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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} \dots

2) 1 \dots

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 \dots

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 + 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 \dots