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Q. If $z = \frac{-1}{2}+i \frac{\sqrt{3}}{2}$, then $8 + 10z + 7z^{2}$ is equal to

KEAMKEAM 2012

Solution:

Given, $z=-\frac{1}{2}+i \frac{\sqrt{3}}{2}$
$\therefore z^{2}=\frac{1}{4}-\frac{3}{4}-2 i \times \frac{1}{2} \frac{\sqrt{3}}{2}=-\frac{1}{2}-i \frac{\sqrt{3}}{2}$
$\therefore 8+10 z+7 z^{2}=8+10\left(-\frac{1}{2}+i \frac{\sqrt{3}}{2}\right)$
$+7\left(-\frac{1}{2}-\frac{i \sqrt{3}}{2}\right)$
$=8-5+i 5 \sqrt{3}-\frac{7}{2}-i \frac{7 \sqrt{3}}{2}=-\frac{1}{2}+\frac{i 3 \sqrt{3}}{2}$