0
$\begingroup$

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.

$\endgroup$
3
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.