Is there any way to compute a static P value in a polynomial equation?

Question

Assume that we have a second order polynomial equation.

$$0 = P\times 2z^2 + P\times 3z - 3 + z + z^2$$

And $z \in ℂ$ plane. But $|z| = 1$. Then I want to determine $P$ when $|z| = 1$. Assume that we are on the dotted grey line, but we don't know where. We know that the absolute value where we are is $|z| = 1$.

My question is: How can I find $P$? If I was on $z = -1 +0i$, then it was easy. But I don't know $real(z)$ and $imag(z)$. Notice that $P > 0$.

Answer 1

Suppose $p\in \mathbb{R}$, $z\in \mathbb{C}$, with $|z|=1$, are such that $$(2p+1)z^2 + (3p+1)z - 3 = 0$$

If $p=-{\large{\frac{1}{2}}}$, the equation reduces to $-{\large{\frac{z}{2}}}-3=0$, which yields $z=-6$, contrary to $|z|=1$.

Hence $p\ne-{\large{\frac{1}{2}}}$, so the equation is quadratic in $z$.

Consider cases . . .

Case $(1)$:$\;z$ is real.

Then $|z|=1$ implies $z = \pm 1$.

If $z=1$, the equation reduces to $5p -1=0$, which yields $p={\large{\frac{1}{5}}}$.

If $z=-1$, the equation reduces to $-p - 5=0$, which yields $p=-5$.

Case $(2)$:$\;z$ is non-real.

Since the coefficients of the quadratic are real, it follows that the roots, $z_1,z_2$ say, are complex conjugates, hence $z_1z_2=1$.

But by Vieta's formulas, $z_1z_2 = {\large{\frac{-3}{2p+1}}}$, hence $1= {\large{\frac{-3}{2p+1}}}$, which yields $p=-2$.

Thus, considering all cases, the possible values of $p$ are $-5,-2,{\large{\frac{1}{5}}}$.

Answer 2

From the equation,

$$P=-\frac{z^2+z-3}{2z^2+3z}$$ and we can set $z=e^{i\theta}$ to enforce the constraint $|z|=1$. Now, express that the imaginary part of

$$P=-\frac{e^{i2\theta}+e^{i\theta}-3}{2e^{i2\theta}+3e^{i\theta}}$$ is zero, while the real part is positive.