When will the quadratic equation $z^2+z_1z+z_0=0$ have a root lie on the unit circle Very similar to this question, but what if the coefficients are complex?
Is there a necessary and sufficient condition to guarentee that there is at least a root on unit circle for $z^2+z_1z+z_0=0$ where $z_{0,1} \in \mathbb{C}$ ?
 A: $z^2 + z_1z + z_0$ has solutions with $|z| = 1$ if and only if the following equation does:
$$
z + z_1 + z_0z^* = 0.
$$
Taking complex conjugates and doing some linear algebra on them gives
$$
(|z_0|^2-1)z = z_1 - z_0z_1^*.
$$
If $|z_0| =  1$, we immediately see that $z_1 = z_1^* z_0$, or equivalently $z_1 = \pm|z_1|z_0^{1/2}$. Putting this into the original equation and simplifying gives
$$
2\mathrm{Re}[zz_0^{-1/2}] \pm |z_1| = 0,
$$
which has solutions when $|z_1| \le 2$.
For $|z_0|\ne 1$, we can just solve for $z$ to get
$$
z = \frac{z_1-z_0z_1^*}{|z_0|^2 - 1}.
$$
This will give $|z| = 1$ provided that
$$
(z_1-z_0z_1^*)(z_1^* - z_0^*z_1) = |z_1|^2(1+|z_0
|^2)-2\mathrm{Re}[z_0^*z_1^2] = (1-|z_0|^2)^2.
$$
Thus, there are two categories of equations $z^2 + z_1z + z_0 = 0$ that have solutions on the unit circle. The first is of the form
$$
z^2 + 2\cos\theta e^{i\phi}z + e^{2i\phi} = 0
$$
for real $\theta,\phi$, which has both solutions on the unit circle: $z = e^{i(\phi\pm\theta)}$. Second is those whose coefficients satisfy
$$
(1-|z_0|^2)^2 + 2\mathrm{Re}[z_0^*z_1^2] = |z_1|^2(1+|z_0|^2)
$$
with $|z_0| \ne 1$, which have only one solution on the unit circle: $z = (z_1-z_0z_1^*)/(|z_0|^2 - 1)$.
