SOLUTION: will you please heal me Find the cube roots of 8i. Write the answer in a + bi form? I honestly have no idea what I'm doing. I'm just trying to pass trigonometry to graduate.

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Question 1116783: will you please heal me Find the cube roots of 8i. Write the answer in a + bi form? I honestly have no idea what I'm doing. I'm just trying to pass trigonometry to graduate.
Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
This is 8i=0%2B8i , with a=0 and b=8 ,
plotted in the complex plane,
where you plot the real part of a%2Bbi=0%2B8i, a=0 as x ,
and the imaginary part, b=8 as y :


The cube roots of 8i=8%28cos%2890%5Eo%29%2Bi%2Asin%2890%5Eo%29%29
are the three solutions to
%28a%2Bib%29%5E3%22=+%5B%22r%28cos%28theta%29%2Bi%2Asin%28theta%29%29%22%5D%22%5E3%22=%228i

With complex numbers written in the form r%28cos%28theta%29%2Bi%2Asin%28theta%29%29 ,
to multiply them, you multiply the r parts (called modulus or absolute value),
and you add the theta angle part, (called argument).
As a consequence,
%22%5B%22r%28cos%28theta%29%2Bi%2Asin%28theta%29%29%22%5D%22%5E3%22=%22r%5E3%28cos%283theta%29%2Bi%2Asin%283theta%29%29 .
So, what you have to solve is
r%5E3=8 and 3theta=90%5Eo%2Bk%2A360%5Eo for k=%220+%2C+1+%2C+2%22 .
The result is r=2 and theta=30%5Eo%2Bk%2A120%5Eo for k=%220+%2C+1+%2C+2%22 .
So, the three values for theta are
30%5Eo , with cos%2830%5Eo%29=sqrt%283%29%2F2=about0.866 and sin%2830%5Eo%29=1%2F2=0.5 ,
150%5Eo , with cos%28150%5Eo%29=-sqrt%283%29%2F2=about0.866 and sin%28150%5Eo%29=1%2F2=0.5 , and
270%5Eo , with cos%28270%5Eo%29=0 and sin%28270%5Eo%29=-1 ,
and the three cube roots of 8i are
2%28%28sqrt%283%29%2F2%29%2Bi%2A%281%2F2%29%29=highlight%28sqrt%283%29%2Bi%29 ,
2%28%28-sqrt%283%29%2F2%29%2Bi%2A%281%2F2%29%29=highlight%28-sqrt%283%29%2Bi%29 , and
2%280%2Bi%2A%28-1%29%29=-2i .
They can be plotted, as three equally spaced points,
at distance 2 from the origin, like this: