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


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 $z$.


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.

  • $\begingroup$ 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/… $\endgroup$ – anonymous Nov 29 '18 at 2:24

Your Answer

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

Browse other questions tagged or ask your own question.