Let z _ 0 be a root of quadratic equation, x ^ 2 + x + 1 = 0 . If z = 3 + 6i z _ 0 ^ 81 - 3i z _ 0 ^ 93 , then arg (z) is equal to:

Correct answer: B. (π)/(4) z _ 0 = w or w ^ 2 When z _ 0 = w z = 3 + 6i × w ^ 81 - 3i × w ^ 93 = 3 + 6i - 3i = 3 + 3i When z _ 0 = w ^ 2 z = 3 + 6i × ( w ^ 2 ) ^ 81 - 3i × ( w ^ 2 ) ^ 93 = 3 + 6i - 3i = 3 + 3i ∴ arg (z) = ^ - 1 ((Im(z))/(Re(z))) = ^ - 1 ((3)/(3)) = (π)/(4)

JEE Main 20199 JanuaryEvening Shift

Exam: JEE Main
Year: 2019
Date: 9 January
Shift: Evening Shift
Subject: Mathematics
Chapter: Complex Number

JEE Main 2019 · 9 January · Evening Shift · Mathematics · Complex Number

Let z _ 0 be a root of quadratic equation, x ^ 2 + x + 1 = 0 . If z...

Let

{z}_{0}

be a root of quadratic equation,

{x}^{2} + x + 1 = 0 .

z = 3 + 6i{z}_{0}^{81} - 3i{z}_{0}^{93}

, then

arg \left(z\right) is equal to:

Options

A $0$
B $\dfrac{\pi}{4}$
C $\dfrac{\pi}{6}$
D $\dfrac{\pi}{3}$

Correct answer

B. $\dfrac{\pi}{4}$

Step-by-step solution

{z}_{0} = w

{w}^{2}

When

{z}_{0} = w

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

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

When

{z}_{0} = {w}^{2}

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

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

∴ arg \left(z\right) = \tan^{- 1}\left(\dfrac{Im\left(z\right)}{Re\left(z\right)}\right) = \tan^{- 1} \left(\dfrac{3}{3}\right) = \dfrac{\pi}{4}

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