Q38 / 90 · JEE Main 2019 (09 Jan Shift 2) · Mathematics

Let $z_{0}$ be a root of quadratic equation, $x^{2} + x + 1 = 0 .$ If $z = 3 + 6 i z_{0}^{81} - 3 i z_{0}^{93}$ , then $a r g$ $(z)$ is equal to:

A
$0$
Correct answer
B
$\frac{π}{4}$
C
$\frac{π}{6}$
D
$\frac{π}{3}$

Solution

z_{0} = w

or

w^{2}

When

z_{0} = w

z = 3 + 6 i \times w^{81} - 3 i \times w^{93}

= 3 + 6 i - 3 i = 3 + 3 i

When

z_{0} = w^{2}

z = 3 + 6 i \times {(w^{2})}^{81} - 3 i \times {(w^{2})}^{93}

= 3 + 6 i - 3 i = 3 + 3 i

∴ \arg (z) = \tan^{- 1} (\frac{I m (z)}{R e (z)}) = \tan^{- 1} (\frac{3}{3}) = \frac{π}{4}

Hence, the correct answer is option: 'B'.

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