Question

Which expression is a cube root of -1+i sqrt 3?

I will attach choices....

Answer 1

@marissalovescats do u know this one?

Answer 2

No idea. Lol

Answer 3

any number in the form x+y*i can be turned into the form

r*[ cos(theta) + i*sin(theta) ]

where

r = sqrt(x^2 + y^2)
theta = arctan(y/x)

Answer 4

once you get the number into r*[ cos(theta) + i*sin(theta) ] form, you would use a variation of De Moivre's Theorem to evaluate the cube root

Answer 5

how do you do that? do you know the answer?

Answer 6

yes I do, but it's against the rules to give out the answers

Answer 7

give it a shot and see how far you get

Answer 8

I did. I am stuck and cannot go any further.

Answer 9

what is -1+i*sqrt( 3 ) in r*[ cos(theta) + i*sin(theta) ] form?

Answer 10

2[cos sqrt 3 +sin sqrt 3]????

Answer 11

r = sqrt((-1)^2 + (sqrt(3))^2 )

r = sqrt(1 + 3)

r = sqrt(4)

r = 2

so that much is correct

Answer 12

theta is NOT sqrt(3) though

Answer 13

Hold on I have it

2[cos 60 + sin 60]

Answer 14

theta = arctan(y/x)

theta = arctan(sqrt(3)/(-1))

theta = ???

Answer 15

I figured it out! I posted it. Now how do i do the cube root?

Answer 16

@jim_thompson5910 where did u go?

Answer 17

are you sure it's 60?

Answer 18

Tan 60 = sqrt 3

Answer 19

yeah but it's negative

Answer 20

Tangent is negative in Quads 2 and 4

Answer 21

what does arctan(sqrt(3)/(-1)) evaluate to

Answer 22

-5pi/3

Answer 23

Can you tell me the answer!! I need to go to sleep! Please!!!

Answer 24

arctan(sqrt(3)/(-1)) = -60

-60+180 = 120

so theta = 120

Answer 25

now what you do is take the cube root of r

then you divide theta by 3

Answer 26

so which answer is it

Answer 27

D?

Answer 28

@jim_thompson5910

Answer 29

no it's not D

Answer 30

remember you're dividing theta by 3

Answer 31

120/3 = ?

Answer 32

C

Answer 33

yep

Answer 34

thx

Answer 35

np