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Question Number 184084 by HeferH last updated on 02/Jan/23

Answered by Rasheed.Sindhi last updated on 03/Jan/23

$$w^3=1\Rightarrow w=1,\omega ,{\omega }^2$$$$Let{w}_1=\omega \mathrm{\&}{w}_2={\omega }^2$$$$x=a+b$$$$y=a^2\omega +b^2{\omega }^2$$$$z=a{\omega }^2+b\omega$$$$\frac{x^2-z^2+y+2ab}{x^2}$$$$x^2=a^2+2ab+b^2$$$$z^2=a^2{\omega }^4+2ab{\omega }^2\omega +b^2{\omega }^2$$$$=a^2{\omega }^3\omega +2ab{\omega }^3+b^2{\omega }^2$$$$=a^2\omega +2ab+b^2{\omega }^2$$$$y=a^2\omega +b^2{\omega }^2$$$$$$$$\frac{x^2-z^2+y+2ab}{x^2}=1+\frac{-z^2+y+2ab}{x^2}$$$$=1+\frac{-(a^2\omega +2ab+b^2{\omega }^2)+(a^2\omega +b^2{\omega }^2)+2ab}{a^2+2ab+b^2}$$$$=1+\frac{-\cancel{a^2\omega }-\cancel{2ab}-\cancel{b^2{\omega }^2}+\cancel{a^2\omega }+\cancel{b^2{\omega }^2}+\cancel{2ab}}{a^2+2ab+b^2}$$$$=1+0=1$$

Commented by HeferH last updated on 03/Jan/23

$$thankyousir,butwhywesay$$$$w^{2}andwaresolutionsof{w}^3=1,itisa$$$$propertyofcomplexnumbers?$$$$$$

Commented by Rasheed.Sindhi last updated on 03/Jan/23

$$Solving{x}^3=1:$$$$x^3-1=0$$$$(x-1)(x^2+x+1)=0$$$$x=1,x=\frac{-1\pm i\sqrt{3}}{2}$$$$Youcanverifyeasilythat$$$$\frac{-1+i\sqrt{3}}{2}\mathrm{\&}\frac{-1-i\sqrt{3}}{2}aresquareof$$$$oneanother.Soifanyoneofthe$$$$twoisdenotedby\omega thentheother$$$$isequalto{\omega }^2.$$

Commented by HeferH last updated on 03/Jan/23

$$Oh,Ididntrealizethatuntilnow,thank$$$$youverymuch:)$$

Commented by Rasheed.Sindhi last updated on 03/Jan/23

$$Anywaysir,yourknowledgeof$$$$mathismorethanofmine!$$$$\mathcal{BTW}$$$$!\mathrm{RAEY}\mathrm{WEN}\mathrm{YPPAH}$$

Commented by HeferH last updated on 03/Jan/23

$$Happynewyeartoyouaswell!:)$$

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