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 $z$ as parametric equations and eliminate $t$ to find the cartesian equation.

$x = \operatorname{Re}(z) = 2\cos t \implies \cos t = \frac{x}{2}$
$y = \operatorname{Im}(z) = 3\sin t \implies \sin t = \frac{y}{3}$

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

2. Sketch the locus of $z$ on an Argand diagram based on the cartesian equation.

The equation $\frac{x^2}{4} + \frac{y^2}{9} = 1$ describes an ellipse centered at the origin.
On an Argand diagram:

• The intercepts on the real axis ($\operatorname{Re}$) are at $2$ and $-2$.
• The intercepts on the imaginary axis ($\operatorname{Im}$) are at $3$ and $-3$.

As $0 \le t < 2\pi$, the locus traces out the entire ellipse.
(A continuous elliptical curve is drawn passing through $(2,0)$, $(0,3)$, $(-2,0)$, and $(0,-3)$ on the Argand plane.)

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

From part (a), the curve is an ellipse centered at the origin, symmetric about the real and imaginary axes.
The condition $|z_1| = |z_2|$ implies that $z_1$ and $z_2$ are equidistant from the origin, lying on the intersection of a circle and the ellipse.
Since $0 < \arg(z_1) < \frac{\pi}{2}$, $z_1$ lies strictly in the first quadrant. By symmetry, for a distinct point $z_2$, the possible positions are:

1. Reflection across the real axis (4th quadrant): $z_2 = \bar{z}_1$
2. Reflection across the imaginary axis (2nd quadrant): $z_2 = -\bar{z}_1$
3. Reflection through the origin (3rd quadrant): $z_2 = -z_1$

2. Determine $\arg(z_1 + z_2)$ for the reflection across the real axis.

If $z_2 = \bar{z}_1$:
$z_1 + z_2 = z_1 + \bar{z}_1 = 2\operatorname{Re}(z_1)$
Since $z_1$ is in the first quadrant, $\operatorname{Re}(z_1) > 0$.
The sum lies on the positive real axis, so $\arg(z_1 + z_2) = 0$.

3. Determine $\arg(z_1 + z_2)$ for the reflection across the imaginary axis and origin.

If $z_2 = -\bar{z}_1$:
$z_1 + z_2 = z_1 - \bar{z}_1 = 2i\operatorname{Im}(z_1)$
Since $z_1$ is in the first quadrant, $\operatorname{Im}(z_1) > 0$.
The sum lies on the positive imaginary axis, so $\arg(z_1 + z_2) = \frac{\pi}{2}$.

If $z_2 = -z_1$:
$z_1 + z_2 = 0$
The argument of $0$ is undefined.

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

1. Relate the coefficient $a$ to the sum of the roots.

By the sum of roots for the quadratic equation $z^2 + az + \beta = 0$, we have:
$z_1 + z_2 = -a$

2. Use the possible values of $\arg(z_1+z_2)$ from part (b) to determine if $a$ must be real.

From part (b), one possible configuration (when $z_2 = -\bar{z}_1$) gives $\arg(z_1 + z_2) = \frac{\pi}{2}$, which means $z_1 + z_2$ can be a purely imaginary number.
If $z_1 + z_2$ is purely imaginary, then $a = -(z_1 + z_2)$ is also purely imaginary (i.e. not real).
Hence, it is not necessary for $a$ to be real.

1. Use the given modulus to find the exact value of $z_1$.

Given $|z_1| = \frac{\sqrt{26}}{2}$, we have $|z_1|^2 = \frac{26}{4} = \frac{13}{2}$.
Substituting $z_1 = 2\cos t_1 + i(3\sin t_1)$ into the modulus formula:
$|z_1|^2 = (2\cos t_1)^2 + (3\sin t_1)^2 = 4\cos^2 t_1 + 9\sin^2 t_1$
$4(1-\sin^2 t_1) + 9\sin^2 t_1 = \frac{13}{2}$
$4 + 5\sin^2 t_1 = 6.5 \implies 5\sin^2 t_1 = 2.5 \implies \sin^2 t_1 = \frac{1}{2}$
Since $0 < \arg(z_1) < \frac{\pi}{2}$ from part (b), $z_1$ must lie in the first quadrant, so $\sin t_1 = \frac{1}{\sqrt{2}}$ and $\cos t_1 = \frac{1}{\sqrt{2}}$.
Thus, $z_1 = 2\left(\frac{1}{\sqrt{2}}\right) + i\left(3\left(\frac{1}{\sqrt{2}}\right)\right) = \sqrt{2} + \frac{3\sqrt{2}}{2}i$.

2. Determine $z_2$ using the condition that $a$ is not real.

The equation $z^2 + az + \beta = 0$ gives the sum of roots as $z_1 + z_2 = -a$.
Since $a$ is not real, $z_1 + z_2$ must have a non-zero imaginary component (it cannot be purely real).
The other points on the curve with the same modulus of $\frac{\sqrt{26}}{2}$ are $z = \pm \sqrt{2} \pm \frac{3\sqrt{2}}{2}i$.
Testing the possible options for $z_2$:

• If $z_2 = \sqrt{2} - \frac{3\sqrt{2}}{2}i$ (the conjugate), $z_1 + z_2 = 2\sqrt{2}$ (real).
• If $z_2 = -\sqrt{2} - \frac{3\sqrt{2}}{2}i$ (the negative), $z_1 + z_2 = 0$ (real).
• If $z_2 = -\sqrt{2} + \frac{3\sqrt{2}}{2}i$ (reflection across $y$-axis), $z_1 + z_2 = 3\sqrt{2}i$ (imaginary, which satisfies $a$ is not real).
  Hence, $z_2$ must be $-\sqrt{2} + \frac{3\sqrt{2}}{2}i$.

3. Calculate the values of $a$ and $\beta$.

From the sum of the roots:
$-a = z_1 + z_2 = 3\sqrt{2}i \implies a = -3\sqrt{2}i$

From the product of the roots, $\beta = z_1 z_2$:
$\beta = \left(\sqrt{2} + \frac{3\sqrt{2}}{2}i\right)\left(-\sqrt{2} + \frac{3\sqrt{2}}{2}i\right)$
$\beta = \left(\frac{3\sqrt{2}}{2}i\right)^2 - (\sqrt{2})^2 = -\frac{18}{4} - 2 = -4.5 - 2 = -6.5 = -\frac{13}{2}$.