When will the quadratic equation $z^2+z_1z+z_0=0$ have a root lie on the unit circle

Question

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 $z_{0,1} \in \mathbb{C}$ ?

Answer 1

$z^2 + z_1z + z_0$ 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: $z = (z_1-z_0z_1^*)/(|z_0|^2 - 1)$.

Answer 2

If $\,z_0=0\,$ the roots are $\,\{0, -z_1\}\,$, therefore the condition for a root on the unit circle is $\,|z_1|=1\,$. The following will assume $\,z_0 \ne 0\,$.

Let $\,z_1 = -2b, \,z_0 = c, \,b^2 - c = \delta^2\,$, then the equation is $\,z^2 - 2bz+c=0\,$ and the roots $\,b \pm \delta\,$ are the diagonals of the parallelogram with sides $\,b, \delta\,$. By the parallelogram law:

$$ |b-\delta|^2 + |b+\delta|^2 = 2\left(|b|^2 + |\delta|^2\right) $$

One of $\,b \pm \delta\,$ has magnitude $\,1\,$ iff the other one has magnitude $\,|c| \ne 0\,$ by Vieta's relations, so:

$$ 1 + |c|^2 = 2\left(|b|^2 + |\delta|^2\right) \quad\iff\quad 2\,|b^2 - c| = 1 + |c|^2 - 2\,|b|^2 $$

Written in terms of $\,z_1, z_0\,$, the condition is:

$$ |z_1^2 - 4 z_0| \,=\, 2 + 2 |z_0|^2 - |z_1|^2 \tag{1} $$

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[ EDIT ] $\;$ The following translates $\,(1)\,$ into equivalent conditions $\,(2.a), (2.b)\,$ in terms of $\,z_1, z_0\,$ alone, without involving the magnitude of the discriminant $\,|z_1^2-4z_0|\,$.

The RHS of $\,(1)\,$ must be non-negative: $$ |z_1|^2 \le 2 + 2 |z_0|^2 \tag{2.a} $$

The squared magnitudes must match: $$ \require{cancel} \begin{align} |z_1^2 - 4 z_0|^2 &\,=\, (2 + 2 |z_0|^2 - |z_1|^2)^2 \\ (z_1^2 - 4 z_0)(\bar z_1^2 - 4 \bar z_0) &\,=\, (2 + 2 |z_0|^2 - |z_1|^2)^2 \\ \cancel{|z_1|^4} + 16 |z_0|^2 - 8 \text{Re}(z_1^2 \bar z_0) &\,=\, 4 + 4|z_0|^4+\cancel{|z_1|^4} + 8 |z_0|^2 - 4 |z_1|^2 - 4 |z_1|^2|z_0|^2 \end{align} $$ $$ \\ 2 |z_0|^2 - 2 \text{Re}(z_1^2 \bar z_0) \,=\, 1 + |z_0|^4 - |z_1|^2 - |z_1|^2 |z_0|^2 \tag{2.b} $$

$\,(2.b)\,$ is equivalent to the last equality in @eyeballfrog's answer. When $\,|z_0|=1\,$, $\,(2.a)\,$ reduces to $\,|z_1| \le 2\,$, and $\,(2.b)\,$ to $\,\text{Re}(z_1^2 \bar z_0) \,=\, |z_1|^2\,$, or $\,z_1^2 \bar z_0 = |z_1|^2\,$ because the magnitudes are equal on the two sides, which again matches the other answer.

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[ EDIT #2 ] $\;-\;$ For reference, this is the purely algebraic proof of $\,(1)\,$ as previously posted.

With the same notations, a root is on the unit circle iff for (at least) one choice of the $\,\pm\,$ signs:

$$ \begin{align} 1 = |b \pm \delta|^2 &= (b \pm \delta)(\bar b \pm \bar \delta) \\ &= |b|^2 + |\delta|^2 \pm (b \bar \delta + \bar b \delta) \\ \iff\quad \pm (b \bar \delta + \bar b \delta) &= 1 - |b|^2 - |\delta|^2 \\ \iff\quad (b \bar \delta + \bar b \delta)^2 &= (1 - |b|^2 - |\delta|^2)^2 \end{align} $$

Moving everything to one side, expanding and collecting:

$$ \begin{align} 0 &= (1 - |b|^2 - |\delta|^2)^2 - (b \bar \delta + \bar b \delta)^2 \\ &= 1 + |b|^4 + |\delta|^4 - 2|b|^2 - 2|\delta|^2+\cancel{2|b|^2|\delta|^2} - b^2 \bar\delta^2 - \bar b^2 \delta^2 -\cancel{ 2 |b|^2|\delta|^2 } \\ &=1 + |b|^4 + (b^2 - c)(\bar b^2 - \bar c) - 2 |b|^2 - 2 |\delta|^2 - b^2(\bar b^2 - \bar c) - \bar b^2(b^2-c) \\ &= 1 + \cancel{|b|^4} + \cancel{|b|^4} + |c|^2 - \bcancel{b^2 \bar c} - \bcancel{\bar b^2 c} -2 |b|^2 - 2 |\delta|^2 - \cancel{|b|^4} + \bcancel{b^2 \bar c} - \cancel{|b|^4} + \bcancel{\bar b^2 c} \\ &= 1 + |c|^2 - 2 |b|^2 - 2 |\delta|^2 \end{align} $$ $$ \begin{align*} &\iff\quad 2\, |b^2 - c| \,=\, 1 + |c|^2 - 2\, |b|^2 \quad\quad\quad\quad \end{align*} $$