复数 $z$ 满足方程 $\sqrt{2}\,z^{3} = 1 + \dfrac{2}{2a - 1}\,i ,$ 其中 $a$ 是实数。 已知 $\arg\bigl(z^{3}\bigr) = -\dfrac{\pi}{4}$。

1. Isolate $z^3$ and equate the ratio of its imaginary to real part to $\tan\left(-\frac{\pi}{4}\right)$.

$z^3 = \frac{1}{\sqrt{2}} + \frac{2}{\sqrt{2}(2a - 1)}i$
Given $\arg(z^3) = -\frac{\pi}{4}$, and since the real part $\frac{1}{\sqrt{2}}$ is positive:
$\frac{\frac{2}{\sqrt{2}(2a - 1)}}{\frac{1}{\sqrt{2}}} = \tan\left(-\frac{\pi}{4}\right)$

2. Simplify and solve the resulting equation for $a$.

$\frac{2}{2a - 1} = -1$
$2 = -1(2a - 1)$
$2 = -2a + 1$
$2a = -1$
$a = -\frac{1}{2} \quad \text{(shown)}$

1. Substitute $a = -\frac{1}{2}$ into the equation to find $z^3$ in Cartesian form.

Substitute $a = -\frac{1}{2}$ into the equation:
$\sqrt{2}\,z^3 = 1 + \frac{2}{2(-\frac{1}{2}) - 1}\,i$
$\sqrt{2}\,z^3 = 1 + \frac{2}{-2}\,i$
$\sqrt{2}\,z^3 = 1 - i$
$z^3 = \frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}}i$

2. Find the modulus and argument of $z^3$ to express it in polar form.

Find the modulus of $z^3$:
$|z^3| = \sqrt{\left(\frac{1}{\sqrt{2}}\right)^2 + \left(-\frac{1}{\sqrt{2}}\right)^2} = \sqrt{\frac{1}{2} + \frac{1}{2}} = 1$
As given in the question (and verified by the 4th quadrant position), $\arg(z^3) = -\frac{\pi}{4}$.
Thus, $z^3 = 1\left(\cos\left(-\frac{\pi}{4}\right) + i\sin\left(-\frac{\pi}{4}\right)\right)$.

3. Use De Moivre's Theorem to find the roots $z$.

Let $z = r(\cos\theta + i\sin\theta)$. Then $z^3 = r^3(\cos 3\theta + i\sin 3\theta)$.
Comparing the modulus:
$r^3 = 1 \implies r = 1$
Comparing the argument:
$3\theta = -\frac{\pi}{4} + 2k\pi \quad \text{for } k \in \mathbb{Z}$
$\theta = -\frac{\pi}{12} + \frac{2k\pi}{3}$
To find $\theta$ in the principal range $-\pi < \theta \le \pi$, we substitute $k = 0, 1, -1$:
For $k = 0$, $\theta = -\frac{\pi}{12}$
For $k = 1$, $\theta = -\frac{\pi}{12} + \frac{2\pi}{3} = \frac{7\pi}{12}$
For $k = -1$, $\theta = -\frac{\pi}{12} - \frac{2\pi}{3} = -\frac{9\pi}{12} = -\frac{3\pi}{4}$

4. Express the three solutions in the required polar form.

The solutions in the form $r(\cos\theta + i\sin\theta)$ are:
$z_1 = 1\left(\cos\left(-\frac{\pi}{12}\right) + i\sin\left(-\frac{\pi}{12}\right)\right)$
$z_2 = 1\left(\cos\left(\frac{7\pi}{12}\right) + i\sin\left(\frac{7\pi}{12}\right)\right)$
$z_3 = 1\left(\cos\left(-\frac{3\pi}{4}\right) + i\sin\left(-\frac{3\pi}{4}\right)\right)$

1. Establish the relationship between $w$ and $z$ using the equation from part (i).

From part (i), with $a = -\dfrac{1}{2}$, the equation for $z$ simplifies to:
$\sqrt{2}\,z^{3} = 1 + \dfrac{2}{2(-1/2) - 1}\,i = 1 - i$

The equation for $w$ is given as:
$\sqrt{2}\left(\dfrac{w}{1 + i}\right)^{3} = 1 - i$

Comparing the two equations, the roots must satisfy:
$\dfrac{w}{1 + i} = z \implies w = (1 + i)z$

2. Determine the modulus and argument of the multiplier $1+i$ to deduce the geometric effects.

For the complex number $1 + i$:
$\text{Modulus } |1+i| = \sqrt{1^2 + 1^2} = \sqrt{2}$
$\text{Argument } \arg(1+i) = \tan^{-1}\left(\dfrac{1}{1}\right) = \dfrac{\pi}{4}$

Using the properties of multiplication for complex numbers in polar form, for $w = (1+i)z$:
$|w| = \sqrt{2}|z| \quad \text{and} \quad \arg(w) = \arg(z) + \dfrac{\pi}{4}$

3. State the corresponding series of geometric transformations.

The change in modulus and argument corresponds directly to standard geometric transformations on the Argand diagram.

Therefore, the point representing $w$ is obtained from the point representing $z$ by:

1. A rotation about the origin by $\dfrac{\pi}{4}$ radians anticlockwise.
2. An enlargement with centre the origin and scale factor $\sqrt{2}$.