Complex Numbers 5 Question 3
3. Let $z _0$ be a root of the quadratic equation, $x^{2}+x+1=0$, If $z=3+6 i z _0^{81}-3 i z _0^{93}$, then $\arg z$ is equal to
(a) $\frac{\pi}{4}$
(b) $\frac{\pi}{6}$
(c) 0
(d) $\frac{\pi}{3}$
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Answer:
Correct Answer: 3. (a)
Solution:
- Given, $x^{2}+x+1=0$
$\Rightarrow \quad x=\frac{-1 \pm \sqrt{3} i}{2}$
$\left[\because\right.$ Roots of quadratic equation $a x^{2}+b x+c=0$
are given by $x=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}$ ]
$\Rightarrow z _0=\omega, \omega^{2}\left[\right.$ where $\omega=\frac{-1+\sqrt{3} i}{2}$ and
$$ \omega^{2}=\frac{-1-\sqrt{3} i}{2} $$
are the cube roots of unity and $\omega, \omega^{2} \neq 1$ )
Now consider $z=3+6 i z _0^{81}-3 i z _0^{93}$
$$ \begin{aligned} & =3+6 i-3 i \quad\left(\because \omega^{3 n}=\left(\omega^{2}\right)^{3 n}=1\right) \\ & =3+3 i=3(1+i) \end{aligned} $$
If ’ $\theta$ ’ is the argument of $z$, then
$$ \begin{aligned} \tan \theta & =\frac{\operatorname{Im}(z)}{\operatorname{Re}(z)} \quad[\because z \text { is in the first quadrant }] \\ & =\frac{3}{3}=1 \Rightarrow \quad \theta=\frac{\pi}{4} \end{aligned} $$
