If z= cos θ + isin θ , 0<θ<(π/6) , then prove that argument of 1−z^4 = 2θ − (π/2) .


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Question Number 43363 by rahul 19 last updated on 10/Sep/18

$If z = \cos θ + isin θ, 0 < θ < \frac{π}{6}, then prove$ $that argument of 1 - z^{4} = 2 θ - \frac{π}{2} .$
Commented bymaxmathsup by imad last updated on 10/Sep/18
$we have z =e^(iθ) ⇒1−z^4 =1−e^(4iθ) =1−cos(4θ)−isin(4θ) =2 sin^2 (2θ) −2isin(2θ)cos(2θ) =−2isin(2θ){cos(2θ) +isin(2θ)} =2 sin(2θ)(−i) e^(i(2θ)) we have 0<2θ<(π/3) ⇒sin(2θ)>0 ⇒∣1−z^4 ∣ =2 sin(2θ) and arg(1−z^4 ) ≡arg(−i) +2θ [2π] ⇒arg(1−z^4 )≡ −(π/2) +2θ[2π] .$
$w e h a v e z = e^{i θ} \Rightarrow 1 - z^{4} = 1 - e^{4 i θ} = 1 - c o s (4 θ) - i s i n (4 θ)$ $= 2 {s i n}^{2} (2 θ) - 2 i s i n (2 θ) c o s (2 θ) = - 2 i s i n (2 θ) {c o s (2 θ) + i s i n (2 θ)}$ $= 2 s i n (2 θ) (- i) e^{i (2 θ)} w e h a v e 0 < 2 θ < \frac{π}{3} \Rightarrow s i n (2 θ) > 0 \Rightarrow∣ 1 - z^{4} ∣ = 2 s i n (2 θ)$ $a n d a r g (1 - z^{4}) \equiv a r g (- i) + 2 θ [2 π] \Rightarrow a r g (1 - z^{4}) \equiv - \frac{π}{2} + 2 θ [2 π] .$
	Answered by ajfour last updated on 10/Sep/18

	$1 - z^{4} = 1 - {(x + i y)}^{4}$ $= (1 - x^{4} - y^{4} + 6 x^{2} y^{2}) + 4 i ({x y}^{3} - x^{3} y)$ $\tan ϕ = \frac{4 x y (y^{2} - x^{2})}{1 - {(x^{2} - y^{2})}^{2} + 4 x^{2} y^{2}}$ $= \frac{4 m (m^{2} - 1)}{{(1 + m^{2})}^{2} - {(1 - m^{2})}^{2} + 4 m^{2}}$ $(i f m = \tan θ)$ $\Rightarrow \tan ϕ = \frac{m^{2} - 1}{2 m} = - \cot 2 θ$ $= \tan (2 θ - \frac{π}{2})$ $\Rightarrow ϕ (a r g u m e n t o f 1 - z^{4}) = 2 θ - \frac{π}{2} .$
	Commented byrahul 19 last updated on 10/Sep/18

	$thanks sir .$
	Answered by tanmay.chaudhury50@gmail.com last updated on 10/Sep/18

	$z = c o s θ + i s i n θ = e^{i θ}$ $1 - z^{4}$ $= 1 - e^{i 4 θ}$ $= 1 - (c o s 4 θ + i s i n 4 θ)$ $= 1 - c o s 4 θ - i s i n 4 θ$ $= 2 {s i n}^{2} 2 θ - i \times 2 s i n 2 θ . c o s 2 θ$ $s o t a n α = \frac{- 2 s i n 2 θ c o s 2 θ}{2 {s i n}^{2} 2 θ} = - c o t 2 θ$ $t a n \propto= - t a n (\frac{Π}{2} - 2 θ)$ $t a n α = t a n (2 θ - \frac{Π}{2})$ $α = 2 θ - \frac{Π}{2}$
	Commented byrahul 19 last updated on 10/Sep/18

	$thanks sir .$
	Answered by tanmay.chaudhury50@gmail.com last updated on 10/Sep/18

	$z = c o s θ + i s i n θ = e^{i θ}$ $1 - z^{4}$ $= 1 - e^{i 4 θ}$ $= 1 - (c o s 4 θ + i s i n 4 θ)$ $= 1 - c o s 4 θ - i s i n 4 θ$ $= 2 {s i n}^{2} 2 θ - i \times 2 s i n 2 θ . c o s 2 θ$ $s o t a n α = \frac{- 2 s i n 2 θ c o s 2 θ}{2 {s i n}^{2} 2 θ} = - c o t 2 θ$ $t a n \propto= - t a n (\frac{Π}{2} - 2 θ)$ $t a n α = t a n (2 θ - \frac{Π}{2})$ $α = 2 θ - \frac{Π}{2}$
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