Answer:
The fourth roots of 16(cos 200° + i sin 200°) are [tex]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})[/tex].
Step-by-step explanation:
The given expression is
[tex]z=16(\cos 200^{\circ}+i\sin 200^{\circ})[/tex]
Using deMoivre's Theorem
[tex](\cos \theta+i\sin \theta)^n=\cos n\theta+i\sin n\theta[/tex]
[tex](z)^{\frac{1}{4}}=[16(\cos 200^{\circ}+i\sin 200^{\circ})]^{\frac{1}{4}}[/tex]
[tex](z)^{\frac{1}{4}}=2(\cos (200\times \frac{1}{4})^{\circ}+i\sin (200\times \frac{1}{4})^{\circ})[/tex]
[tex](z)^{\frac{1}{4}}=2(\cos 50^{\circ}+i\sin 50^{\circ})[/tex]
The four roots are in the form of
[tex](z)^{\frac{1}{4}}=2(\cos (50+90n)^{\circ}+i\sin (50+90n)^{\circ})[/tex]
For n=1,
[tex](z)^{\frac{1}{4}}=2(\cos 140^{\circ}+i\sin 140^{\circ})[/tex]
For n=2,
[tex](z)^{\frac{1}{4}}=2(\cos 230^{\circ}+i\sin 230^{\circ})[/tex]
For n=3,
[tex](z)^{\frac{1}{4}}=2(\cos 320^{\circ}+i\sin 320^{\circ})[/tex]
Therefore the fourth roots of 16(cos 200° + i sin 200°) are [tex]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})[/tex].
