I think what I would do is write the equation as the sum of two cubes:
Factor:
Now you can easily get all six roots in rectangular form.
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
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 Z^{2} +2 = 0
I then minus -2 to both side
then sub z=x+iY
(x+iY^{2}) = -2
X^{2 }+ 2XYi -Y^{2}= -2
(X^{2} + Y^{2}) - 2XYi = -2
I now have the answer in Cartesian form, however how do I find all the solutions?
thank