Question

let z^3 = -8i

Find all z in the form a + bi

I need someone to tell me if I'm doing this correctly,

I got r = 8, theta = 3pi/2 = arcsin(-8/8) = arcsin(-1)

Then I got the roots,

1 + (3)^(1/2)i
2
1-(3)^(1/2)i

Did I make a mistake? Should I have set theta to -pi/2 instead of 3pi/2 I'm kind of confused in regards to that

Answer 1

r=2.

Answer 2

r = (-8^2)^(1/2) = (64)^(1/2) = 8

Answer 3

I dont think you are right in regards to r = 2

Answer 4

it's r^3=8

Answer 5

oh yeah I know

Answer 6

I gave the roots I got

Answer 7

my general solution is,

\[z_0 = 2(\cos(\frac{\pi}{3} + \frac{2\pi k}{3}) + \sin(\frac{\pi}{3} + \frac{2\pi k}{3})i)\]
where, k = 0, 1, 2

Answer 8

Still need help?

Answer 9

My question is should I have used -pi/2 instead of 3pi/2 they both equal the same thing but does it make a difference in regards to this question are my roots correct?

Answer 10

One moment, I'm going to check

Answer 11

Is there a way to check if your roots are correct make sense?

Answer 12

@Australopithecus shouldn't it be Pi/2 instead of Pi/3 in your answer?

Answer 13

thank you I really appreciate the help

Answer 14

I just finished the problem. I'm going to attach the file.

Answer 15

http://i.imgur.com/Dkf54kX.jpg

Answer 16

You're welcome! I hope you understand it :) Good luck!

Answer 17

@Australopithecus The angle you choose has to be in between 0 and 2Pi. So 3Pi/2 is the right choice.

Answer 18

well,

(3pi/2)*1/3 = pi/2

Answer 19

good eye lol :) yeah you are rght nipunmalhotra93

Answer 20

thanks nipunmalho and jesstho wish I could give multiple medals :S

Answer 21

Lol you're welcome