Find the equation of the locus of points satisfied by the…

Find the equation of the locus of points satisfied by the complex conjugate roots of the equations: $$ 1^{\circ}\ \ z^2-2pz+6p-9=0; \qquad 2^{\circ}\ \ z^2-4pz+2p^2-3p+2=0; \qquad 3^{\circ}\ \ z^2-2pz+4p=0. $$

Explanation of Method

A quadratic with real coefficients has genuine complex-conjugate roots exactly when its discriminant is negative. Writing a root as $z=x+iy$, Vieta's formulas give $2x = $ (sum of roots, in terms of $p$) and $x^2+y^2=$ (product of roots, in terms of $p$). Eliminating the parameter $p$ between these two relations produces an equation in $x,y$ alone — the locus — and the discriminant condition tells you which portion of that curve is actually achieved.

Part 1° · $z^2-2pz+6p-9=0$

Discriminant: $$ \Delta = (2p)^2-4(6p-9) = 4p^2-24p+36 = 4(p-3)^2 $$

For conjugate roots we need $\Delta<0$, i.e. $(p-3)^2<0$ — impossible for any real $p$, since a square is never negative. At $p=3$, $\Delta=0$ and the equation becomes $(z-3)^2=0$, a repeated real root, not a conjugate pair.

Cross-check via Vieta: root $z=x+iy$ gives $2x=2p\Rightarrow x=p$, and $x^2+y^2=6p-9$. Substituting: $y^2 = 6p-9-p^2 = -(p-3)^2 \le 0$, so real $y\ne0$ is impossible.

Conclusion: no value of $p$ gives this equation genuine complex-conjugate roots — there is no locus.

Part 2° · $z^2-4pz+2p^2-3p+2=0$

Step 1 · Discriminant condition

$$ \Delta = 16p^2-4(2p^2-3p+2) = 8p^2+12p-8 = 4(2p^2+3p-2) $$

Solving $2p^2+3p-2=0$: $p=\dfrac{-3\pm5}{4} = \dfrac12$ or $-2$. Since the parabola opens upward, $\Delta<0$ exactly for:

$$ -2 < p < \frac12 $$

Step 2 · Eliminate $p$ via Vieta

For a conjugate root $z=x+iy$: sum $2x=4p \Rightarrow x=2p$, product $x^2+y^2=2p^2-3p+2$. Substituting $p=\dfrac x2$:

$$ x^2+y^2 = 2\left(\frac x2\right)^2-3\left(\frac x2\right)+2 = \frac{x^2}{2}-\frac{3x}{2}+2 $$

Multiplying by $2$ and rearranging:

$$ x^2+3x+2y^2 = 4 $$

Step 3 · Complete the square

$$ \left(x+\frac32\right)^2+2y^2 = 4+\frac94 = \frac{25}{4} $$

an ellipse centered at $\left(-\dfrac32,0\right)$ with semi-axes $\dfrac52$ (along the real axis) and $\dfrac{5}{2\sqrt2}$ (along the imaginary axis).

Step 4 · Match the range to the discriminant condition

$-2entire ellipse, with only its two real-axis vertices $(-4,0)$ and $(1,0)$ excluded (these correspond to the boundary cases $\Delta=0$, i.e. repeated real roots).

Locus for 2°: the full ellipse $\left(x+\frac32\right)^2+2y^2=\frac{25}{4}$, minus its two real-axis vertices (open circles).

Part 3° · $z^2-2pz+4p=0$

Step 1 · Discriminant condition

$$ \Delta = 4p^2-16p = 4p(p-4) $$

$\Delta<0 \iff p(p-4)<0 \iff 0

Step 2 · Eliminate $p$

Vieta: $2x=2p \Rightarrow x=p$; $x^2+y^2=4p=4x$. So:

$$ x^2-4x+y^2=0 \quad\Longrightarrow\quad (x-2)^2+y^2 = 4 $$

a circle centered at $(2,0)$ with radius $2$.

Step 3 · Match the range

$0entire circle except its two real-axis points $(0,0)$ and $(4,0)$.

Locus for 3°: the full circle $(x-2)^2+y^2=4$, minus $(0,0)$ and $(4,0)$.

Result

$$ \begin{aligned} 1^{\circ}\ &\text{No locus — no real } p \text{ gives genuine conjugate roots.} \\ 2^{\circ}\ &\left(x+\tfrac32\right)^2+2y^2=\tfrac{25}{4} \quad \text{(full ellipse, excluding } (-4,0),(1,0)\text{)} \\ 3^{\circ}\ &(x-2)^2+y^2=4 \quad \text{(full circle, excluding } (0,0),(4,0)\text{)} \end{aligned} $$

In both 2° and 3°, the discriminant's valid range for $p$ turns out to match the curve's entire natural extent along the real axis — so the excluded points are just the two vertices where the "conjugate pair" degenerates into a repeated real root, not a genuine sub-arc.