The sum of the products of the non-conjugate | vedclass

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(B) Let $z = i^{1/4}$,which implies $z^4 = i = e^{i\pi/2}$.The equation is $z^4 - i = 0$.The roots of this equation are $z_k = e^{i\left(\frac{\pi/2 +

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$0$

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(B) Let $z = i^{1/4}$,which implies $z^4 = i = e^{i\pi/2}$.The equation is $z^4 - i = 0$.The roots of this equation are $z_k = e^{i\left(\frac{\pi/2 + 2k\pi}{4}\right)}$ for $k = 0, 1, 2, 3$.These roots are $e^{i\pi/8}, e^{i5\pi/8}, e^{i9\pi/8}, e^{i13\pi/8}$.None of these roots are conjugates of each other.According to Vieta's formulas for the polynomial $z^4 + 0z^3 + 0z^2 + 0z - i = 0$,the sum of the products of the roots taken two at a time is the coefficient of $z^2$,which is $0$.Therefore,the sum is $0$.

The sum of the products of the non-conjugate roots of $i^{1/4}$ taken two at a time is

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