Find the complex number z in the form x+iy, where x, y \in \mathbb{R} such that $\dfrac{3i z}{z^{*}-1+2i} = -i$. [Note: $z^{*}$ is the conjugate of z.] The two roots $z_{1}, z_{2}$ of the equation $z^{2}+z+p=0$, where $p \in \mathbb{R}$, are such that $z_{1}=a+ib$ where a, b \in \mathbb{R}, b \ne 0$and$|z_{1}-z_{2}|=\sqrt{3}$. Find the value of p.

1. Let $z = x+iy$ and substitute it into the given equation.

Let $z = x+iy$, which means the complex conjugate is $z^{*} = x-iy$. Substituting these into the given equation yields:

$\frac{3i(x+iy)}{x-iy-1+2i} = -i$

2. Multiply by the denominator and expand both sides.

Multiply both sides by $(x-iy-1+2i)$ to remove the fraction:

$3i(x+iy) = -i(x - iy - 1 + 2i)$

Expand the brackets:

$3ix + 3i^2y = -ix + i^2y + i - 2i^2$

Substitute $i^2 = -1$:

$-3y + 3ix = -ix - y + i + 2$

Group the real and imaginary components together:

$-3y + i(3x) = (2 - y) + i(1 - x)$

3. Equate the real and imaginary parts to solve for $x$ and $y$.

Equating the real parts:
$-3y = 2 - y$
$-2y = 2 \implies y = -1$

Equating the imaginary parts:
$3x = 1 - x$
$4x = 1 \implies x = \frac{1}{4}$

Therefore, the complex number is $z = \frac{1}{4} - i$.

1. Identify the relationship between the two complex roots.

Since the quadratic equation $z^2+z+p=0$ has real coefficients ($1, 1, p \in \mathbb{R}$) and a complex root $z_1 = a+ib$ (with $b \ne 0$), the roots must occur in conjugate pairs.
Therefore, $z_2 = z_1^* = a - ib$.

2. Use the difference of the roots to find $b^2$.

Substitute $z_1$ and $z_2$ into the given condition $|z_1 - z_2| = \sqrt{3}$:
$|(a+ib) - (a-ib)| = \sqrt{3}$
$|2ib| = \sqrt{3}$
$2|b| = \sqrt{3} \implies |b| = \frac{\sqrt{3}}{2}$
Squaring both sides gives:
$b^2 = \frac{3}{4}$

3. Use the sum of roots to find $a$.

Using the relationship for the sum of roots ($\alpha + \beta = -\frac{b}{a}$):
$z_1 + z_2 = -1$
$(a+ib) + (a-ib) = -1$
$2a = -1 \implies a = -\frac{1}{2}$

4. Use the product of roots to find $p$.

Using the relationship for the product of roots ($\alpha\beta = \frac{c}{a}$):
$z_1 z_2 = p$
$(a+ib)(a-ib) = p$
$a^2 + b^2 = p$
Substitute $a = -\frac{1}{2}$ and $b^2 = \frac{3}{4}$:
$p = \left(-\frac{1}{2}\right)^2 + \frac{3}{4}$
$p = \frac{1}{4} + \frac{3}{4} = 1$