Question

Find the cube roots of 125(cos 288° + i sin 288°).

Answer 1

cos+ i sin = (re^theta)1/3 = r^1*3 * e^(theta/3)
5*(cos96+i sin 96)

Answer 2

\[125^\frac{1}{3}(\cos(\frac{288^o+360^on}{3})+i \sin(\frac{288^o+360^on}{3}))\]
where one solution is given by n=0
another by n=1
and the last by n=2

Answer 3

So is the first cube root 0?
The second cube root 5((cos( 216) + i sin(216))?
And the third cube root 5((cos( 432) + i sin(432))?
@freckles

Answer 4

I solved it that way first but I didn't know what the cube roots were @Catch.me

Answer 5

no the first cube root isn't 0

Answer 6

If I multiply everything by n=0 then what would it be?

Answer 7

why would you multiply everything by 0?

Answer 8

and also I don't see how you got the 432

Answer 9

\[\text{ when } n=0 \\ \text{ you have } \\ \text{ one cube root is given by } \\ 5 (\cos(\frac{288^o+360^o \cdot 0}{3})+i \sin(\frac{288^o+360^o+0}{3})) \\ \text{ simplifying a bit we see this is } \\ 5 (\cos(96^o)+i \sin(96^o))\]
simplifyin it more or putting it in a calculator
this will not give you 0 when n=0

Answer 10

When I did the third cube root I did 288+360=648(2)= 1296/ 3=432

Answer 11

your order of operations is way off

Answer 12

\[\frac{288+360 \cdot 2}{3}=\frac{288+720}{3}=\frac{1008}{3}=336\]

Answer 13

first we do the top
we have 288+360*2
order of operations tells us to do the multiplication then the addition

Answer 14

the top=288+720=1008
and then the bottom of course is 3
1008/3=336

Answer 15

oh ok sorry stupid mistake so the third cube root is 5((cos( 336) + i sin(336))?

Answer 16

yeah and or you suppose to approximate that or leave it like that?

Answer 17

I don't know, can that answer be simplified?

Answer 18

well you can approximate it or leave it in its exact form

Answer 19

Ok thanks for your help @freckles

Answer 20

np

Answer 21

by the way say we want to find the nth roots of
\[r(\cos(\theta)+i \sin(\theta)) \\ \text{ that will be given by } r^\frac{1}{n}(\cos(\frac{\theta+360^o k}{n})+i \sin(\frac{\theta+360^ok}{n})) \\ \text{ where } k=0,1,2,3,...,(n-1)\]
like that will give n amount of nth roots