Question

find all of the fourth roots of 81(cos(3pi/8) + i sin (3pi/8) express your answer in polar (r(cos theta+ i sin theta)) form

Answer 1

r = 81^1/4=3

Answer 2

i will be with you in a moment just got a call...

Answer 3

recall ...
if
\[w = r(\cos(\theta)+isin(\theta))\]
is a complex number
then in order to obtain the nth root that yields to
\[\sqrt[n]{w}=r^{1/n}[\cos(\frac{\theta+2k \pi}{n})+ isin(\frac{\theta+2k \pi}{n})]\]
where k=1,2,3...n-1

Answer 4

that would be the answer or would I have to do something else?

Answer 5

sorry k=0,1,2,...n-1

Answer 6

that would give you the exact answer to your question with no hassle...

Answer 7

but i will help you solve it don't worry

Answer 8

choose n=4, then k=n-1, so k=3

Answer 9

your theta=3pi/8

Answer 10

and you are done, so give the values k=0,1,2, and 3

Answer 11

if you can't follow me then tell me to solve the whole thing ...

Answer 12

do I need to plug in something into the first equation you gave ?

Answer 13

yes i told you plug k=0, then k=1, then k=2, then k=3, n =4 always

Answer 14

n=4 because you want to find the 4th root

Answer 15

wheredo I plug in the k variables?

Answer 16

\[\sqrt[n]{w}=r^{1/n}[\cos(\frac{\theta+2k \pi}{n})+ isin(\frac{\theta+2k \pi}{n})]\]

Answer 17

let me solve the first root for you as an example then you may generate the rest...ok!

Answer 18

\[\sqrt[4]{w}=r^{1/4}[\cos(\frac{\theta+2(0) \pi}{4})+ isin(\frac{\theta+2(0) \pi}{4})]\]

Answer 19

now notice for every n i substituted 4, and for every k i pluged in 0

Answer 20

and thata would be 3pi/8?

Answer 21

let me simplify this

Answer 22

\[\sqrt[4]{w}=r^{1/4}[\cos(\frac{\frac{3\pi}{8}}{4}+\frac{2\pi (0)}{4})+ isin(\frac{\frac{3\pi}{8}}{4}+\frac{2\pi (0)}{4})]\]

Answer 23

do you want me to simpliy this even more...

Answer 24

yes please because I dont get what I would have to do next after this step

Answer 25

\[\sqrt[4]{w}=r^{1/4}[\cos(\frac{3\pi}{32})+ isin(\frac{3\pi}{32})]\]

Answer 26

how is that now?

Answer 27

you see when k was zero we got rid of 2pik/4

Answer 28

yes because it was zero.. so that would be the first root?

Answer 29

the next root let k=1

Answer 30

in this case 3pi/32+2pi/4

Answer 31

all you worry about is what is inside the cos and sin

Answer 32

how did you get 2pi/4?

Answer 33

i got it from k=1, then 2(1)pi/4

Answer 34

which is 2kpi/n

Answer 35

3pi/32+2pi/4=19pi/32

Answer 36

oh okay I got it now! Thank you so much!

Answer 37

you are welcome but you need to find it till k=3

Answer 38

yes and now I understand how to do it so I'll do it on my own Thank you

Answer 39

this method is a secured method if you can memorize the formula, it will reduce the chance of making errors

Answer 40

ok take care