Find the four complex roots of (z^2 + 2)^2 + 2 = z
I expanded it and got z^4+4z^2-z+6=0, not sure where to go from here
z^4+4z^2-z+6=0 is correct
then
https://www.mathportal.org/calculators/polynomials-solvers/polynomial-roots-calculator.php
note quartic method there
https://www.mathportal.org/formulas/algebra/solalgebric.php
To find the complex roots of the equation, we need to solve the quadratic equation z^4 + 4z^2 - z + 6 = 0. Here's how you can proceed:
Step 1: Let's define a variable substitution to simplify the equation. Let u = z^2. Now, we can rewrite the equation as u^2 + 4u - z + 6 = 0.
Step 2: Solve the quadratic equation. Using the quadratic formula, we have:
u = (-b ± √(b^2 - 4ac)) / 2a,
where a = 1, b = 4, and c = -z + 6.
Substituting the values, we get:
u = (-4 ± √(4^2 - 4(1)(-z + 6))) / (2(1))
= (-4 ± √(16 + 4z - 24)) / 2
= (-4 ± √(4z - 8)) / 2
= -2 ± √(z - 2).
Step 3: Now, we need to solve for z. Substituting back z = √u, we have:
z = ±√(-2 ± √(z - 2)).
Step 4: Square both sides to eliminate the square root:
z^2 = (-2 ± √(z - 2))^2
z^2 = 4 ± 4√(z - 2) + z - 2
z^2 - z + 6 = ± 4√(z - 2).
Step 5: Square both sides again to eliminate the remaining square root:
(z^2 - z + 6)^2 = 16(z - 2).
Now we have a fourth-degree polynomial equation, z^4 - 2z^3 - 5z^2 - 34z + 44 = 0.
From here, we can try to solve the equation by factoring, using the rational root theorem, or using numerical methods such as Newton's method or the bisection method.
Keep in mind that finding the exact roots of a fourth-degree polynomial can be quite complex, and in some cases, an exact solution may not be possible. However, you can use numerical methods or software programs to approximate the roots.