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Q.
The sum of the products of the non-conjugate roots of $i^{1 / 4}$ taken two at a time is
TS EAMCET 2019
Solution:
Let $z=i^{1 / 4} $
$\Rightarrow z^{4}=i=e^{i \pi / 2}$
$i\left(\frac{\pi}{8}\right), e^{i\left(\frac{5 \pi}{8}\right)}, e^{i\left(\frac{9 \pi}{8}\right)}$ and $e^{i\left(\frac{13 \pi}{8}\right)}$
and its roots are and there is no pair of conjugate roots.
So, sum of products of the roots (which are non-conjugate) of equation $z^{4}-i=0$ is zero.