已知 $z = \dfrac{\lambda - 4i}{1 - \lambda i}$，$\lambda \in \mathbb{R}$ 且 $\arg(z) = \pi$，求 $z$ 的值。 解答本题时请勿使用计算器。 复数 $z$ 和 $w$ 由下式给出： $z = \dfrac{1 + i}{1 - i}$ 且 $w = \dfrac{\sqrt{2}}{1 + i}$。 将 $z$ 和 $w$ 表示为精确的极坐标形式 $re^{i\theta}$，其中 $r > 0$ 且 $-\pi < \theta \le \pi$。

1. Express $z$ in Cartesian form by multiplying the numerator and denominator by the conjugate of the denominator.

$z = \frac{\lambda - 4i}{1 - \lambda i} \times \frac{1 + \lambda i}{1 + \lambda i}$
$z = \frac{\lambda + \lambda^2 i - 4i - 4\lambda i^2}{1^2 + \lambda^2}$
$z = \frac{5\lambda + (\lambda^2 - 4)i}{1 + \lambda^2}$

2. Identify the conditions for the real and imaginary parts given that the argument of $z$ is $\pi$.

Since $\arg(z) = \pi$, $z$ must lie on the negative real axis.
Therefore, the imaginary part must be $0$ and the real part must be strictly negative ($\text{Im}(z) = 0$, $\text{Re}(z) < 0$).

3. Equate the imaginary part to zero to find the possible values of $\lambda$.

$\frac{\lambda^2 - 4}{1 + \lambda^2} = 0$
$\lambda^2 - 4 = 0$
$\lambda = 2 \quad \text{or} \quad \lambda = -2$

4. Test the possible values of $\lambda$ against the real part condition to find the correct value of $z$.

When $\lambda = 2$:
$\text{Re}(z) = \frac{5(2)}{1 + 2^2} = \frac{10}{5} = 2$
Since $\text{Re}(z) > 0$, $\arg(z) = 0 \neq \pi$, so we reject $\lambda = 2$.

When $\lambda = -2$:
$\text{Re}(z) = \frac{5(-2)}{1 + (-2)^2} = \frac{-10}{5} = -2$
Since $\text{Re}(z) < 0$ and $\text{Im}(z) = 0$, $\arg(z) = \pi$.
Thus, the correct value is $z = -2$.

1. Simplify $z$ into Cartesian form and convert it to polar form.

$z = \frac{1 + i}{1 - i} \times \frac{1 + i}{1 + i}$
$z = \frac{1 + 2i + i^2}{1^2 + 1^2} = \frac{1 + 2i - 1}{2} = \frac{2i}{2} = i$
For $z = 0 + 1i$:
Modulus $|z| = \sqrt{0^2 + 1^2} = 1$
Argument $\arg(z) = \frac{\pi}{2}$ (since it lies on the positive imaginary axis)
Therefore, in polar form, $z = 1e^{i\frac{\pi}{2}} = e^{i\frac{\pi}{2}}$.

2. Simplify $w$ into Cartesian form and convert it to polar form.

$w = \frac{\sqrt{2}}{1 + i} \times \frac{1 - i}{1 - i}$
$w = \frac{\sqrt{2}(1 - i)}{1^2 + 1^2} = \frac{\sqrt{2}}{2} - i\frac{\sqrt{2}}{2}$
For $w$:
Modulus $|w| = \sqrt{\left(\frac{\sqrt{2}}{2}\right)^2 + \left(-\frac{\sqrt{2}}{2}\right)^2} = \sqrt{\frac{2}{4} + \frac{2}{4}} = \sqrt{1} = 1$
Argument $\arg(w) = \tan^{-1}\left(\frac{-\sqrt{2}/2}{\sqrt{2}/2}\right) = \tan^{-1}(-1) = -\frac{\pi}{4}$ (since it lies in the 4th quadrant)
Therefore, in polar form, $w = 1e^{-i\frac{\pi}{4}} = e^{-i\frac{\pi}{4}}$.