Problem on solving $z^4 - 2z + 4 = 0$

Question

I'm having a hard time solving this equation $z^4 - 2z +4 =0$. If you could help me how to solve it that would be amazing and also give me some references of how to solve equations like this generally

Answer 1

A standard method with quartics is to try to express them as a difference of two squares - doing this carefully in general leads to a cubic equation which needs to be solved first. In general you would change the variable to eliminate the term in $z^3$, but here that has already been done.

So we write $$p(z)=z^4-2z+4=(z^2+a)^2-(bz+c)^2$$ so that comparing coefficients gives the three equations $2a=b^2, 2bc=2, a^2-c^2=4$.

Then we can square the second to obtain $b^2c^2=1$, use the first to give us $2ac^2=1$ and multiply the third by $2a$ to obtain $$8a=2a^3-2ac^2=2a^3-1$$ and this gives the cubic $$2a^3-8a-1=0$$

Solving this cubic gives $a$, which determines $b$ and $c$ up to sign (we can change the sign of both without affecting anything).

The difference of two squares then allows us to write $$p(z)=(z^2+bz+a+c)(z^2-bz+a-c)$$and we get the roots by solving the quadratics.

The three different solutions for $a$ coming from the cubic correspond to three different ways of pairing the roots of the quartic to give quadratic factors. The sign change for $b$ and $c$ simply exchanges the two factors.