Question

trig... same stuff I have been doing... can some one check my work? I dont know where I went wrong....

Answer 1

where is the question?

Answer 2

you want 4 roots so
first realize sin and cos have period 2pi
\[z=81(\cos(\frac{4\pi}{3}+2 npi)+i \sin(\frac{4\pi}{3}+2 n \pi))\]
now you want the 4th root of this so you will do the whole n=0,1,2,3 thing
where \[z^\frac{1}{4}=81^\frac{1}{4}(\cos(\frac{1}{4}[\frac{4\pi}{3}+2n \pi])+i \sin[ \frac{1}{4}[\frac{4\pi}{3}+2n \pi]))\]

Answer 3

let me look at your work

Answer 4

so n=0 we have
\[3(\cos(\frac{\pi}{3})+i \sin(\frac{\pi}{3}))\]
you did good there
for n=1 we have
\[3(\cos(\frac{5\pi}{6})+i \sin(\frac{5\pi}{6}))\]
that one also looks good
now for n=2 we have that looks good
I think your n=3 is off just a hair

Answer 5

you change 3 to 2?

Answer 6

also
\[3(\frac{\sqrt{3}}{2}-\frac{1}{2}i)=\frac{3 \sqrt{3}}{2}-\frac{3}{2}i\]
is this what you have for n=3?

Answer 7

you did really good
I just think you got tired of all that writing :p

Answer 8

ok let me see if it works..

Answer 9

nope something is wrong.. because the software keeps marking it wrong

Answer 10

everything on your paper is good up until that last line you have

Answer 11

like you corrected just the last line right?

Answer 12

oh wait i think i put it in wrong.

Answer 13

i wasnt putting in the answer in the right order! silly silly mistakes