Question

Last Question on study guide Need Help Bad.
3root Sqrt(8(cos(4pi/5)+i sin(4pi/5))

Answer 1

wolframalpha.com

Answer 2

does say find the cube roots of such and such?

Answer 3

No it wants me to use the de morive theorem

Answer 4

so it isn't cube root ?
it is square root?

Answer 5

I thought it was cube root because of the 3
but should I ignore the 3?

Answer 6

I'm just asking what does 3root sqrt mean?
If it doesn't mean cube root or square root, I'm totally lost.

Answer 7

No my bad it is cube im sorry im just confused

Answer 8

\[\text{ Let } z=r(\cos(\theta)+i \sin(\theta)) \\ \\ \text{ But } z \text{ actually can be represented in infinitely many ways } \\ \\ \text{ that is, it can be written as } z=r (\cos(\theta+2 n \pi)+i \sin(\theta+2 n \pi))\]
\[\text{ Now if wanted to find the } k-\text{th root of } z\]
\[\text{ there are actually } k \text{ amount of } k-\text{th roots of } z\]
Then I have
\[z^\frac{1}{k}=r^{\frac{1}{k}} (\cos(\frac{ \theta+ 2 n \pi}{k})+i \sin (\frac{\theta+2n \pi}{k})) \\ n=0,1,...,k-1\]

Answer 9

Your book should say something like this

Answer 10

You will wind up with 3 roots

Answer 11

Just use the formula in the theorem

Answer 12

your theta is 4pi/5

Answer 13

your k is 3

Answer 14

This is the formula that i have for it (cos θ + i sin θ)n = cos nθ + i sin nθ.

Answer 15

\[(\cos(\theta) +i \sin(\theta))^n=\cos(n \theta)+i \sin (n \theta)\]
this is correct
except here we have n=1/k
\[(\cos(\theta)+i \sin(\theta))^\frac{1}{k}=\cos(\frac{1}{k} \theta)+i \sin(\frac{1}{k} \theta) \\ (\cos(\theta)+i \sin( \theta))^\frac{1}{k} = \cos(\frac{\theta}{k})+i \sin (\frac{\theta}{k})\]

Answer 16

were your k is 3 in this case

Answer 17

ok so it would be (cos(4pi/5)+i sin(4pi/4)1/3=cos(-.81/3)+i sin(.59/3)?

Answer 18

you will also need to use the period is 2pi to find the other 2 cube roots

I'm not sure how 4pi/5 turned into -.81 and .59 at the same time :p

Answer 19

I found the exact value of cos and sin (4pi/5)

Answer 20

\[(\cos(\frac{4 \pi}{5})+ i \sin(\frac{4\pi}{5}))^\frac{1}{3} = \cos(\frac{1}{3}\frac{4\pi}{5})+i \sin(\frac{1}{3} \frac{4 \pi}{5})\]

Answer 21

I should put a multiplication sign there between the fractions
\[(\cos(\frac{4 \pi}{5})+ i \sin(\frac{4\pi}{5}))^\frac{1}{3} = \cos(\frac{1}{3} \cdot \frac{4\pi}{5})+i \sin(\frac{1}{3} \cdot \frac{4 \pi}{5})\]

Answer 22

anyways this will be one cube root
there are still two others
this goes back to using the period is 2pi

Answer 23

Ok i see where i messed up

Answer 24

\[z=8 (\cos(\frac{4 \pi}{5}+2 n \pi)+i \sin( \frac{4\pi}{5}+2 n \pi)) \\ z^\frac{1}{3}=[8 ( \cos(\frac{4 \pi}{5}+2n \pi)+i \sin(\frac{4\pi}{5}+2 n \pi)] ^\frac{1}{3} \\ z^\frac{1}{3}=8^\frac{1}{3} (\cos(\frac{1}{3}(\frac{4 \pi}{5}+2 n \pi))+ i \sin(\frac{1}{3}(\frac{4 \pi}{5}+2 n \pi)))\]

Answer 25

Then would i multiply from there or am i still missing a step?

Answer 26

yes distribute the 1/3 inside both of those trig function things

Answer 27

and then to find the 3 cube roots just enter in n=0,1,2

Answer 28

Ok i would where would i enter n=0 in place of the z?

Answer 29

do you see the n?

Answer 30

that is where you replace n at...

Answer 31

Oh ok im sorry

Answer 32

\[z=8 (\cos(\frac{4 \pi}{5}+2 n \pi)+i \sin( \frac{4\pi}{5}+2 n \pi)) \\ z^\frac{1}{3}=[8 ( \cos(\frac{4 \pi}{5}+2n \pi)+i \sin(\frac{4\pi}{5}+2 n \pi)] ^\frac{1}{3} \\ z^\frac{1}{3}=8^\frac{1}{3} (\cos(\frac{1}{3}(\frac{4 \pi}{5}+2 \color{red}{(0)} \pi))+ i \sin(\frac{1}{3}(\frac{4 \pi}{5}+2 \color{red}{(0)} \pi)))\]

Answer 33

and you would do this for n=1 and n=2

Answer 34

i got z=20.19+6.99i for the 3 cube

Answer 35

n=2 i got 14.25+4.93i

Answer 36

how did you those

Answer 37

i replaced n and the multiply cos(1/3)(4pi/5+2(3)pi)

Answer 38

\[z_n^\frac{1}{3}=8^\frac{1}{3}( \cos(\frac{1}{3}(\frac{4 \pi}{5}+2 n \pi))+i \sin(\frac{1}{3} (\frac{4\pi}{5}+2 n \pi)) \\ z_0^\frac{1}{3}=2 (\cos(\frac{4\pi}{15})+ i \sin(\frac{4\pi}{15})) \\ z_1^\frac{1}{3}=2 ( \cos(\frac{4\pi}{15}+\frac{2 \pi}{3})+ i \sin(\frac{4\pi}{15} + \frac{ 2 \pi}{3}) ) \\ z_1^\frac{1}{3}=2( \cos(\frac{14 \pi}{15})+ i \sin(\frac{14 \pi}{15})) \]

ok I will leave the last cube root to you (the simplifying of fractions part)
but anyways try putting these into your calculator again

Answer 39

Ok thank you i understand this alot better now ill let you know if i need help again

Answer 40

k

Answer 41

and I see what you did
also you entered in n=3 from some reason (this will be the same answer at n=0)

you did cos(1/3) *(4pi/5+2*3pi)

instead of
cos(1/3(4pi/5+2*3pi))

the whole thing needs to be inside the cos not just the 1/3

Answer 42

OK thanks again

Answer 43

np
good luck ryan