Question

Find all the values of the given root :
$(-64a^4{)}^{1/4}(a\in R)$

Answer 1

$(-64a^4{)}^{1/4}=(2\sqrt{2}\left)a\right(-1{)}^{1/4}$

We know that $-1=\cos \pi +i\sin \pi$

Now put $-1=r\cos \theta ,0=r\sin \theta$

$\therefore (-64a^4{)}^{1/4}=2\sqrt{2}a[\cos \pi +i\sin \pi {]}^{1/4}$

$=2\sqrt{2}a\left[\cos \right(2n\pi +\pi )+i\sin (2n\pi +\pi ){]}^{1/4}$

$=2\sqrt{2}a[\cos \frac{2n\pi +\pi }{4}+i\sin \frac{2n\pi +\pi }{4}]$

where n =0,1,2,and .

Hence the required roots are

$2\sqrt{2}a\left[\cos \right(\pi /4)+i\sin (\pi /4\left)\right]$

$2\sqrt{2}a\left[\cos \right(3\pi /4)+i\sin (3\pi /4\left)\right]$

$2\sqrt{2}a\left[\cos \right(5\pi /4)+i\sin (5\pi /4\left)\right]$

$2\sqrt{2}a\left[\cos \right(7\pi /4)+i\sin (7\pi /4\left)\right]$

Thus the roots on putting the values are

$2\sqrt{2}a(1/\sqrt{2}+i/\sqrt{2}),2\sqrt{2}a(-1/\sqrt{2}+i/\sqrt{2}),$

$2\sqrt{2}a(-1/\sqrt{2}-i/\sqrt{2}),2\sqrt{2}a(1/\sqrt{2}-i/\sqrt{2}),$

Hence the roots are$\pm 2a(1\pm i)$