible}\:{values} \\ $$$${of}\:{n}\:\in\:\mathbb{N}\:{such}\:{such}\:{that}\:{a}^{{n}} \:\in\:\mathbb{R}\:{and} \\ $$$${b}^{{n}} \:\in\:{i}\mathbb{R}\:\left({pure}\:{imaginary}\right). \\ $$$$ \\ $$$$ \\ $$$$ \\ $$
Commented by malwan last updated on 01/Jan/21
$${a}=\sqrt{\mathrm{3}}\left(\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\:+\:\frac{{i}}{\mathrm{2}}\right)\:=\:\left[\sqrt{\mathrm{3}}\:,\:\frac{\pi}{\mathrm{6}}\right] \\ $$$${a}^{{n}} \:=\:\left[\left(\sqrt{\mathrm{3}}\right)^{{n}} \:,\:\frac{{n}\pi}{\mathrm{6}}\right]\in\mathbb{R} \\ $$$$\Rightarrow\frac{{n}\pi}{\mathrm{6}}\:=\:{m}\pi\:;\:{m}\in\mathbb{N} \\ $$$$\Rightarrow{n}\:=\:\mathrm{6}{m}\:;\:{m}\in\mathbb{N} \\ $$$$ \\ $$$${b}\:=\:\mathrm{1}+{i}\:=\:\left[\sqrt{\mathrm{2}}\:,\:\frac{\pi}{\mathrm{4}}\right] \\ $$$$\Rightarrow{b}^{{n}} \:=\:\left[\left(\sqrt{\mathrm{2}}\right)^{{n}} \:,\:\frac{{n}\pi}{\mathrm{4}}\right]\in{i}\mathbb{R} \\ $$$$\Rightarrow\frac{{n}\pi}{\mathrm{4}}\:=\:\left(\mathrm{2}{m}\:+\mathrm{1}\right)\frac{\pi}{\mathrm{2}} \\ $$$$\Rightarrow{n}=\:\mathrm{4}{m}\:+\:\mathrm{2}\:;\:{m}\in\mathbb{N} \\ $$
Answered by Ar Brandon last updated on 30/Dec/20
$$\mathrm{a}=\sqrt{\mathrm{2}}\mathrm{e}^{\frac{\pi}{\mathrm{6}}\mathrm{i}} \Rightarrow\:\mathrm{a}^{\mathrm{n}} =\sqrt{\mathrm{2}^{\mathrm{n}} }\mathrm{e}^{\frac{\pi\mathrm{n}}{\mathrm{6}}\mathrm{i}} \\ $$$$\mathrm{a}^{\mathrm{n}} \in\mathbb{R}\:\Rightarrow\:\frac{\pi\mathrm{n}}{\mathrm{6}}=\mathrm{k}\pi,\:\mathrm{k}\in\mathbb{Z}\:\Rightarrow\:\mathrm{n}=\mathrm{6k} \\ $$$$\mathrm{b}=\sqrt{\mathrm{2}}\mathrm{e}^{\frac{\pi}{\mathrm{4}}\mathrm{i}} \Rightarrow\mathrm{b}^{\mathrm{n}} =\sqrt{\mathrm{2}^{\mathrm{n}} }\mathrm{e}^{\frac{\pi\mathrm{n}}{\mathrm{4}}\mathrm{i}} \\ $$$$\mathrm{b}^{\mathrm{n}} \in\mathrm{i}\mathbb{R}\:\Rightarrow\frac{\pi\mathrm{n}}{\mathrm{4}}=\frac{\left(\mathrm{2k}+\mathrm{1}\right)\pi}{\mathrm{2}},\:\mathrm{k}\in\mathbb{Z}\:\Rightarrow\mathrm{n}=\mathrm{4k}+\mathrm{2} \\ $$
Answered by mr W last updated on 31/Dec/20
$${a}=\sqrt{\mathrm{2}}\left(\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}{i}\right)=\sqrt{\mathrm{2}}\left(\mathrm{cos}\:\frac{\pi}{\mathrm{6}}+{i}\:\mathrm{sin}\:\frac{\pi}{\mathrm{6}}\right) \\ $$$${a}^{{n}} =\sqrt{\mathrm{2}^{{n}} }\left(\mathrm{cos}\:\frac{{n}\pi}{\mathrm{6}}+{i}\:\mathrm{sin}\:\frac{{n}\pi}{\mathrm{6}}\right) \\ $$$${such}\:{that}\:{a}^{{n}} \in{R}, \\ $$$$\mathrm{sin}\:\frac{{n}\pi}{\mathrm{6}}=\mathrm{0} \\ $$$$\Rightarrow\frac{{n}\pi}{\mathrm{6}}={k}\pi,\:{k}\in\mathbb{Z} \\ $$$$\Rightarrow{n}=\mathrm{6}{k}\:\:\:…\left(\mathrm{1}\right) \\ $$$$ \\ $$$${b}=\mathrm{1}+{i}=\sqrt{\mathrm{2}}\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}+\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}{i}\right)=\sqrt{\mathrm{2}}\left(\mathrm{cos}\:\frac{\pi}{\mathrm{4}}+{i}\:\mathrm{sin}\:\frac{\pi}{\mathrm{4}}\right) \\ $$$${b}^{{n}} =\sqrt{\mathrm{2}^{{n}} }\left(\mathrm{cos}\:\frac{{n}\pi}{\mathrm{4}}+{i}\:\mathrm{sin}\:\frac{{n}\pi}{\mathrm{4}}\right) \\ $$$${such}\:{that}\:{b}^{{n}} \in{iR}, \\ $$$$\mathrm{cos}\:\frac{{n}\pi}{\mathrm{4}}=\mathrm{0},\:\mathrm{sin}\:\frac{{n}\pi}{\mathrm{4}}\neq\mathrm{0}\:{automatically} \\ $$$$\Rightarrow\frac{{n}\pi}{\mathrm{4}}=\frac{\left(\mathrm{2}{k}+\mathrm{1}\right)}{\mathrm{2}}\pi,\:{k}\in\mathbb{Z} \\ $$$$\Rightarrow{n}=\mathrm{2}\left(\mathrm{2}{k}+\mathrm{1}\right)\:\:\:…\left(\mathrm{2}\right) \\ $$$$ \\ $$$${for}\:{both}\:\left(\mathrm{1}\right)\:{and}\:\left(\mathrm{2}\right): \\ $$$$\Rightarrow{n}=\mathrm{6}\left(\mathrm{2}{k}+\mathrm{1}\right)\:{with}\:{k}\in\mathbb{Z} \\ $$