1)Find the area of the triangle formed by roots of cubic equation (z+αb)^3 =α^3_ (α≠0). 2) Find product of all possible values of ((1/2)+(((√3)i)/2))^(3/4) .


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Question Number 49200 by rahul 19 last updated on 04/Dec/18

$1) F i n d t h e a r e a o f t h e t r i a n g l e f o r m e d$ $b y r o o t s o f c u b i c e q u a t i o n$ ${(z + α b)}^{3} = α^{3_{}} (α \neq 0) .$ $2) F i n d p r o d u c t o f a l l p o s s i b l e v a l u e s$ $o f {(\frac{1}{2} + \frac{\sqrt{3} i}{2})}^{\frac{3}{4}} .$
Commented by rahul 19 last updated on 04/Dec/18

$A n s \to 1) 1.3 α^{2} . . .$ $2) 1 . . .$
	Answered by tanmay.chaudhury50@gmail.com last updated on 04/Dec/18

	$2) a n o t h e r a p o r o a c h . .$ ${(\frac{1}{2} + \frac{\sqrt{3}}{2} i)}^{\frac{3}{4}} = {(e^{i \frac{π}{3}})}^{\frac{3}{4}} = {(e^{i π})}^{\frac{1}{4}} = {(c o s 2 k π + i s i n 2 k π)}^{\frac{1}{4}}$ $= (c o s \frac{2 k π}{4} + i s i n \frac{2 k π}{4}) = e^{\frac{i 2 k π}{4}}$ $s o r e q u i r e d a n s i s$ $e^{i \times \frac{2 \times 0 \times π}{4}} \times e^{i \frac{2 \times 1 \times π}{4}} \times e^{i \frac{2 \times - 1 \times π}{4}} \times e^{i \frac{2 \times 2 \times π}{4}}$ $= e^{0} \times e^{0} \times e^{i π} = 1 \times 1 \times (c o s π + i s i n π) = 1$
	Commented by rahul 19 last updated on 05/Dec/18

	$\cos π = - 1, r i g h t ?$ $S o y o u r a n s w e r c o m e s o u t t o b e - 1 . . .$ $T h a n k y o u s i r f o r p r o v i d i n g h i n t i^{'} v e$ $g o t t h e c o r r e c t a n s . n o w :)$
	Commented by rahul 19 last updated on 05/Dec/18

	$Y o u g e t$ $z = {(e^{i π})}^{\frac{1}{4}} = {(- 1)}^{\frac{1}{4}}$ $\Rightarrow z^{4} + 1 = 0$ $∴ P r o d u c t = 1 .$
	Commented by tanmay.chaudhury50@gmail.com last updated on 05/Dec/18

	$y e s c o s π = - 1 t y p o . .$
	Answered by tanmay.chaudhury50@gmail.com last updated on 04/Dec/18
	$1)(z+αb)^3 =α^3 ×1 (z+αb)^3 =α^3 ×e^(i2kπ) z+αb=α×e^((i2kπ)/3) z_1 +αb=α×e^((i×0×π)/3) =α(cos0^o +isin0^o ) (x_1 +iy_1 )+αb=α(1+i×0) (x_1 ,y_1 )={(α−αb),0}←A z_2 +αb=α×e^((i×2×1×π)/3) =α(cos120^o +isin120^o ) (x_2 +iy_2 )=−αb+α(((−1)/2)+((i×(√3))/2)) (x_2 ,y_2 )={(−αb−(α/2)),((α(√3))/2)}←B z_3 +αb=α×e^((i×2×−1×π)/3) =α(cos120^o −isin120^o ) (x_3 +iy_3 )+αb=α(((−1)/2)−((i(√3))/2)) (x_3 ,y_3 )={(−αb−(α/2)),((−α(√3))/2)}←C BC=α(√3) perpendicular distance from A to BC is α−αb+αb+(α/2)=((3α)/2) so area ABC=(1/2)×α(√3) ×((3α)/2)=((3(√3) α^2 )/4)=1.30α^2$
	$1) {(z + α b)}^{3} = α^{3} \times 1$ ${(z + α b)}^{3} = α^{3} \times e^{i 2 k π}$ $z + α b = α \times e^{\frac{i 2 k π}{3}}$ $z_{1} + α b = α \times e^{\frac{i \times 0 \times π}{3}} = α (c o s 0^{o} + i s i n 0^{o})$ $(x_{1} + {i y}_{1}) + α b = α (1 + i \times 0)$ $(x_{1}, y_{1}) = {(α - α b), 0} \leftarrow A$ $z_{2} + α b = α \times e^{\frac{i \times 2 \times 1 \times π}{3}} = α (c o s 120^{o} + i s i n 120^{o})$ $(x_{2} + {i y}_{2}) = - α b + α (\frac{- 1}{2} + \frac{i \times \sqrt{3}}{2})$ $(x_{2}, y_{2}) = {(- α b - \frac{α}{2}), \frac{α \sqrt{3}}{2}} \leftarrow B$ $z_{3} + α b = α \times e^{\frac{i \times 2 \times - 1 \times π}{3}} = α (c o s 120^{o} - i s i n 120^{o})$ $(x_{3} + {i y}_{3}) + α b = α (\frac{- 1}{2} - \frac{i \sqrt{3}}{2})$ $(x_{3}, y_{3}) = {(- α b - \frac{α}{2}), \frac{- α \sqrt{3}}{2}} \leftarrow C$ $B C = α \sqrt{3}$ $p e r p e n d i c u l a r d i s t a n c e f r o m A t o B C i s$ $α - α b + α b + \frac{α}{2} = \frac{3 α}{2}$ $s o a r e a A B C = \frac{1}{2} \times α \sqrt{3} \times \frac{3 α}{2} = \frac{3 \sqrt{3} α^{2}}{4} = 1.30 α^{2}$
	Commented by rahul 19 last updated on 05/Dec/18
	thank you sir!�� colourful solñ!
	Commented by tanmay.chaudhury50@gmail.com last updated on 05/Dec/18

	$y o u r p o s t s a r e u n i q u e . . .$
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