Let $\mathrm{z}_{0}$ be a root of the quadratic equation, $\mathrm{x}^{2}+\mathrm{x}+1=0$. If $\mathrm{z}=3+6 \mathrm{iz}_{0}^{81}-3 \mathrm{iz}_{0}^{93}$, then $\arg \mathrm{z}$ is equal to

JEE Main · 2019 · 09 Jan 2019 · Shift 2 · Complex Numbers

Let

z_{0}

be a root of the quadratic equation,

x^{2} + x + 1 = 0

. If

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

, then

ar g z

is equal to

Answer: (A)

The roots of

x^{2} + x + 1 = 0

are the non-real cube roots of unity, so take

z_{0} = ω

with

ω^{3} = 1

. Then

ω^{81} = (ω^{3})^{27} = 1

and

ω^{93} = (ω^{3})^{31} = 1

, so

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

. This lies on the line

y = x

in the first quadrant, so

ar g z = tan^{- 1} \frac{3}{3} = \frac{π}{4}

JEE Main · 2019 · 09 Jan 2019 · Shift 2 · Single correct option