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

$$\mathrm{Determine}\mathrm{the}\mathrm{roots}\mathrm{of}\mathrm{the}\mathrm{equation}{\mathrm{x}}^3+64=0\mathrm{in}\mathrm{the}\mathrm{polar}\mathrm{form}\mathrm{a}+\mathrm{jb},$$$$\mathrm{Where}\mathrm{a}\mathrm{and}\mathrm{b}\mathrm{are}\mathrm{real}.$$

Answered by mrW1 last updated on 03/Jun/17

$$x=r(\cos \theta +i\sin \theta )$$$$x^3=r^3(\cos 3\vartheta +i\sin 3\theta )=-64=64(-1+0i)$$$$\Rightarrow r^3=64\Rightarrow r={}^3\sqrt{64}=4$$$$\Rightarrow \cos 3\theta =-1$$$$\Rightarrow \sin 3\theta =0$$$$\Rightarrow 3\theta =(2n-1)\pi ,n=1,2,3$$$$\Rightarrow {\theta }_1=\frac{\pi }{3}$$$$\Rightarrow {\theta }_2=\pi$$$$\Rightarrow {\theta }_3=\frac{5\pi }{3}$$$$x_1=4(\cos \frac{\pi }{3}+i\sin \frac{\pi }{3})=2+2\sqrt{3}i$$$$x_2=4(\cos \pi +i\sin \pi )=-4+0i$$$$x_3=4(\cos \frac{5\pi }{3}+i\sin \frac{5\pi }{3})=2-2\sqrt{3}i$$

Commented by RasheedSoomro last updated on 04/Jun/17

$$\mathcal{T}hisshortwayisnewforme!$$

Commented by mrW1 last updated on 03/Jun/17

$$or$$$$x^3=-64$$$${\left(\frac{x}{-4}\right)}^3=1$$$$\Rightarrow \frac{x}{-4}=1,\omega ,{\omega }^2$$$$\Rightarrow x=-4,-4\omega ,-4{\omega }^2$$

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

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}.$$

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

$$\mathrm{Then}\mathrm{what}\mathrm{is}\mathrm{w}\mathrm{sir}$$

Commented by Tinkutara last updated on 04/Jun/17

$$\mathrm{It}\mathrm{is}\mathrm{not}\mathrm{w};\mathrm{they}\mathrm{are}\omega ,\mathrm{cube}\mathrm{roots}\mathrm{of}$$$$\mathrm{unity}.$$$$x^3-1=0$$$$(x-1)(x^2+x+1)=0$$$$\mathrm{Roots}\mathrm{of}{x}^2+x+1\mathrm{are}\omega \mathrm{and}{\omega }^2.$$

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

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}.$$

Commented by mrW1 last updated on 04/Jun/17

$$\omega and{\omega }^{2}standfor-\frac{1}{2}\pm \frac{\sqrt{3}}{2}i$$

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

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}.$$

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

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}.$$

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

$$\mathrm{To}\mathrm{dear}\mathrm{tawa}.$$$$\mathrm{To}\mathrm{find}\mathrm{the}\mathrm{qube}\mathrm{root}\mathrm{of}1\mathrm{put}{\mathrm{x}}^3=1$$$$\therefore {\mathrm{x}}^3-1=0$$$$\mathrm{or},(\mathrm{x}-1)({\mathrm{x}}^2+\mathrm{x}+1)=0$$$$\therefore (\mathrm{x}-1)=0\Rightarrow \mathrm{x}=1$$$$\mathrm{or},({\mathrm{x}}^2+\mathrm{x}+1)=0$$$$\therefore \mathrm{x}=\frac{-1\pm \sqrt{1-4.1.1}}{2}=\frac{-1\pm \mathrm{i}\sqrt{3}}{2}$$$$\mathrm{It}\mathrm{is}\mathrm{denoted}\mathrm{by}\omega \mathrm{and}{\omega }^2.$$$$\mathrm{Where}\omega =\frac{-1+\mathrm{i}\sqrt{3}}{2}$$$${\omega }^2=\frac{-1-\mathrm{i}\sqrt{3}}{2}\left[\mathrm{You}\mathrm{may}\mathrm{varify}\mathrm{it}\right]$$$$\mathrm{Properties}:$$$$\left(\mathrm{a}\right){\omega }^3=1$$$$\therefore {\omega }^4={\omega }^3.\omega =\omega$$$$\left(\mathrm{b}\right)1+\omega +{\omega }^2=0$$

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