Fun with Complex

Geometry Level 4

If α \alpha and β \beta are roots to the equation, t 2 2 t + 2 = 0 t^2-2t +2= 0 and ( x + α ) n ( x β ) n α β = sin n θ sin n θ \dfrac{(x+\alpha)^n - (x-\beta)^n}{\alpha - \beta} = \dfrac{\sin n \theta}{\sin^n \theta} . Find the value of x x .

1 cot θ 1-\cot { \theta } csc θ 1 \csc { \theta } -1 cot θ 1 \cot { \theta } -1 1 csc θ 1-\csc { \theta }

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1 solution

Soumya Shrivastva
Jan 13, 2016

α = 1 + i , β = 1 i x + α = x + 1 + i = r e i Ψ w h e r e tan Ψ = 1 x + 1 . . . . . . ( 1 ) a n d r = ( x + 1 ) 2 + 1 = cot 2 Ψ + 1 = csc Ψ . . . . . . . . ( 2 ) x + β = x + 1 i = r e i Ψ P u t i n t h e g i v e n e q u a t i o n , r n ( e i n Ψ e i n Ψ ) 2 i = r n ( 2 i sin n Ψ ) 2 i = r n sin n Ψ = sin n Ψ sin n Ψ . . . . . [ f r o m ( 2 ) ] Ψ = θ cot Ψ = cot θ x + 1 = cot θ . . . . . [ f r o m ( 1 ) ] x = cot θ 1 \alpha =1+i,\quad \beta =1-i\\ x+\alpha =x+1+i=r{ e }^{ i\Psi }\quad where\quad \tan { \Psi } =\frac { 1 }{ x+1 } \quad \quad ......(1)\\ and\quad r=\sqrt { \left( x+1 \right) ^{ 2 }+1 } =\sqrt { \cot ^{ 2 }{ \Psi } +1 } =\csc { \Psi } \quad ........(2)\\ x+\beta =x+1-i=r{ e }^{ -i\Psi }\\ Put\quad in\quad the\quad given\quad equation,\\ \frac { { r }^{ n }\left( { e }^{ in\Psi }-{ e }^{ -in\Psi } \right) }{ 2i } =\frac { { r }^{ n }\left( 2i\sin { n\Psi } \right) }{ 2i } ={ r }^{ n }\sin { n\Psi } =\frac { \sin { n\Psi } }{ \sin ^{ n }{ \Psi } } \quad \quad .....\left[ from\left( 2 \right) \right] \\ \Longrightarrow \Psi =\theta \quad \quad \quad \quad \Rightarrow \cot { \Psi } =\cot { \theta } \\ \Rightarrow x+1=\cot { \theta } \quad \quad .....\left[ from\left( 1 \right) \right] \\ \Rightarrow x=\cot { \theta } -1

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