I feel ridiculous asking this, its something I should be able to do, however I shall ask anyway. I am doing a calculation that requires me to find the roots of the equation $\frac{1}{4}(z^{-2}+2z-z^2)$. Using wolfram alpha you get two real roots z = -0.717, 2.10692 and two complex roots, z = 0.30 -0.755i and z = 0.30 + 0.755i.
How do I get the complex solutions? Thanks for any help.
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As P Vanchinathan answered, you can reduce the problem to the roots of $$f(z)=z^4-2z^3-1=0$$ Quartic equation have analytical solutions with radicals. I accept that this is not the most pleasnt task to achieve.
So, you could first compute the real roots using Newton method which, starting from a reasonable guess $z_0$ will update it according to $$z_{n+1}=z_n-\frac{f(z_n)}{f'(z_n)}$$ Applied to your case, this gives $$z_{n+1}=\frac{3 z_n^4-4 z_n^3+1}{2 z_n^2 (2 z_n-3)}$$ A quick look at the plot of $f(z)$ shows roots close to $-0.7$ and $2$. Applying twice Newton method with these starting points, we find (very few iterations required for ten decimal places) $$z_1=-0.7166727493$$ $$z_2=+2.1069193404$$ Looking again the quartic polynomial we know that $z_1+z_2+z_3+z_4=2$ and that $z_1\times z_2\times z_3\times z_4=-1$.
So, since $f(z)=(z-z_1)(z-z_2)(z^2+a z+b)$ we know that $a=-(z_3+z_4)$ and $b=z_3\times z_4$; then $a$ and $b$ and a quadratic with no real roots to solve.
I am sure that you can take from here.
Call your equation $f(z)=0$. Their roots are the same as the roots of $z^2f(z)=0$. Now this equation, rewritten as $z^4-2z^3-1=0$ is a polynomial equation of degree $4$. This has two real roots, $a,b$. Then, in theory you can divide this polynomial by $(x-a)(x-b)$ and get a quadratic equation with real coefficients. This has negative discriminant leading to two complex roots of your equation.