Question

Find the fourth roots of 16(cos 200° + i sin 200°).

Answer 1

y=2(cos(50=90n)=isin(50+90n))
take n=0 ,y=2(cos(50)+isin(50))

take n=1,y=2(cos(140)+ isin (140))

take n=2,y=2(cos(230)+ isin (230))

take n=3,y=2(cos(320)+isin(320))

Answer 2

The fourth roots of 16(cos 200° + i sin 200°) are
`2(\cos 50^(\circ)+i\sin 50^(\circ)), 2(\cos 140^(\circ)+i\sin 140^(\circ)),2(\cos 230^(\circ)+i\sin 230^(\circ)), 2(\cos 320^(\circ)+i\sin 320^(\circ))`.

Explanation:

The given expression is

`z=16(\cos 200^(\circ)+i\sin 200^(\circ))`

Using deMoivre's Theorem

`(\cos \theta+i\sin \theta)^n=\cos n\theta+i\sin n\theta`

`(z)^{(1)/(4)}=[16(\cos 200^(\circ)+i\sin 200^(\circ))]^{(1)/(4)}`

`(z)^{(1)/(4)}=2(\cos (200* (1)/(4))^(\circ)+i\sin (200* (1)/(4))^(\circ))`

`(z)^{(1)/(4)}=2(\cos 50^(\circ)+i\sin 50^(\circ))`

The four roots are in the form of

`(z)^{(1)/(4)}=2(\cos (50+90n)^(\circ)+i\sin (50+90n)^(\circ))`

For n=1,

`(z)^{(1)/(4)}=2(\cos 140^(\circ)+i\sin 140^(\circ))`

For n=2,

`(z)^{(1)/(4)}=2(\cos 230^(\circ)+i\sin 230^(\circ))`

For n=3,

`(z)^{(1)/(4)}=2(\cos 320^(\circ)+i\sin 320^(\circ))`

Therefore the fourth roots of 16(cos 200° + i sin 200°) are
`2(\cos 50^(\circ)+i\sin 50^(\circ)), 2(\cos 140^(\circ)+i\sin 140^(\circ)),2(\cos 230^(\circ)+i\sin 230^(\circ)), 2(\cos 320^(\circ)+i\sin 320^(\circ))`.