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

a=((√6)/2)+((√2)/2)i ; b=1+i. Determinate possible values of n ∈ N such such that a^n ∈ R and b^n ∈ iR (pure imaginary).

$${a}=\frac{\sqrt{\mathrm{6}}}{\mathrm{2}}+\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}{i}\:;\:{b}=\mathrm{1}+{i}. \\ $$$${Determinate}\:{possible}\:{values} \\ $$$${of}\:{n}\:\in\:\mathbb{N}\:{such}\:{such}\:{that}\:{a}^{{n}} \:\in\:\mathbb{R}\:{and} \\ $$$${b}^{{n}} \:\in\:{i}\mathbb{R}\:\left({pure}\:{imaginary}\right). \\ $$$$ \\ $$$$ \\ $$$$ \\ $$

Commented by malwan last updated on 01/Jan/21

a=(√3)(((√3)/2) + (i/2)) = [(√3) , (π/6)] a^n = [((√3))^n , ((nπ)/6)]∈R ⇒((nπ)/6) = mπ ; m∈N ⇒n = 6m ; m∈N b = 1+i = [(√2) , (π/4)] ⇒b^n = [((√2))^n , ((nπ)/4)]∈iR ⇒((nπ)/4) = (2m +1)(π/2) ⇒n= 4m + 2 ; m∈N

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

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

a=(√2)e^((π/6)i) ⇒ a^n =(√2^n )e^(((πn)/6)i) a^n ∈R ⇒ ((πn)/6)=kπ, k∈Z ⇒ n=6k b=(√2)e^((π/4)i) ⇒b^n =(√2^n )e^(((πn)/4)i) b^n ∈iR ⇒((πn)/4)=(((2k+1)π)/2), k∈Z ⇒n=4k+2

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

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

a=(√2)(((√3)/2)+(1/2)i)=(√2)(cos (π/6)+i sin (π/6)) a^n =(√2^n )(cos ((nπ)/6)+i sin ((nπ)/6)) such that a^n ∈R, sin ((nπ)/6)=0 ⇒((nπ)/6)=kπ, k∈Z ⇒n=6k ...(1) b=1+i=(√2)((1/( (√2)))+(1/( (√2)))i)=(√2)(cos (π/4)+i sin (π/4)) b^n =(√2^n )(cos ((nπ)/4)+i sin ((nπ)/4)) such that b^n ∈iR, cos ((nπ)/4)=0, sin ((nπ)/4)≠0 automatically ⇒((nπ)/4)=(((2k+1))/2)π, k∈Z ⇒n=2(2k+1) ...(2) for both (1) and (2): ⇒n=6(2k+1) with k∈Z

$${a}=\sqrt{\mathrm{2}}\left(\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}{i}\right)=\sqrt{\mathrm{2}}\left(\mathrm{cos}\:\frac{\pi}{\mathrm{6}}+{i}\:\mathrm{sin}\:\frac{\pi}{\mathrm{6}}\right) \\ $$$${a}^{{n}} =\sqrt{\mathrm{2}^{{n}} }\left(\mathrm{cos}\:\frac{{n}\pi}{\mathrm{6}}+{i}\:\mathrm{sin}\:\frac{{n}\pi}{\mathrm{6}}\right) \\ $$$${such}\:{that}\:{a}^{{n}} \in{R}, \\ $$$$\mathrm{sin}\:\frac{{n}\pi}{\mathrm{6}}=\mathrm{0} \\ $$$$\Rightarrow\frac{{n}\pi}{\mathrm{6}}={k}\pi,\:{k}\in\mathbb{Z} \\ $$$$\Rightarrow{n}=\mathrm{6}{k}\:\:\:...\left(\mathrm{1}\right) \\ $$$$ \\ $$$${b}=\mathrm{1}+{i}=\sqrt{\mathrm{2}}\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}+\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}{i}\right)=\sqrt{\mathrm{2}}\left(\mathrm{cos}\:\frac{\pi}{\mathrm{4}}+{i}\:\mathrm{sin}\:\frac{\pi}{\mathrm{4}}\right) \\ $$$${b}^{{n}} =\sqrt{\mathrm{2}^{{n}} }\left(\mathrm{cos}\:\frac{{n}\pi}{\mathrm{4}}+{i}\:\mathrm{sin}\:\frac{{n}\pi}{\mathrm{4}}\right) \\ $$$${such}\:{that}\:{b}^{{n}} \in{iR}, \\ $$$$\mathrm{cos}\:\frac{{n}\pi}{\mathrm{4}}=\mathrm{0},\:\mathrm{sin}\:\frac{{n}\pi}{\mathrm{4}}\neq\mathrm{0}\:{automatically} \\ $$$$\Rightarrow\frac{{n}\pi}{\mathrm{4}}=\frac{\left(\mathrm{2}{k}+\mathrm{1}\right)}{\mathrm{2}}\pi,\:{k}\in\mathbb{Z} \\ $$$$\Rightarrow{n}=\mathrm{2}\left(\mathrm{2}{k}+\mathrm{1}\right)\:\:\:...\left(\mathrm{2}\right) \\ $$$$ \\ $$$${for}\:{both}\:\left(\mathrm{1}\right)\:{and}\:\left(\mathrm{2}\right): \\ $$$$\Rightarrow{n}=\mathrm{6}\left(\mathrm{2}{k}+\mathrm{1}\right)\:{with}\:{k}\in\mathbb{Z} \\ $$

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