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

1. Express the real and imaginary parts of the complex number as parametric equations.

Let $x = \mathrm{Re}(z)$ and $y = \mathrm{Im}(z)$.
From $z = 2\cos t + i(3\sin t)$, we have:
$x = 2\cos t$
$y = 3\sin t$

2. Eliminate the parameter $t$ to obtain the Cartesian equation.

Rearranging the equations gives:
$\cos t = \frac{x}{2} \quad \text{and} \quad \sin t = \frac{y}{3}$
Using the trigonometric identity $\cos^2 t + \sin^2 t = 1$:
$\left(\frac{x}{2}\right)^2 + \left(\frac{y}{3}\right)^2 = 1$
$\frac{x^2}{4} + \frac{y^2}{9} = 1$
Since $0 \le t < 2\pi$, this draws a complete closed curve.

3. Describe the sketch of the locus on the Argand diagram.

The Cartesian equation $\frac{x^2}{4} + \frac{y^2}{9} = 1$ represents an ellipse centered at the origin.

Sketch requirements:

• Draw a set of axes, labelling the horizontal axis as $\mathrm{Re}$ and the vertical axis as $\mathrm{Im}$.
• Draw a complete, single ellipse centred at $(0, 0)$.
• Mark the intersections on the $\mathrm{Re}$-axis at $2$ and $-2$.
• Mark the intersections on the $\mathrm{Im}$-axis at $3$ and $-3$.

1. Determine the possible geometric relationships between $z_1$ and $z_2$ based on the curve's symmetry.

From part (a), the locus of $z$ is an ellipse centred at the origin, represented by $\frac{x^2}{4} + \frac{y^2}{9} = 1$.
For $|z_1| = |z_2|$, the points representing the complex numbers must be equidistant from the origin. Visually, a circle centred at the origin intersects an ellipse at up to 4 points which are symmetric about the axes.
Since $0 < \arg(z_1) < \frac{\pi}{2}$, $z_1$ lies in the first quadrant.
Because $z_1$ and $z_2$ are distinct, $z_2$ must be a reflection of $z_1$ about the real axis, the imaginary axis, or the origin.

2. Evaluate $\arg(z_1 + z_2)$ for each valid position of $z_2$.

Case 1: $z_2$ is the reflection of $z_1$ about the real axis ($z_2 = \bar{z}_1$).
By the parallelogram law of vector addition, $z_1 + z_2$ is a vector strictly along the positive real axis.
$\therefore \arg(z_1 + z_2) = 0$

Case 2: $z_2$ is the reflection of $z_1$ about the imaginary axis ($z_2 = -\bar{z}_1$).
$z_1 + z_2$ is a vector strictly along the positive imaginary axis.
$\therefore \arg(z_1 + z_2) = \frac{\pi}{2}$

Case 3: $z_2$ is the reflection of $z_1$ about the origin ($z_2 = -z_1$).
$z_1 + z_2 = 0$, and the argument of $0$ is undefined.

Thus, the possible values are $0$ and $\frac{\pi}{2}$.

1. State the relationship between $a$ and the roots $z_1, z_2$.

Using the sum of roots for the given quadratic equation, we have:
$z_1 + z_2 = -a$

2. Provide a valid case where $a$ is not real to conclude it is not necessary.

As established in part (b), $z_2$ can be the reflection of $z_1$ across the imaginary axis, meaning $z_2 = -\overline{z_1}$. In this case:
$-a = z_1 - \overline{z_1} = 2i \operatorname{Im}(z_1)$
Since $0 < \arg(z_1) < \frac{\pi}{2}$, $\operatorname{Im}(z_1) > 0$. Thus, $-a$ is a non-zero purely imaginary number, which means $a$ is not real.
Therefore, it is not necessary for $a$ to be real.

1. Relate $a$ to the sum of roots and determine the correct $z_2$ using the given condition.

For the quadratic equation $z^2 + az + \beta = 0$, the sum of the roots is $z_1 + z_2 = -a$.
Since $a$ is not real, $z_1 + z_2$ must be a non-real complex number.
From the symmetry of the ellipse, the points with the same modulus $|z_2| = |z_1|$ correspond to $z_1$, $\bar{z_1}$, $-z_1$, and $-\bar{z_1}$.
If $z_2 = \bar{z_1}$, $z_1 + z_2 = 2\text{Re}(z_1)$ (real).
If $z_2 = -z_1$, $z_1 + z_2 = 0$ (real).
If $z_2 = -\bar{z_1}$, $z_1 + z_2 = 2i\text{Im}(z_1)$ (purely imaginary, not real).
Hence, for $a$ to be not real, we must have $z_2 = -\bar{z_1}$.

2. Evaluate $\beta$ using the product of the roots.

The product of the roots is $\beta = z_1 z_2$.
Substitute $z_2 = -\bar{z_1}$:
$\beta = z_1(-\bar{z_1}) = -|z_1|^2$
Given $|z_1| = \dfrac{\sqrt{26}}{2}$, we have:
$\beta = -\left(\dfrac{\sqrt{26}}{2}\right)^2 = -\dfrac{26}{4} = -6.5 = -\dfrac{13}{2}$

3. Use the cartesian equation from part (a) to find the imaginary part of $z_1$.

Let $z_1 = x_1 + iy_1$. From part (a), $z_1$ lies on the ellipse $\dfrac{x^2}{4} + \dfrac{y^2}{9} = 1$.
We are given $|z_1|^2 = x_1^2 + y_1^2 = \left(\dfrac{\sqrt{26}}{2}\right)^2 = \dfrac{26}{4} = 6.5$, so $x_1^2 = 6.5 - y_1^2$.
Substituting this into the ellipse equation:
$\dfrac{6.5 - y_1^2}{4} + \dfrac{y_1^2}{9} = 1$
$9(6.5 - y_1^2) + 4y_1^2 = 36 \implies 58.5 - 5y_1^2 = 36$
$5y_1^2 = 22.5 \implies y_1^2 = 4.5 = \dfrac{9}{2}$
Since $0 < \arg(z_1) < \dfrac{\pi}{2}$, $z_1$ lies in the first quadrant, so $y_1 > 0$.
Thus, $\text{Im}(z_1) = y_1 = \sqrt{\dfrac{9}{2}} = \dfrac{3\sqrt{2}}{2}$.

4. Evaluate $a$ using the sum of the roots.

Using the previously established relation $z_1 + z_2 = 2i\text{Im}(z_1)$:
$z_1 + z_2 = 2i\left(\dfrac{3\sqrt{2}}{2}\right) = 3\sqrt{2}i$
Since $a = -(z_1 + z_2)$, we have:
$a = -3\sqrt{2}i$