已知 $z=\frac{\lambda-4i}{1-\lambda i}$，其中 $\lambda\in\mathbb{R}$ 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： $\displaystyle z=\frac{1+i}{1-i}\quad\text{和}\quad w=\frac{\sqrt{2}}{1+i}.$ 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 $a + ib$ by multiplying the numerator and denominator by the complex conjugate of the denominator.

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

2. Interpret the given argument geometrically to establish conditions for the real and imaginary parts.

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

3. Solve for $\lambda$ by equating the imaginary part to zero, then verify against the real part to find the value of $z$.

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

Checking the real part:
If $\lambda = 2$, $z = \frac{5(2)}{1 + 2^2} = 2 > 0$ (rejected, as $\arg(2) = 0$).
If $\lambda = -2$, $z = \frac{5(-2)}{1 + (-2)^2} = -2 < 0$ (accepted).

1. Express $z$ in Cartesian form by rationalising the denominator

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

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

$|z| = |i| = 1$
$\arg(z) = \arg(i) = \frac{\pi}{2}$
$\therefore z = e^{i\frac{\pi}{2}}$

3. Express $w$ in Cartesian form by rationalising the denominator

$w = \frac{\sqrt{2}}{1+i} \times \frac{1-i}{1-i}$
$w = \frac{\sqrt{2}(1-i)}{1^2 - i^2}$
$w = \frac{\sqrt{2}(1-i)}{2} = \frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2}i$

4. Find the modulus and argument of $w$ to express it in polar form

$|w| = \sqrt{\left(\frac{\sqrt{2}}{2}\right)^2 + \left(-\frac{\sqrt{2}}{2}\right)^2} = \sqrt{\frac{2}{4} + \frac{2}{4}} = 1$
$\arg(w) = \tan^{-1}\left(\frac{-\frac{\sqrt{2}}{2}}{\frac{\sqrt{2}}{2}}\right) = \tan^{-1}(-1) = -\frac{\pi}{4}$
$\therefore w = e^{-i\frac{\pi}{4}}$