复数 $z$ 随 $t$ 变化，满足 $z = 2\cos t + \mathrm{i}(3\sin t), \quad \text{where } 0 \le t < 2\pi.$

1. Express trigonometric functions in terms of the real and imaginary parts of $z$.

Given $z = 2\cos t + \mathrm{i}(3\sin t)$, we have:
$x = \mathrm{Re}(z) = 2\cos t \implies \cos t = \frac{x}{2}$
$y = \mathrm{Im}(z) = 3\sin t \implies \sin t = \frac{y}{3}$

2. Find the Cartesian equation by eliminating the parameter $t$.

Using the trigonometric identity $\cos^2 t + \sin^2 t = 1$:
$\left(\frac{x}{2}\right)^2 + \left(\frac{y}{3}\right)^2 = 1$
$\frac{x^2}{4} + \frac{y^2}{9} = 1$

3. Sketch the curve on an Argand diagram.

The equation $\frac{x^2}{4} + \frac{y^2}{9} = 1$ represents an ellipse centred at the origin $(0,0)$.
On the Argand diagram (with axes $\mathrm{Re}(z)$ and $\mathrm{Im}(z)$), the ellipse intersects the real axis at $2$ and $-2$, and the imaginary axis at $3$ and $-3$.
Since $0 \le t < 2\pi$, exactly one full continuous ellipse is traced.
(Produce a sketch reflecting these intercepts and shape)

1. Identify possible positions of $z_2$ using the symmetry of the ellipse.

From the Argand diagram in part (a), the curve $\frac{x^2}{4} + \frac{y^2}{9} = 1$ is an ellipse which is symmetric about both the real and imaginary axes.
Since $z_1$ and $z_2$ lie on the ellipse and are equidistant from the origin ($|z_1| = |z_2|$), they must lie on the same circle centered at the origin. Thus, $z_2$ must be a reflection of $z_1$ across the axes or the origin.
Let $z_1 = x_1 + \mathrm{i}y_1$. Since $0 < \arg(z_1) < \frac{\pi}{2}$, $z_1$ lies strictly in the first quadrant, meaning $x_1 > 0$ and $y_1 > 0$.
The possible distinct values for $z_2$ (which corresponds to different quadrants) are:

1. $z_2 = -x_1 + \mathrm{i}y_1$ (reflection across the imaginary axis)
2. $z_2 = x_1 - \mathrm{i}y_1$ (reflection across the real axis)
3. $z_2 = -x_1 - \mathrm{i}y_1$ (reflection across the origin)

2. Evaluate $z_1 + z_2$ and find its argument for each possible position.

Case 1: $z_2 = -x_1 + \mathrm{i}y_1$
$z_1 + z_2 = (x_1 + \mathrm{i}y_1) + (-x_1 + \mathrm{i}y_1) = 2\mathrm{i}y_1$
Since $y_1 > 0$, the sum $z_1 + z_2$ lies on the positive imaginary axis, so $\arg(z_1 + z_2) = \frac{\pi}{2}$.

Case 2: $z_2 = x_1 - \mathrm{i}y_1$
$z_1 + z_2 = (x_1 + \mathrm{i}y_1) + (x_1 - \mathrm{i}y_1) = 2x_1$
Since $x_1 > 0$, the sum $z_1 + z_2$ lies on the positive real axis, so $\arg(z_1 + z_2) = 0$.

Case 3: $z_2 = -x_1 - \mathrm{i}y_1$
$z_1 + z_2 = (x_1 + \mathrm{i}y_1) + (-x_1 - \mathrm{i}y_1) = 0$
The argument of $0$ is undefined.

Therefore, the possible values for $\arg(z_1 + z_2)$ are $0$ and $\frac{\pi}{2}$.

1. Express the coefficient $\alpha$ in terms of the roots $z_1$ and $z_2$.

Since $z_1$ and $z_2$ are roots of $z^2 + \alpha z + \beta = 0$, we have:
$(z - z_1)(z - z_2) = 0$
$z^2 - (z_1 + z_2)z + z_1 z_2 = 0$
By comparing coefficients, we obtain:
$\alpha = -(z_1 + z_2)$

2. Use a valid counter-example from the properties of the Argand diagram to show that $\alpha$ can be non-real.

From part (b), we know that for distinct roots with $|z_1| = |z_2|$ on the given ellipse, one possible scenario is that $z_2$ is the reflection of $z_1$ across the imaginary axis, i.e., $z_2 = -\bar{z}_1$.
In this case:
$z_1 + z_2 = z_1 - \bar{z}_1 = 2\mathrm{i}\,\mathrm{Im}(z_1)$
Given $0 < \arg(z_1) < \frac{\pi}{2}$, we know $\mathrm{Im}(z_1) > 0$.
Thus, $z_1 + z_2$ is a non-zero purely imaginary number.
This makes $\alpha = -2\mathrm{i}\,\mathrm{Im}(z_1)$ purely imaginary (not real).
Therefore, it is not necessary for $\alpha$ to be real.

1. Determine the relationship between $z_1$ and $z_2$ to find $\beta$.

Given $z_1$ and $z_2$ lie on the ellipse and $|z_1| = |z_2|$, the points lie on the intersection of the ellipse and a circle centered at the origin. This yields symmetric points, so $z_2$ must be a reflection of $z_1$.
For the quadratic equation $z^2 + \alpha z + \beta = 0$, the sum of roots is $z_1 + z_2 = -\alpha$.
For $\alpha$ to not be real, $z_1 + z_2$ cannot be purely real.
If $z_2 = \overline{z_1}$ (reflection across the $x$-axis) or $z_2 = -z_1$ (reflection across the origin), $z_1 + z_2$ would be real.
Therefore, $z_2$ must be the reflection of $z_1$ across the $y$-axis, meaning $z_2 = -\overline{z_1}$.
The product of roots is $\beta = z_1 z_2 = z_1(-\overline{z_1}) = -|z_1|^2$.
Given $|z_1| = \frac{\sqrt{26}}{2}$, we have:
$\beta = -\left(\frac{\sqrt{26}}{2}\right)^2 = -\frac{26}{4} = -\frac{13}{2}$

2. Calculate the coordinates of $z_1$ to find $\alpha$.

Let $z_1 = x + \mathrm{i}y$. Since $z_1$ lies on the ellipse from part (a),
$\frac{x^2}{4} + \frac{y^2}{9} = 1 \implies 9x^2 + 4y^2 = 36$
We also know $x^2 + y^2 = |z_1|^2 = \frac{13}{2} \implies x^2 = \frac{13}{2} - y^2$.
Substituting into the ellipse equation:
$9\left(\frac{13}{2} - y^2\right) + 4y^2 = 36$
$\frac{117}{2} - 5y^2 = 36 \implies 5y^2 = \frac{117}{2} - \frac{72}{2} = \frac{45}{2} \implies y^2 = \frac{9}{2}$
Since $0 < \arg(z_1) < \frac{\pi}{2}$, $y > 0$, so $y = \frac{3}{\sqrt{2}} = \frac{3\sqrt{2}}{2}$.
With $z_2 = -x + \mathrm{i}y$, the sum of roots gives:
$\alpha = -(z_1 + z_2) = -(x + \mathrm{i}y - x + \mathrm{i}y) = -2y\mathrm{i} = -3\sqrt{2}\mathrm{i}$