If-z-cos-isin-0-lt-lt-pi-6-then-prove-that-argument-of-1-z-4-2-pi-2-

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 by maxmathsup by imad last updated on 10/Sep/18

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 by rahul 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 by rahul 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}