# Thread: need help with complex number

1. ## need help with complex number

Hi, the question I'm struggle with is (z^2 + 1)^3 = -1
I have to find all the solutions for this question and the answer should be in Cartesian form

I start the question by powering both side by 1/3
thus the equation become

Z^2 + 1 = -1^(1/3)
since i^2 = -1
Z^2 +1 = (i^2)^(1/3)
from here I sub in Z = X + iY
so the equation become
X^2 + 2XYi - Y^2 = (i^2)^(1/3) + 1

I'm pretty much stuck now and unable to continue the question. Can you guy tell me how to continue this question and get the solution in cartesian form?
(I have a feeling that I did something wrong in the step above

2. ## Re: need help with complex number

I think what I would do is write the equation as the sum of two cubes:

$\displaystyle (z^2+1)^3+1^3=0$

Factor:

$\displaystyle ((z^2+1)+1)((z^2+1)^2-(z^2+1)+1)=0$

$\displaystyle (z^2+2)(z^4+z^2+1)=0$

$\displaystyle (z^2+2)(z^2+z+1)(z^2-z+1)=0$

Now you can easily get all six roots in rectangular form.

3. ## Re: need help with complex number

I have another tries with the question, thank to your reply I got an idea to move 1 to the other side first

I started by +1 to both side of the equation then powered both side by (1/3)

the equation become Z2 +2 = 0
I then minus -2 to both side
then sub z=x+iY
(x+iY2) = -2
X2 + 2XYi -Y2= -2
(X2 + Y2) - 2XYi = -2
I now have the answer in Cartesian form, however how do I find all the solutions?

thank