Question

Let principle argument of complex number be re-defined between $\left(\pi ,3\pi \right)$, then sum of principle arguments of roots of equation $z^6+z^3+1=0$ is

• A. 0
• B. $3\pi$
• C. $6\pi$
• D. $12\pi$

Answer 1

Answer: (D)

$12\pi$

$z^6+z^3+1=0$
$let{z}^3=t$
$t^2+t+1=0$
$\Rightarrow t=\frac{-1\pm \sqrt{1-4}}{2}$
$=\frac{-1\pm \sqrt{3}i}{2}$
$=e^{i\frac{2\pi }{3}},e^{i\frac{4\pi }{3}}$
$weknow$
$z=r{e}^{i\theta }$
$\Rightarrow z^3=r^3{e}^{3i\theta }$
$comparing$
$e^{3i\theta }=e^{i\frac{2\pi }{3}},e^{3i\theta }=e^{i\frac{4\pi }{3}}$
$3\theta =2n\pi +\frac{2\pi }{3},3\theta =2n\pi +\frac{4\pi }{3}$
$forn=0$
$3\theta =\frac{2\pi }{3},3\theta =\frac{4\pi }{3}$
$\Rightarrow \theta =\frac{2\pi }{9}=40,\theta =\frac{2\pi }{9}=80$
$forn=1$
$3\theta =\frac{8\pi }{3},3\theta =\frac{10\pi }{3}$
$\Rightarrow \theta =\frac{8\pi }{9}=160,\theta =\frac{10\pi }{9}=200$
$forn=2$
$3\theta =\frac{14\pi }{3},3\theta =\frac{16\pi }{3}$
$\Rightarrow \theta =\frac{14\pi }{9}=280,\theta =\frac{16\pi }{9}=320$
$forn=3$
$3\theta =\frac{20\pi }{3},3\theta =\frac{22\pi }{3}$
$\Rightarrow \theta =\frac{20\pi }{9}=400,\theta =\frac{22\pi }{9}=440$
$forn=4$
$3\theta =\frac{26\pi }{3},3\theta =\frac{28\pi }{3}$
$\Rightarrow \theta =\frac{26\pi }{9}=520,\theta =\frac{28\pi }{9}=560$
$forn=5$
$3\theta =\frac{32\pi }{3},3\theta =\frac{34\pi }{3}$
$\Rightarrow \theta =\frac{32\pi }{9}=640,\theta =\frac{34\pi }{9}=680$
$\theta \in (\pi ,3\pi )$
$=\theta \in (180,540)$
$Hence$
$sum=280+400+510+200+320+440$
$=2160\times \frac{\pi }{180}$
$=12\pi$