Question

If $x^{6}=(4-3 i)^{5}$, then the product of all of its roots is (where $\left.\theta=-\tan ^{-1}(3 / 4)\right)$

• A. $5^{5}(\cos 5 \,\theta+i \sin 5 \, \theta)$
• B. $-5^{5}(\cos 5 \,\theta+i \sin 5 \,\theta)$
• C. $5^{5}(\cos 5 \,\theta-i \sin 5 \, \theta)$
• D. $-5^{5}(\cos 5 \, \theta-i \sin 5 \, \theta)$

Answer 1

Answer: (B)

$x^{6}=(4-3 i)^{5}$
$x^{6}=5^{6}\left(\frac{4}{5}-\frac{3 i}{5}\right)=5^{5}(\cos \theta+i \sin \theta)^{5}$
where $\theta=-\tan ^{-1}\left(\frac{3}{4}\right)=5^{5}(\cos 5 \theta+i \sin 5 \theta)$
$x=5^{5 / 6}(\cos 5\, \theta+i \sin 5\, \theta)^{1 / 6}$
$=5^{5 / 6}\left[\cos \left(\frac{2 k \pi+5 \,\theta}{6}\right)+i \sin \left(\frac{2 k \pi+5 \,\theta}{6}\right)\right]$
$x_{1} x_{2} \ldots x_{6}=5^{5}[\cos (5 \pi+5 \,\theta)+i \sin (5 \pi+5 \,\theta)]$
$=5^{5}(-\cos 5\, \theta-i \sin 5 \,\theta)$
$=-5^{5}(\cos 5 \,\theta+i \sin 5 \,\theta)$