Question

find all the values of the given root :
$(2-2i{)}^{1/3}$

Answer 1

Put 2 = r cos$\theta$;2 = r sin$\theta$; then r = $2\sqrt{2}$ and $\theta =\pi /4$
Hence $(2-2i{)}^{1/3}=(rcos\theta -risin\theta {)}^{1/3}$
$=r^{1/3}(cos\theta -isin\theta {)}^{1/3}$
$=r^{1/3}\left[cos\right(2n\pi +\theta )-isin(2n\pi +\theta ){]}^{1/3}$
$=\left(2\sqrt{2}{)}^{1/3}\right[cos(2n\pi +\pi /4)-isin(2n\pi +\pi /4){]}^{1/3}$
$=\sqrt{2}\left[cos\frac{1}{3}\right(2n\pi +\pi /4)-isin\frac{1}{3}(2n\pi +\pi /4){]}^{1/3}$
putting n =0,1,2, the required roots are
$\sqrt{2}\left[cos\right(\pi /12)-isin(\pi /12\left)\right]$
$\sqrt{2}\left[cos\right(3\pi /4)-isin(3\pi /4\left)\right]$
and $\sqrt{2}\left[cos\right(17\pi /12)-isin(17\pi /12\left)\right]$
Now $cos\frac{3\pi }{4}=-\frac{1}{\sqrt{2}},sin\frac{3\pi }{4}=\frac{1}{\sqrt{2}},$
$cos\frac{17\pi }{12}=cos(\frac{3\pi }{2}-\frac{\pi }{12})=-sin\frac{\pi }{12}$
and $sin\frac{17\pi }{12}=sin(\frac{3\pi }{2}-\frac{\pi }{12})=cos\frac{\pi }{12}$
Hence the roots are :
$\sqrt{2}(cos\frac{\pi }{12}-isin\frac{\pi }{12}),-1-i$
$\sqrt{2}(sin\frac{\pi }{12}+icos\frac{\pi }{12})$