The complex number $z$ is such that $az^2 + bz + a = 0$ where $a$ and $b$ are real constants. It is given that $z = z_0$ is a solution to this equation where $\mathrm{Im}(z_0) \ne 0$. Take $\mathrm{Im}(z_0)=\frac{1}{2}$ for the rest of the question.

1. Verify $z = \dfrac{1}{z_0}$ is a solution by substituting it into the equation.

Substitute $z = \dfrac{1}{z_0}$ into the left-hand side of $az^2 + bz + a = 0$:
$a\left(\frac{1}{z_0}\right)^2 + b\left(\frac{1}{z_0}\right) + a = \frac{a}{z_0^2} + \frac{b}{z_0} + a$
$= \frac{a + bz_0 + az_0^2}{z_0^2}$
Since $z_0$ is a given solution, $az_0^2 + bz_0 + a = 0$.
$= \frac{0}{z_0^2} = 0$
Thus, $z = \dfrac{1}{z_0}$ satisfies the equation and is the other solution.

2. Relate the verified other solution to the complex conjugate.

Since the coefficients $a$ and $b$ are real and $\mathrm{Im}(z_0) \neq 0$, the roots of the quadratic equation must occur in complex conjugate pairs. Therefore, the other solution is $z_0^*$.
$z_0^* = \frac{1}{z_0}$

3. Evaluate the modulus to complete the proof.

Rearranging the equation gives:
$z_0 z_0^* = 1$
Using the property $z_0 z_0^* = |z_0|^2$:
$|z_0|^2 = 1$
Since modulus is non-negative ($|z_0| > 0$), we have:
$|z_0| = 1 \text{ (shown)}$

1. Set up an equation for the real part of $z_0$ using its modulus and given imaginary part.

Let $z_0 = x + \frac{1}{2}\mathrm{i}$, where $x \in \mathbb{R}$.
From part (i), we know $|z_0| = 1$.
$|x + \frac{1}{2}\mathrm{i}| = 1$
$x^2 + \left(\frac{1}{2}\right)^2 = 1^2$

2. Solve for the real part and state the possible values for $z_0$.

$x^2 + \frac{1}{4} = 1$
$x^2 = \frac{3}{4}$
$x = \pm \frac{\sqrt{3}}{2}$
Therefore, the possible complex numbers are $z_0 = \frac{\sqrt{3}}{2} + \frac{1}{2}\mathrm{i}$ and $z_0 = -\frac{\sqrt{3}}{2} + \frac{1}{2}\mathrm{i}$.

1. Determine the real part of $z_0$.

From part (i), we know $|z_0| = 1$. Given that $\mathrm{Im}(z_0) = \frac{1}{2}$, we can find the real part:
$(\mathrm{Re}(z_0))^2 + \left(\frac{1}{2}\right)^2 = |z_0|^2 = 1$
$(\mathrm{Re}(z_0))^2 = 1 - \frac{1}{4} = \frac{3}{4}$
Since it is given that $\mathrm{Re}(z_0) > 0$, we have:
$\mathrm{Re}(z_0) = \frac{\sqrt{3}}{2}$

2. Apply the sum of roots formula to find $b$ in terms of $a$.

Since $a$ and $b$ are real coefficients, the non-real roots of the quadratic equation $az^2 + bz + a = 0$ must occur in conjugate pairs. Thus, the roots are $z_0$ and $z_0^*$.
The sum of the roots is:
$z_0 + z_0^* = 2\mathrm{Re}(z_0) = 2\left(\frac{\sqrt{3}}{2}\right) = \sqrt{3}$
Using the coefficients of the quadratic equation, the sum of roots is also given by $-\frac{b}{a}$.
Equating the two expressions:
$-\frac{b}{a} = \sqrt{3}$
$b = -\sqrt{3}a$