Hi,
Here is another example (with nice root values and
detailed explanations), which is closer to the original problem
that was proposed: Solve z^4 - 4z^2 + 16 = 0, by considering the
substitution y = z^2.
First, the equation becomes y^2 - 4y + 16 = 0. By
completing the square,
(y - 2)^2 + 12 = 0
=> y = 2 +/- 2*sqrt(3) i.
Let z = x + iy, so that z^2 = (x^2 - y^2) + 2xyi.
For y = 2 + 2*sqrt(3) i, we compare real and imaginary
parts to obtain two equations:
x^2 - y^2 = 2 -- (1)
2xy = 2*sqrt(3) => xy = sqrt(3) => y = sqrt(3) / x
-- (2).
Substituting (2) into (1),
x^2 - 3/x^2 = 2
=> x^2 - 2 - 3/x^2 = 0
=> (x - 3/x)(x + 1/x) = 0
=> x = +/- sqrt(3), so y = +/- 1.
Two roots are z = sqrt(3) + i and z = -sqrt(3) - i.
For y = 2 - 2*sqrt(3) i, the same method applies for us
to find x, y values.
One will get the roots z = sqrt(3) - i and z = -sqrt(3) +
i.
Thanks.
Cheers,
Wen Shih