Question

find the n, nth roots of the complex number. Leave answer in polar form.

32i n=5

Answer 1

Have you considered DeMoivre's formula?

Answer 2

Yes but I am unsure how to apply it.

Answer 3

Okay, write it out and let's see where it goes.

Answer 4

(r(cos(theta)+i*sin(theta)))^5=(32(cos(90,450,810,1160,1520) +i*sin(90,450,810,1160,1520))

Answer 5

That's a little funny looking. First, it should be 1/5, not 5.

You have 32i = 32cis(pi/2) <== This is excellent. Getting this far is great.

Expand this to the periodic nature of the cosine and sine.

32i = 32[cis(pi/2 + 2kpi)]

Does that make sense?

Answer 6

No that is not what my teacher told me to do so I am confused

Answer 7

Well, the fact that you are here asking about it suggests to me that what your teacher told you could be of greater value. :-)

Can you share what you teacher told you?

Answer 8

apparently @Samantha_Rose is working in degrees and not radians
unfortunate, but it might make the arithmetic easier

Answer 9

the real fifth root of 32 is 2, so that part is taken care of
then take the angle, which is 90 degrees, and divide it by 5

Answer 10

this gives one answer of \(2\left(\cos(18)+i\sin(18)\right)\) as \(\frac{90}{5}=18\)

Answer 11

We were told that n eguals the power to raise it by and then we were to find as many answers as the power stated so we had to find 5 values of beta

Answer 12

ok so we have one right?

Answer 13

\[2\left(\cos(18)+i\sin(18)\right)\] is one fifth root

Answer 14

now there are a couple different methods to find the other 4
they all amount to the same thing really

Answer 15

you can add 360 to 90 and get 450, then divide by 5 again to get \(\frac{450}{5}=90\) giving a second answer of
\[2\left(\cos(90)+i\sin(90)\right)\]

Answer 16

then continue

Answer 17

\(90+360+360=810\) and \(\frac{810}{5}=162\) so another answer is
\[2\left(\cos(162)+i\sin(162)\right)\]

Answer 18

now we have 3, i bet you can find the other two