# Analytically find roots of equation with anti-symmetric, complex coefficients.

I've probably killed a couple of trees by now trying to solve the following equation over the complex plane analytically and not numerically:

$$z^4+2z_0z^3-2\bar{z_0}z-1=0$$

Where $$z_0=p+iq$$, a complex number made of $$p$$ and $$q$$ being reals derived from parameters in the original problem.

It has some implications in geometry and physics I'm interested in exploring.

If you divide the equation by $$z^4$$ and take the complex conjugate of the whole equation, you end up with the original equation. This necessarily implies the conjugate of the reciprocal of z is 1, correct? That would make sense, because the context in which I derived it requires solutions on the unit circle.

What I've tried: Multiply by $$\bar{z}^2$$ on both sides and rearrange a bit: $$z^2+2z_0z=2\bar{z}\bar{z_0}+\bar{z}^2$$

Take the conjugate of both sides and you end up with the same equation. I think the means either side is real.

1st attempt:

Set $$z^2+2{z_o}z=p_0$$ where $$p_0$$ is some real constant. Is this jumping the gun, assuming $$p_0$$ is constant?

The quadratic formula gives you something, but I'm not sure what to make of the result: $$z=-z_0+\sqrt{z_0^2+p_0}$$.

You can also multiply the original quadratic equation by $$\bar{z}$$ to get $$z+2z_0=p_0\bar{z}$$ with $$\bar{z}+2\bar{z_o}=p_0z$$ being the conjugate.

Now $$(z+\bar{z})$$ and $$(z-\bar{z})$$ can be solved for independently to get the real and imaginary parts of $$z$$.

$$z=-2\frac{\bar{z_0}+z_0}{1-p_0}-2\frac{z_0-\bar{z_0}}{1+p_0}$$

Inserting p and q from above: $$z=-4\frac{p}{1-p_0}-4i\frac{q}{1+p_0}$$

I know I want $$z$$ on the unit circle, this implies equating the magnitude to 1 getting: $$0=p_0^4-2p_0^2(1+8p^2+8q^2)-32p_0(p^2-q^2)+(1-16p^2-16q^2)$$.

I guess I could use the quartic formula here, but I think this is reducible. Graphing the original equation has a hyperbola intersecting the unit circle in 2 places.

Alternative approach:

Plug in $$z=x+iy$$ into the equation quadratic in $$z$$ and $$\bar{z}$$. Also let $$z_0=p+iq$$.

You get $$z^2+2z_0z=(x^2-y^2+2px-2qy)+i(xy+py+qx)$$.

Interestingly, having set the imaginary term equal to zero is what generated the original 4th order polynomial. Not sure what to make of that, but I don't think its a coincidence.

• The roots $z$ are rather messy. I don't have an analytic solution, but Wolfram Mathematica shows that the result is a mess: wolframalpha.com/input/… – anonymous Nov 29 '18 at 2:24