STD 11 - 4.1 complex nubers — JEE कक्षा 11 साइन्स — Question

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 $

$\sqrt{2}\left(\sin \frac{\pi}{12}+i \cos \frac{\pi}{12}\right)$