Determine-the-roots-of-the-equation-x-3-64-0-in-the-polar-form-a-jb-Where-a-and-b-are-real-

Question Number 14587 by tawa tawa last updated on 02/Jun/17

Determine the roots of the equation x^{3} + 64 = 0 in the polar form a + jb,

Where a and b are real .

Answered by mrW1 last updated on 03/Jun/17

x = r (\cos θ + i \sin θ)

x^{3} = r^{3} (\cos 3 ϑ + i \sin 3 θ) = - 64 = 64 (- 1 + 0 i)

\Rightarrow r^{3} = 64 \Rightarrow r =^{3} \sqrt{64} = 4

\Rightarrow \cos 3 θ = - 1

\Rightarrow \sin 3 θ = 0

\Rightarrow 3 θ = (2 n - 1) π, n = 1, 2, 3

\Rightarrow θ_{1} = \frac{π}{3}

\Rightarrow θ_{2} = π

\Rightarrow θ_{3} = \frac{5 π}{3}

x_{1} = 4 (\cos \frac{π}{3} + i \sin \frac{π}{3}) = 2 + 2 \sqrt{3} i

x_{2} = 4 (\cos π + i \sin π) = - 4 + 0 i

x_{3} = 4 (\cos \frac{5 π}{3} + i \sin \frac{5 π}{3}) = 2 - 2 \sqrt{3} i

Commented by RasheedSoomro last updated on 04/Jun/17

T h i s s h o r t w a y i s n e w f o r m e!

Commented by mrW1 last updated on 03/Jun/17

o r

x^{3} = - 64

{(\frac{x}{- 4})}^{3} = 1

\Rightarrow \frac{x}{- 4} = 1, ω, ω^{2}

\Rightarrow x = - 4, - 4 ω, - 4 ω^{2}

Commented by tawa tawa last updated on 04/Jun/17

God bless you sir .

Commented by tawa tawa last updated on 04/Jun/17

Then what is w sir

Commented by Tinkutara last updated on 04/Jun/17

It is not w; they are ω, cube roots of

unity .

x^{3} - 1 = 0

(x - 1) (x^{2} + x + 1) = 0

Roots of x^{2} + x + 1 are ω and ω^{2} .

Commented by tawa tawa last updated on 04/Jun/17

God bless you sir .

Commented by mrW1 last updated on 04/Jun/17

ω a n d ω^{2} s t a n d f o r - \frac{1}{2} \pm \frac{\sqrt{3}}{2} i

Commented by tawa tawa last updated on 04/Jun/17

God bless you sir .

Commented by tawa tawa last updated on 04/Jun/17

God bless you sir .

Commented by arnabpapu550@gmail.com last updated on 13/Jun/17

To dear tawa .

To find the qube root of 1 put x^{3} = 1

∴ x^{3} - 1 = 0

or, (x - 1) (x^{2} + x + 1) = 0

∴ (x - 1) = 0 \Rightarrow x = 1

or, (x^{2} + x + 1) = 0

∴ x = \frac{- 1 \pm \sqrt{1 - 4.1.1}}{2} = \frac{- 1 \pm i \sqrt{3}}{2}

It is denoted by ω and ω^{2} .

Where ω = \frac{- 1 + i \sqrt{3}}{2}

ω^{2} = \frac{- 1 - i \sqrt{3}}{2} [You may varify it]

Properties :

(a) ω^{3} = 1

∴ ω^{4} = ω^{3} . ω = ω

(b) 1 + ω + ω^{2} = 0