Hence $(2-2 i)^{1 / 3}=(r \cos \theta-r i \sin \theta)^{1 / 3}$
$ =r^{1 / 3}(\cos \theta-i \sin \theta)^{1 / 3} $
$ =r^{1 / 3}[\cos (2 n \pi+\theta)-i \sin (2 n \pi+\theta)]^{1 / 3} $
$=(2 \sqrt{2})^{1 / 3}[\cos (2 n \pi+\pi / 4)-i \sin (2 n \pi+\pi / 4)]^{1 / 3} $
$=\sqrt{2}\left[\cos \frac{1}{3}(2 n \pi+\pi / 4)-i \sin \frac{1}{3}(2 n \pi+\pi / 4)\right]^{1 / 3}$
putting $n =0,1,2$, the required roots are
$\sqrt{2}[\cos (\pi / 12)-i \sin (\pi / 12)] $
$ \sqrt{2}[\cos (3 \pi / 4)-i \sin (3 \pi / 4)]$
and $\sqrt{2}[\cos (17 \pi / 12)-i \sin (17 \pi / 12)]$
Now $\cos \frac{3 \pi}{4}=-\frac{1}{\sqrt{2}}, \sin \frac{3 \pi}{4}=\frac{1}{\sqrt{2}}$,
$\cos \frac{17 \pi}{12}=\cos \left(\frac{3 \pi}{2}-\frac{\pi}{12}\right)=-\sin \frac{\pi}{12}$
and $\sin \frac{17 \pi}{12}=\sin \left(\frac{3 \pi}{2}-\frac{\pi}{12}\right)=\cos \frac{\pi}{12}$
Hence the roots are :
$\sqrt{2}\left(\cos \frac{\pi}{12}-i \sin \frac{\pi}{12}\right),-1-i $
$\sqrt{2}\left(\sin \frac{\pi}{12}+i \cos \frac{\pi}{12}\right)$
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
$1.$ The coordinates of $\mathrm{A}$ and $\mathrm{B}$ are
$(A)$ $(3,0)$ and $(0,2)$
$(B)$ $\left(-\frac{8}{5}, \frac{2 \sqrt{161}}{15}\right)$ and $\left(-\frac{9}{5}, \frac{8}{5}\right)$
$(C)$ $\left(-\frac{8}{5}, \frac{2 \sqrt{161}}{15}\right)$ and $(0,2)$
$(D)$ $(3,0)$ and $\left(-\frac{9}{5}, \frac{8}{5}\right)$
$2.$ The orthocentre of the triangle $\mathrm{PAB}$ is
$(A)$ $\left(5, \frac{8}{7}\right)$ $(B)$ $\left(\frac{7}{5}, \frac{25}{8}\right)$
$(C)$ $\left(\frac{11}{5}, \frac{8}{5}\right)$ $(D)$ $\left(\frac{8}{25}, \frac{7}{5}\right)$
$3.$ The equation of the locus of the point whose distances from the point $\mathrm{P}$ and the line $\mathrm{AB}$ are equal, is
$(A)$ $9 x^2+y^2-6 x y-54 x-62 y+241=0$
$(B)$ $x^2+9 y^2+6 x y-54 x+62 y-241=0$
$(C)$ $9 x^2+9 y^2-6 x y-54 x-62 y-241=0$
$(D)$ $x^2+y^2-2 x y+27 x+31 y-120=0$
Give the answer question $1,2$ and $3.$