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Question Number 118984 by 675480065 last updated on 21/Oct/20

$$\mathrm{z}=\mathrm{x}+\mathrm{iy}$$$$\mathrm{and}{\mathrm{z}}^5=1$$$$\mathrm{Show}\mathrm{that}4\mathrm{x}({\mathrm{y}}^4-{\mathrm{x}}^4)=1$$$$\mathrm{Pls}\mathrm{help}$$

Answered by Dwaipayan Shikari last updated on 21/Oct/20

$${(x+iy)}^5=1$$$$x^5+5x^{4}iy-10x^3{y}^2-i10x^2{y}^3+5{xy}^4+{iy}^5=1+0i$$$$i(5x^{4}y-10x^2{y}^3+y^5)=0$$$$★5x^{4}y+y^5-10x^2{y}^3=0\Rightarrow 5x^5+{xy}^4-10x^3{y}^2=0.\frac{x}{y}=0\to \left(a\right)$$$$★Mutiplybothsidesby\frac{x}{y}(x,y\ne 0)$$$$x^5-10x^3{y}^2+5{xy}^4=1\to \left(b\right)\left(Realpart\right)$$$$\left(b\right)-\left(a\right)$$$$-4x^5+4{xy}^4=1-0$$$$4x(y^4-x^4)=1$$

Commented by 675480065 last updated on 21/Oct/20

$$\mathbf{T}\mathrm{Hankz}$$

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