Question

Find the fourth roots of 16(cos280 + i sin280). Answer in trigonometric form.

Answer 1

for 4th root
\[= \sqrt[4]{r}(\cos (\frac{\theta +2\pi k}{4}) +i \sin (\frac{\theta +2\pi k}{4}))\]
where k = 0,1,2,3

Answer 2

sorry you are in degrees so you would use 360 instead of 2pi

Answer 3

What does k represent?

Answer 4

just a scalar factor used to get all 4 roots

Answer 5

by adding 360 you dont change the initial angle but each time when you divide by 4 you get a distinct root

Answer 6

\[\let z=rcis \theta be a complex number \omega=scis \alpha be its nth \root then \omega ^{n}=z \]

Answer 7

I guess a more drawn out way to explain what they're trying to say is for your first root, simply take the 4th root of your radius of 16 and then divide your beginning angles by 4. This is the first of your 4th roots. After this, take 360 and divide this by 4. Once you find that value, your 2nd root comes from adding (360/4) to the angle of your 1st root. The 3rd root comes from adding (360/4) to your 2nd root. And your final root is found by adding (360/4) to your 3rd root. Throughout this process your radius r still remains the same, as 4th root of 16. The reason we chose 4 the whole time is because we want to use the number that corresponds to the root you want. If you needed 5th roots, we would have taken the 5th root of your radius and divided the angles by 5, etc. Once you understand the process, using the formula @dumbcow posted is a great way to go about it whenever you need to find the roots of a complex number.

Answer 8

@Psymon , thanks for explaining that better :)

Answer 9

Yep, yep ^_^

Answer 10

Thanks