复数 $a$ 和 $b$ 由 $a = -(1 + \sqrt{3}\text{i})$ and $b = \frac{1}{2}(1 - \text{i})$ 给出。

1. Calculate the value of $a^2$.

$a^2 = \left( -(1 + \sqrt{3}\text{i}) \right)^2$
$= (1 + \sqrt{3}\text{i})^2$
$= 1^2 + 2(\sqrt{3}\text{i}) + (\sqrt{3}\text{i})^2$
$= 1 + 2\sqrt{3}\text{i} - 3$
$= -2 + 2\sqrt{3}\text{i}$

2. Multiply $a^2$ by $b$ and simplify into the form $x + \text{i}y$.

$a^2b = (-2 + 2\sqrt{3}\text{i}) \left[ \frac{1}{2}(1 - \text{i}) \right]$
$= (-1 + \sqrt{3}\text{i})(1 - \text{i})$
$= -1 + \text{i} + \sqrt{3}\text{i} - \sqrt{3}\text{i}^2$
$= -1 + \text{i} + \sqrt{3}\text{i} - \sqrt{3}(-1)$
$= (\sqrt{3} - 1) + (\sqrt{3} + 1)\text{i}$

1. Find the modulus and argument of $a$.

$|a| = \sqrt{(-1)^2 + (-\sqrt{3})^2} = 2$

Since $a = -1 - \sqrt{3}\text{i}$ is in the third quadrant:
$\arg(a) = -\pi + \tan^{-1}\left(\frac{\sqrt{3}}{1}\right) = -\pi + \frac{\pi}{3} = -\frac{2\pi}{3}$

2. Find the modulus and argument of $b$.

$|b| = \sqrt{\left(\frac{1}{2}\right)^2 + \left(-\frac{1}{2}\right)^2} = \sqrt{\frac{1}{4} + \frac{1}{4}} = \frac{1}{\sqrt{2}}$

Since $b = \frac{1}{2} - \frac{1}{2}\text{i}$ is in the fourth quadrant:
$\arg(b) = \tan^{-1}\left(\frac{-1/2}{1/2}\right) = \tan^{-1}(-1) = -\frac{\pi}{4}$

3. Calculate the modulus of $a^2b$.

Using the property $|z_1 z_2| = |z_1| |z_2|$ and $|z^n| = |z|^n$:
$|a^2b| = |a|^2|b| = 2^2 \times \frac{1}{\sqrt{2}} = \frac{4}{\sqrt{2}} = 2\sqrt{2}$

4. Calculate the argument of $a^2b$.

Using the property $\arg(z_1 z_2) = \arg(z_1) + \arg(z_2)$ and $\arg(z^n) = n\arg(z)$:
$\arg(a^2b) = 2\arg(a) + \arg(b)$
$= 2\left(-\frac{2\pi}{3}\right) + \left(-\frac{\pi}{4}\right)$
$= -\frac{4\pi}{3} - \frac{\pi}{4}$
$= -\frac{16\pi}{12} - \frac{3\pi}{12} = -\frac{19\pi}{12}$

Adjusting to the principal argument range $(-\pi, \pi]$:
$\arg(a^2b) = -\frac{19\pi}{12} + 2\pi = \frac{5\pi}{12}$

1. Concisely evaluate $a^2b$ in both Cartesian form (from part i) and modulus-argument form (from part ii) as they are required to solve this part.

From part (i):
$a^2b = \left( -(1 + \sqrt{3}\text{i}) \right)^2 \left( \frac{1}{2}(1 - \text{i}) \right)$
$= (-2 + 2\sqrt{3}\text{i}) \left( \frac{1}{2} - \frac{1}{2}\text{i} \right)$
$= -1 + \text{i} + \sqrt{3}\text{i} + \sqrt{3}$
$= (\sqrt{3} - 1) + (\sqrt{3} + 1)\text{i}$

From part (ii):
$|a| = 2$, $\arg(a) = -\frac{2\pi}{3}$
$|b| = \frac{1}{\sqrt{2}}$, $\arg(b) = -\frac{\pi}{4}$
$|a^2b| = |a|^2 |b| = 2^2 \left( \frac{1}{\sqrt{2}} \right) = 2\sqrt{2}$
$\arg(a^2b) = 2\arg(a) + \arg(b) = 2\left(-\frac{2\pi}{3}\right) - \frac{\pi}{4} = -\frac{19\pi}{12}$
Principal argument $= -\frac{19\pi}{12} + 2\pi = \frac{5\pi}{12}$

2. Express $a^2b$ using its modulus and principal argument, and equate the imaginary parts.

Using the modulus and argument, $a^2b$ can be written in trigonometric form:
$a^2b = 2\sqrt{2}\left(\cos\frac{5\pi}{12} + \text{i}\sin\frac{5\pi}{12}\right)$

Equating the imaginary part of this polar form to the imaginary part of the Cartesian form from part (i):
$2\sqrt{2}\sin\frac{5\pi}{12} = \sqrt{3} + 1$

Dividing both sides by $2\sqrt{2}$:
$\sin\frac{5\pi}{12} = \frac{\sqrt{3} + 1}{2\sqrt{2}}$

1. Set up an equation relating $a$, $b$, and $c$ using the geometric properties of the isosceles right-angled triangle.

Since $\triangle ABC$ is an isosceles right-angled triangle with $AB = BC$ and a right angle at $B$, the vector $\overrightarrow{BC}$ is obtained by rotating the vector $\overrightarrow{BA}$ by exactly $90^\circ$ (or $\frac{\pi}{2}$ radians) either clockwise or anticlockwise about point $B$.

In the Argand diagram, multiplying a complex number by $\text{i}$ results in an anticlockwise rotation of $90^\circ$, and multiplying by $-\text{i}$ results in a clockwise rotation. Therefore, we can write the relationship as:
$c - b = \pm \text{i}(a - b)$

2. Evaluate the two possible complex numbers for $c$ by substituting the given values of $a$ and $b$.

Substitute $a = -1 - \sqrt{3}\text{i}$ and $b = \frac{1}{2}(1 - \text{i})$ into the rotation equations:

Case 1: Anticlockwise rotation ($+\text{i}$)
$c - b = \text{i}(a - b) \implies c = b(1 - \text{i}) + a\text{i}$
$c = \frac{1}{2}(1 - \text{i})^2 + \text{i}(-1 - \sqrt{3}\text{i})$
$c = \frac{1}{2}(1 - 2\text{i} - 1) - \text{i} - \sqrt{3}(\text{i}^2)$
$c = -\text{i} - \text{i} + \sqrt{3} = \sqrt{3} - 2\text{i}$

Case 2: Clockwise rotation ($-\text{i}$)
$c - b = -\text{i}(a - b) \implies c = b(1 + \text{i}) - a\text{i}$
$c = \frac{1}{2}(1 - \text{i})(1 + \text{i}) - \text{i}(-1 - \sqrt{3}\text{i})$
$c = \frac{1}{2}(1^2 - \text{i}^2) + \text{i} + \sqrt{3}(\text{i}^2)$
$c = \frac{1}{2}(2) + \text{i} - \sqrt{3} = 1 - \sqrt{3} + \text{i}$

3. Determine the correct value for $c$ using the quadrant placement shown in the given diagram.

The diagram indicates that point $C$ lies in the second quadrant, meaning its real part must be negative and its imaginary part positive.

For Case 1, $c = \sqrt{3} - 2\text{i}$ has a positive real part ($\approx 1.732$) and a negative imaginary part ($-2$), which would place it in the fourth quadrant. Thus, we reject this case.

For Case 2, $c = 1 - \sqrt{3} + \text{i}$ has a negative real part ($1 - \sqrt{3} \approx -0.732$) and a positive imaginary part ($1$), correctly placing it in the second quadrant.

Therefore, the exact value is:
$c = 1 - \sqrt{3} + \text{i}$