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

	Answered by aleks041103 last updated on 06/Aug/22

	$\Rightarrow z = {({a e}^{i α})}^{n} = a^{n} e^{i n α}$ $r e (z) = 0 \Rightarrow c o s (n α) = 0$ $\Rightarrow n α = (k + \frac{1}{2}) π, k \in Z$ $\Rightarrow α = \frac{π}{n} (k + \frac{1}{2})$ $t g α = \frac{c o s θ}{1 - s i n θ} = t g (\frac{π}{n} (k + \frac{1}{2})) = p$ $\Rightarrow \frac{1 - s^{2}}{{(1 - s)}^{2}} = p^{2}, s = s i n θ, s \neq 1$ $1 - s^{2} = p^{2} + p^{2} s^{2} - 2 p^{2} s$ $\Rightarrow (p^{2} + 1) s^{2} - 2 p^{2} s + (p^{2} - 1) = 0$ $s_{1, 2} = \frac{2 p^{2} \pm \sqrt{4 p^{4} - 4 (p^{4} - 1)}}{2 (p^{2} + 1)} = \frac{p^{2} \pm 1}{p^{2} + 1} = 1; 1 - \frac{2}{p^{2} + 1}$ $\Rightarrow s i n θ = 1 - 2 {c o s}^{2} (\frac{π}{n} (k + \frac{1}{2})) f o r k \in Z$ $\Rightarrow s i n θ = - c o s (\frac{(2 k + 1) π}{n}) = s i n (\frac{(2 k + 1) π}{n} - \frac{π}{2})$ $i f s i n θ = 1, z = 0 w h i c h i s a l w a y s a s o l .$ $\Rightarrow A n s . :$ $θ = \frac{π}{2} + 2 k π, k \in Z$ $θ = (\frac{2 k + 1}{n} - \frac{1}{2}) π, k \in Z$
	Commented by Tawa11 last updated on 06/Aug/22

	$Great sir$
	Answered by Mathspace last updated on 06/Aug/22
	$z=(1−cos((π/2)−θ)+isin((π/2)−θ))^n =(2sin^2 ((π/4)−(θ/2))+2isin((π/4)−(θ/2))cos((π/4)−(θ/2)))^n =(2i)^n sin^n ((π/4)−(θ/2)){cos((π/4)−(θ/2))−isin((π/4)−(θ/2))}^n =(2i)^n sin^n ((π/4)−(θ/2))e^(i(((nπ)/4)−((nθ)/2))) =2^n sin^n ((π/4)−(θ/2))e^(i((nπ)/2)+i(((nπ)/4)−((nθ)/2))) =2^n sin^n ((π/4)−(θ/2))e^(i(((3nπ)/4)−((nθ)/2))) Res(z)=0 ⇒ 2^n sin^n ((π/4)−(θ/2))cos(((3nπ)/4)−((nθ)/2))=0 ⇒ sin((π/4)−(θ/2))=0 aur cos(((3nπ)/4)−((nθ)/2))=0 ⇒ (π/4)−(θ/2)=kπ aur ((3nπ)/4)−((nθ)/2)=(π/2)+kπ ⇒ (π/2)−θ=2kπ aur ((3nπ)/2)−nθ =(2k+1)π ⇒θ=(π/2)+2kπ aur nθ=((3nπ)/2)−(2k+1)π ⇒ θ=(π/2)+2kπ aur θ=((3π)/2)−((2k+1)π)/n)$
	$z = {(1 - c o s (\frac{π}{2} - θ) + i s i n (\frac{π}{2} - θ))}^{n}$ $= {(2 {s i n}^{2} (\frac{π}{4} - \frac{θ}{2}) + 2 i s i n (\frac{π}{4} - \frac{θ}{2}) c o s (\frac{π}{4} - \frac{θ}{2}))}^{n}$ $= {(2 i)}^{n} {s i n}^{n} (\frac{π}{4} - \frac{θ}{2}) {c o s (\frac{π}{4} - \frac{θ}{2}) - i s i n (\frac{π}{4} - \frac{θ}{2})}^{n}$ $= {(2 i)}^{n} {s i n}^{n} (\frac{π}{4} - \frac{θ}{2}) e^{i (\frac{n π}{4} - \frac{n θ}{2})}$ $= 2^{n} {s i n}^{n} (\frac{π}{4} - \frac{θ}{2}) e^{i \frac{n π}{2} + i (\frac{n π}{4} - \frac{n θ}{2})}$ $= 2^{n} {s i n}^{n} (\frac{π}{4} - \frac{θ}{2}) e^{i (\frac{3 n π}{4} - \frac{n θ}{2})}$ $R e s (z) = 0 \Rightarrow$ $2^{n} {s i n}^{n} (\frac{π}{4} - \frac{θ}{2}) c o s (\frac{3 n π}{4} - \frac{n θ}{2}) = 0 \Rightarrow$ $s i n (\frac{π}{4} - \frac{θ}{2}) = 0 a u r c o s (\frac{3 n π}{4} - \frac{n θ}{2}) = 0 \Rightarrow$ $\frac{π}{4} - \frac{θ}{2} = k π a u r \frac{3 n π}{4} - \frac{n θ}{2} = \frac{π}{2} + k π \Rightarrow$ $\frac{π}{2} - θ = 2 k π a u r \frac{3 n π}{2} - n θ = (2 k + 1) π$ $\Rightarrow θ = \frac{π}{2} + 2 k π a u r n θ = \frac{3 n π}{2} - (2 k + 1) π \Rightarrow$ $θ = \frac{π}{2} + 2 k π a u r θ = \frac{3 π}{2} - \frac{2 k + 1) π}{n}$
	Commented by peter frank last updated on 06/Aug/22

	$thnk you$
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