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Question Number 174619 by cortano1 last updated on 06/Aug/22

Answered by aleks041103 last updated on 06/Aug/22

⇒z=(ae^(iα) )^n =a^n e^(inα) re(z)=0⇒cos(nα)=0 ⇒nα=(k+(1/2))π, k∈Z ⇒α=(π/n)(k+(1/2)) tgα=((cosθ)/(1−sinθ))=tg((π/n)(k+(1/2)))=p ⇒((1−s^2 )/((1−s)^2 ))=p^2 , s=sinθ, s≠1 1−s^2 =p^2 +p^2 s^2 −2p^2 s ⇒(p^2 +1)s^2 −2p^2 s+(p^2 −1)=0 s_(1,2) =((2p^2 ±(√(4p^4 −4(p^4 −1))))/(2(p^2 +1)))=((p^2 ±1)/(p^2 +1))=1;1−(2/(p^2 +1)) ⇒sin θ = 1−2cos^2 ((π/n)(k+(1/2))) for k∈Z ⇒sinθ=−cos((((2k+1)π)/n))=sin((((2k+1)π)/n)−(π/2)) if sinθ=1, z=0 which is always a sol. ⇒Ans.: θ=(π/2)+2kπ, k∈Z θ=(((2k+1)/n)−(1/2))π, k∈Z

$$\Rightarrow{z}=\left({ae}^{{i}\alpha} \right)^{{n}} ={a}^{{n}} {e}^{{in}\alpha} \\ $$$${re}\left({z}\right)=\mathrm{0}\Rightarrow{cos}\left({n}\alpha\right)=\mathrm{0} \\ $$$$\Rightarrow{n}\alpha=\left({k}+\frac{\mathrm{1}}{\mathrm{2}}\right)\pi,\:{k}\in\mathbb{Z} \\ $$$$\Rightarrow\alpha=\frac{\pi}{{n}}\left({k}+\frac{\mathrm{1}}{\mathrm{2}}\right) \\ $$$${tg}\alpha=\frac{{cos}\theta}{\mathrm{1}−{sin}\theta}={tg}\left(\frac{\pi}{{n}}\left({k}+\frac{\mathrm{1}}{\mathrm{2}}\right)\right)={p} \\ $$$$\Rightarrow\frac{\mathrm{1}−{s}^{\mathrm{2}} }{\left(\mathrm{1}−{s}\right)^{\mathrm{2}} }={p}^{\mathrm{2}} ,\:{s}={sin}\theta,\:{s}\neq\mathrm{1} \\ $$$$\mathrm{1}−{s}^{\mathrm{2}} ={p}^{\mathrm{2}} +{p}^{\mathrm{2}} {s}^{\mathrm{2}} −\mathrm{2}{p}^{\mathrm{2}} {s} \\ $$$$\Rightarrow\left({p}^{\mathrm{2}} +\mathrm{1}\right){s}^{\mathrm{2}} −\mathrm{2}{p}^{\mathrm{2}} {s}+\left({p}^{\mathrm{2}} −\mathrm{1}\right)=\mathrm{0} \\ $$$${s}_{\mathrm{1},\mathrm{2}} =\frac{\mathrm{2}{p}^{\mathrm{2}} \pm\sqrt{\mathrm{4}{p}^{\mathrm{4}} −\mathrm{4}\left({p}^{\mathrm{4}} −\mathrm{1}\right)}}{\mathrm{2}\left({p}^{\mathrm{2}} +\mathrm{1}\right)}=\frac{{p}^{\mathrm{2}} \pm\mathrm{1}}{{p}^{\mathrm{2}} +\mathrm{1}}=\mathrm{1};\mathrm{1}−\frac{\mathrm{2}}{{p}^{\mathrm{2}} +\mathrm{1}} \\ $$$$\Rightarrow{sin}\:\theta\:=\:\mathrm{1}−\mathrm{2}{cos}^{\mathrm{2}} \left(\frac{\pi}{{n}}\left({k}+\frac{\mathrm{1}}{\mathrm{2}}\right)\right)\:{for}\:{k}\in\mathbb{Z} \\ $$$$\Rightarrow{sin}\theta=−{cos}\left(\frac{\left(\mathrm{2}{k}+\mathrm{1}\right)\pi}{{n}}\right)={sin}\left(\frac{\left(\mathrm{2}{k}+\mathrm{1}\right)\pi}{{n}}−\frac{\pi}{\mathrm{2}}\right) \\ $$$${if}\:{sin}\theta=\mathrm{1},\:{z}=\mathrm{0}\:{which}\:{is}\:{always}\:{a}\:{sol}. \\ $$$$\Rightarrow{Ans}.: \\ $$$$\theta=\frac{\pi}{\mathrm{2}}+\mathrm{2}{k}\pi,\:{k}\in\mathbb{Z} \\ $$$$\theta=\left(\frac{\mathrm{2}{k}+\mathrm{1}}{{n}}−\frac{\mathrm{1}}{\mathrm{2}}\right)\pi,\:{k}\in\mathbb{Z} \\ $$

Commented by Tawa11 last updated on 06/Aug/22

Great sir

$$\mathrm{Great}\:\mathrm{sir} \\ $$

Answered by Mathspace last updated on 06/Aug/22

z=(1−cos((π/2)−θ)+isin((π/2)−θ))^n =(2sin^2 ((π/4)−(θ/2))+2isin((π/4)−(θ/2))cos((π/4)−(θ/2)))^n =(2i)^n sin^n ((π/4)−(θ/2)){cos((π/4)−(θ/2))−isin((π/4)−(θ/2))}^n =(2i)^n sin^n ((π/4)−(θ/2))e^(i(((nπ)/4)−((nθ)/2))) =2^n sin^n ((π/4)−(θ/2))e^(i((nπ)/2)+i(((nπ)/4)−((nθ)/2))) =2^n sin^n ((π/4)−(θ/2))e^(i(((3nπ)/4)−((nθ)/2))) Res(z)=0 ⇒ 2^n sin^n ((π/4)−(θ/2))cos(((3nπ)/4)−((nθ)/2))=0 ⇒ sin((π/4)−(θ/2))=0 aur cos(((3nπ)/4)−((nθ)/2))=0 ⇒ (π/4)−(θ/2)=kπ aur ((3nπ)/4)−((nθ)/2)=(π/2)+kπ ⇒ (π/2)−θ=2kπ aur ((3nπ)/2)−nθ =(2k+1)π ⇒θ=(π/2)+2kπ aur nθ=((3nπ)/2)−(2k+1)π ⇒ θ=(π/2)+2kπ aur θ=((3π)/2)−((2k+1)π)/n)

$${z}=\left(\mathrm{1}−{cos}\left(\frac{\pi}{\mathrm{2}}−\theta\right)+{isin}\left(\frac{\pi}{\mathrm{2}}−\theta\right)\right)^{{n}} \\ $$$$=\left(\mathrm{2}{sin}^{\mathrm{2}} \left(\frac{\pi}{\mathrm{4}}−\frac{\theta}{\mathrm{2}}\right)+\mathrm{2}{isin}\left(\frac{\pi}{\mathrm{4}}−\frac{\theta}{\mathrm{2}}\right){cos}\left(\frac{\pi}{\mathrm{4}}−\frac{\theta}{\mathrm{2}}\right)\right)^{{n}} \\ $$$$=\left(\mathrm{2}{i}\right)^{{n}} {sin}^{{n}} \left(\frac{\pi}{\mathrm{4}}−\frac{\theta}{\mathrm{2}}\right)\left\{{cos}\left(\frac{\pi}{\mathrm{4}}−\frac{\theta}{\mathrm{2}}\right)−{isin}\left(\frac{\pi}{\mathrm{4}}−\frac{\theta}{\mathrm{2}}\right)\right\}^{{n}} \\ $$$$=\left(\mathrm{2}{i}\right)^{{n}} {sin}^{{n}} \left(\frac{\pi}{\mathrm{4}}−\frac{\theta}{\mathrm{2}}\right){e}^{{i}\left(\frac{{n}\pi}{\mathrm{4}}−\frac{{n}\theta}{\mathrm{2}}\right)} \\ $$$$=\mathrm{2}^{{n}} {sin}^{{n}} \left(\frac{\pi}{\mathrm{4}}−\frac{\theta}{\mathrm{2}}\right){e}^{{i}\frac{{n}\pi}{\mathrm{2}}+{i}\left(\frac{{n}\pi}{\mathrm{4}}−\frac{{n}\theta}{\mathrm{2}}\right)} \\ $$$$=\mathrm{2}^{{n}} {sin}^{{n}} \left(\frac{\pi}{\mathrm{4}}−\frac{\theta}{\mathrm{2}}\right){e}^{{i}\left(\frac{\mathrm{3}{n}\pi}{\mathrm{4}}−\frac{{n}\theta}{\mathrm{2}}\right)} \\ $$$${Res}\left({z}\right)=\mathrm{0}\:\Rightarrow \\ $$$$\mathrm{2}^{{n}} {sin}^{{n}} \left(\frac{\pi}{\mathrm{4}}−\frac{\theta}{\mathrm{2}}\right){cos}\left(\frac{\mathrm{3}{n}\pi}{\mathrm{4}}−\frac{{n}\theta}{\mathrm{2}}\right)=\mathrm{0}\:\Rightarrow \\ $$$${sin}\left(\frac{\pi}{\mathrm{4}}−\frac{\theta}{\mathrm{2}}\right)=\mathrm{0}\:{aur}\:{cos}\left(\frac{\mathrm{3}{n}\pi}{\mathrm{4}}−\frac{{n}\theta}{\mathrm{2}}\right)=\mathrm{0}\:\Rightarrow \\ $$$$\frac{\pi}{\mathrm{4}}−\frac{\theta}{\mathrm{2}}={k}\pi\:\:{aur}\:\frac{\mathrm{3}{n}\pi}{\mathrm{4}}−\frac{{n}\theta}{\mathrm{2}}=\frac{\pi}{\mathrm{2}}+{k}\pi\:\Rightarrow \\ $$$$\frac{\pi}{\mathrm{2}}−\theta=\mathrm{2}{k}\pi\:{aur}\:\frac{\mathrm{3}{n}\pi}{\mathrm{2}}−{n}\theta\:=\left(\mathrm{2}{k}+\mathrm{1}\right)\pi \\ $$$$\Rightarrow\theta=\frac{\pi}{\mathrm{2}}+\mathrm{2}{k}\pi\:{aur}\:{n}\theta=\frac{\mathrm{3}{n}\pi}{\mathrm{2}}−\left(\mathrm{2}{k}+\mathrm{1}\right)\pi\:\Rightarrow \\ $$$$\theta=\frac{\pi}{\mathrm{2}}+\mathrm{2}{k}\pi\:{aur}\:\theta=\frac{\mathrm{3}\pi}{\mathrm{2}}−\frac{\left.\mathrm{2}{k}+\mathrm{1}\right)\pi}{{n}} \\ $$

Commented by peter frank last updated on 06/Aug/22

thnk you

$$\mathrm{thnk}\:\mathrm{you} \\ $$

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