For a non-real root z of the equation z^4+z^2 | vedclass

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Question

(D) Given the equation $z^4+z^2+1=0$. Dividing by $z^2$ (since $z 
eq 0$),we get $z^2+1+\frac{1}{z^2}=0$,which implies $z^2+\frac{1}{z^2}=-1$. Also,$

Vedclass · Accepted Answer

$8$

Vedclass · Answer

(D) Given the equation $z^4+z^2+1=0$. Dividing by $z^2$ (since $z 
eq 0$),we get $z^2+1+\frac{1}{z^2}=0$,which implies $z^2+\frac{1}{z^2}=-1$. Also,$(z+\frac{1}{z})^2 = z^2+\frac{1}{z^2}+2 = -1+2 = 1$,so $z+\frac{1}{z} = \pm 1$. For $z^3+\frac{1}{z^3}$,we use the identity $z^3+\frac{1}{z^3} = (z+\frac{1}{z})(z^2-1+\frac{1}{z^2}) = (z+\frac{1}{z})(-1-1) = -2(z+\frac{1}{z})$. If $z+\frac{1}{z} = 1$,then $z^3+\frac{1}{z^3} = -2(1) = -2$. The expression becomes $(1)^3+(-1)^2+(-2)^3 = 1+1-8 = -6$. If $z+\frac{1}{z} = -1$,then $z^3+\frac{1}{z^3} = -2(-1) = 2$. The expression becomes $(-1)^3+(-1)^2+(2)^3 = -1+1+8 = 8$. Since the question implies a unique value,we check the roots of $z^4+z^2+1=0$,which are $e^{i2\pi/3}, e^{i4\pi/3}, e^{i8\pi/3}, e^{i10\pi/3}$. For these,$z+\frac{1}{z} = 2\cos(2\pi/3) = -1$. Thus,the value is $8$.

For a non-real root $z$ of the equation $z^4+z^2+1=0$,find the value of $\left(z+\frac{1}{z}\right)^3+\left(z^2+\frac{1}{z^2}\right)^2+\left(z^3+\frac{1}{z^3}\right)^3$.

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