Question

Will give medals! and become fan, please help~
use de moivre's theorem to write the complex number in trigonometric form. ((sqrt3)+i)^3

Answer 1

Have you learned any methods to do this?

Answer 2

A. 2(cos(5pi/3) + isin(5pi/3))
B. 2(cos(2pi/3) + isin(2pi/3))
c. 8(cos(pi) + isin(pi))
D. 8(cos(pi/2) + isin(pi/2))

Answer 3

OK. Do you know how to do this?

Answer 4

and just Zn=r^n(cos(ntheta) + isin(ntheta), but other than that i have no clue

Answer 5

Alright.\[\left(\sqrt 3 + i\right)^3\]\[= \left(2 \left(\frac{\sqrt{3}}{2} + \frac{1}{2} i\right)\right)^3\]

Answer 6

okay, im following thus far..

Answer 7

Is the expression\[\frac{\sqrt{3}}{2} + \frac{1}{2}i\]in the form of \(\cos\theta + i\sin\theta\)? Do you know which \(\theta\) in particular?

Answer 8

honestly, im not sure

Answer 9

\[\cos\theta = \dfrac{\sqrt{3}}{2}\]and\[\sin\theta = \dfrac{1}{2}\]Do you happen to know an angle that satisfies both of these?

Answer 10

pi/6?

Answer 11

write ((sqrt3)+i)^3 in trigonometric form, how can i accomplish that?

Answer 12

That's correct.\[\left(2 \left(\frac{\sqrt{3}}{2} + \frac{1}{2} i\right)\right)^3\]\[= 8 \times \left(\cos\left(\frac{\pi}{6}\right)+ i\sin\left(\frac{\pi}{6}\right)\right)^3 \]Does that remind you of De Moivre's Theorem?

Answer 13

yes, thats very familiar!

Answer 14

what does it mean since pi/6 isnt in one of my answer options?

Answer 15

Let me restate De Moivre's Theorem:\[\left(\cos\theta + i\sin\theta\right)^n = \cos(n\theta) + i\sin(n\theta)\]As you can see, \(n = 3\).

Answer 16

because the nth root of a unit for any value is n?

Answer 17

I don't get what you're asking, but...\[= 8 \times \left(\cos\left(\frac{\pi}{6}\right)+ i\sin\left(\frac{\pi}{6}\right)\right)^3\]\[= 8 \times \left(\cos\left(\frac{\pi}{6}\times 3\right) + i\sin\left(\frac{\pi}{6}\times 3\right)\right)\]

Answer 18

my original answer would have been 8(cos(pi/2) + isin(pi/2))

Answer 19

and yeah, thats okay! XD

Answer 20

=)

Answer 21

thank you!

Answer 22

You're welcome :)