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

$\pi/4$
Option 2)

0
Option 3)

$\pi/6$
Option 4)

$\pi/3$

Answers (1)

best_answer

Definition of Argument/Amplitude of z in Complex Numbers -

$\theta =tan^{-1}|\frac{y}{x}|, z\neq 0$

$\boldsymbol{\theta,\pi-\theta,-\pi+\theta,-\theta}$ are Principal Argument if z lies in first, second, third or fourth quadrant respectively.

- wherein

Cube roots of unity -

$z=\left ( 1 \right )^{\frac{1}{3}}\Rightarrow z=\cos \frac{2k\pi }{3}+i\sin \frac{2k\pi }{3}$

k=0,1,2 so z gives three roots

$\Rightarrow 1,\frac{-1}{2}+i\frac{\sqrt{3}}{2}\left ( \omega \right ),\frac{-1}{2}-i\frac{\sqrt{3}}{2}\left ( \omega^{2} \right )$

- wherein

$\omega=\frac{-1}{2} +\frac{i\sqrt{3}}{2},\omega^{2}=\frac{-1}{2} -\frac{i\sqrt{3}}{2},\omega^{3}=1, 1+\omega+\omega^{2}=0$

$1,\omega,\omega^{2}$ are cube roots of unity.

Quadratic Equation

$x^2 + x +1 = 0$ , roots are, $\omega$ and $\omega^2$ where $\omega$ is the cube root of unity.

$z = 3 + 6i(z_0)^{81} - 3i(z_0)^{93}$

$z_0 = \omega \; \textup{and} \; \omega^2$

$z = 3 + 6i(\omega)^{81} - 3i(\omega)^{93}$

$z = 3 + 3i\;\;\;\because \omega^3 =1$

$arg(z) = \frac{\pi}{4}$

Option 1)

$\pi/4$
Option 2)

0

Option 3)

$\pi/6$

Option 4)

$\pi/3$

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