Question

Find the six roots to the equation ... Find the six roots to the equation ... @Mathematics

Answer 1

\[z^{6}=-i\]

Answer 2

Write -i in polar form exp(i3pi/2)^(1/6)
all the answers like equi-spaced on the unit circle

Answer 3

i'm sorry zarkon can i borrow you

Answer 4

when you get done here of course
http://openstudy.com/?F626518757369D0BHIS=_#/updates/4ec0328be4b0caac70006666

Answer 5

I converted -i to polar form:
\[1(\cos \frac{3\pi}{2} + isin \frac{3\pi}{2})\]

Now I'm doing:

\[1(\cos \frac{\frac{3\pi}{2}+2\pi k}{6} + isin \frac{\frac{3\pi}{2}+2\pi k}{6}\]

Where k ranges from 0 to n-1 ( 5 in this case) , am I doing it right?

Answer 6

That is still rectangular form. polar form is
\[\cos \theta + i \sin \theta =e^{i \theta}\]
but you do get all the values doing your way.

Answer 7

alright, thank you

Answer 8

very rough drawing of the roots starting at 3pi/12 (pi/4), incrementing by 4pi/12 (pi/3)

Answer 9

pi/3?

Answer 10

that's what you get from 2pi k/6 for k=0 to 5

Answer 11

If the first root is
\[1(\cos \frac{\frac{3\pi}{2}+2\pi (0)}{6} + isin \frac{\frac{3\pi}{2}+2\pi (0)}{6}\]

That gives cos (pi/4).. i sin(pi/4)

So the first root is

\[\frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2}i\]

The next one would be
\[1(\cos \frac{\frac{3\pi}{2}+2\pi (1)}{6} + isin \frac{\frac{3\pi}{2}+2\pi (1)}{6}\]

Wouldn't that be cos (7pi/12) + i sin (7pi/12)

and that would be the next root?

Answer 12

Yes, or equivalently, e^(i 7pi/12)
The angle is pi/4 + pi/3, or 3pi/12 + 4pi/12.

Answer 13

oh, right I get it now