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： $z = \frac{1 + i}{1 - i}$ 且 $w = \frac{\sqrt{2}}{1 + i}$。 将 $z$ 和 $w$ 表示为精确的极坐标形式 $re^{i\theta}$，其中 $r > 0$ 且 $-\pi < \theta \le \pi$。 在阿甘特图上，分别标出代表 $z$、$w$ 和 $z + w$ 的点 A、B 和 C。指出四边形 OACB 的形状。 由此，通过求 $z + w$ 的辐角，证明 $\tan \frac{\pi}{8} = \sqrt{2} - 1$。

1. Express $z$ in the Cartesian form $x + iy$ by multiplying by the complex conjugate.

$z = \frac{\lambda - 4i}{1 - \lambda i}$
Multiply the numerator and denominator by the complex conjugate of the denominator, $1 + \lambda i$:
$z = \frac{(\lambda - 4i)(1 + \lambda i)}{(1 - \lambda i)(1 + \lambda i)}$
$z = \frac{\lambda + \lambda^2 i - 4i - 4\lambda i^2}{1^2 - (\lambda i)^2}$
Since $i^2 = -1$:
$z = \frac{\lambda + 4\lambda + (\lambda^2 - 4)i}{1 + \lambda^2}$
$z = \frac{5\lambda}{1 + \lambda^2} + \frac{\lambda^2 - 4}{1 + \lambda^2}i$

2. State the condition for $z$ to have an argument of $\pi$.

For a complex number to have an argument of $\pi$, it must lie strictly on the negative real axis.
Therefore, its imaginary part must be zero and its real part must be strictly negative:
$\text{Im}(z) = 0 \quad \text{and} \quad \text{Re}(z) < 0$

3. Solve for $\lambda$ by setting the imaginary part to zero.

$\frac{\lambda^2 - 4}{1 + \lambda^2} = 0$
Since $1 + \lambda^2 > 0$ for all real $\lambda$:
$\lambda^2 - 4 = 0$
$\lambda = 2 \quad \text{or} \quad \lambda = -2$

4. Apply the real part condition to find the valid $\lambda$ and evaluate $z$.

We require $\text{Re}(z) < 0$:
$\frac{5\lambda}{1 + \lambda^2} < 0$
Since the denominator $1 + \lambda^2 > 0$, we must have $5\lambda < 0$, which means $\lambda < 0$.
Thus, we reject $\lambda = 2$ and accept $\lambda = -2$.
Substitute $\lambda = -2$ into the real part to find $z$:
$z = \frac{5(-2)}{1 + (-2)^2} + 0i = \frac{-10}{5} = -2$

1. Rationalise the denominator of $z$ to find its Cartesian form, then convert it to polar form.

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

The modulus is $|z| = 1$ and the argument is $\arg(z) = \frac{\pi}{2}$.
Therefore, $z = e^{i\frac{\pi}{2}}$.

2. Rationalise the denominator of $w$ to find its Cartesian form, then convert it to polar form.

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

The modulus is $|w| = \sqrt{\left(\frac{1}{\sqrt{2}}\right)^2 + \left(-\frac{1}{\sqrt{2}}\right)^2} = \sqrt{\frac{1}{2} + \frac{1}{2}} = 1$.

Since $\operatorname{Re}(w) > 0$ and $\operatorname{Im}(w) < 0$, the complex number lies in the fourth quadrant.
$\arg(w) = -\tan^{-1}\left(\frac{1/\sqrt{2}}{1/\sqrt{2}}\right) = -\frac{\pi}{4}$.
Therefore, $w = e^{-i\frac{\pi}{4}}$.

1. Determine the Cartesian coordinates of points A, B, and C to mark on the Argand diagram.

From part (b)(i), we know $|z| = 1$ and $\arg(z) = \frac{\pi}{2}$, so $z = i$.
We also have $|w| = 1$ and $\arg(w) = -\frac{\pi}{4}$, so $w = \frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2}i$.
The sum is $z + w = \frac{\sqrt{2}}{2} + \left(1 - \frac{\sqrt{2}}{2}\right)i$.
On an Argand diagram with Real and Imaginary axes, we mark:

• Point A at $(0, 1)$
• Point B at $\left(\frac{\sqrt{2}}{2}, -\frac{\sqrt{2}}{2}\right)$
• Point C at $\left(\frac{\sqrt{2}}{2}, 1 - \frac{\sqrt{2}}{2}\right)$

2. Deduce the specific shape of quadrilateral OACB.

In the complex plane, the addition $C = z + w$ means that the quadrilateral OACB forms a parallelogram.
Furthermore, from part (b)(i), we have lengths $OA = |z| = 1$ and $OB = |w| = 1$.
Since the adjacent sides of this parallelogram are equal in length ($OA = OB$), the quadrilateral OACB is a rhombus.

1. Express $z+w$ in Cartesian form using the simplified forms of $z$ and $w$.

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

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

$z+w = \frac{\sqrt{2}}{2} + \left(1 - \frac{\sqrt{2}}{2}\right)i$

2. Deduce $\arg(z+w)$ using the geometric properties of the Argand diagram from part (b)(ii).

From part (b)(ii), quadrilateral $OACB$ is a rhombus since $|z| = 1$ and $|w| = 1$.
Therefore, the diagonal $OC$ (which represents $z+w$) exactly bisects the angle $AOB$ at the origin.

$\arg(z+w) = \frac{\arg(z) + \arg(w)}{2}$

Using the arguments found in (b)(i), $\arg(z) = \frac{\pi}{2}$ and $\arg(w) = -\frac{\pi}{4}$:

$\arg(z+w) = \frac{\frac{\pi}{2} + \left(-\frac{\pi}{4}\right)}{2} = \frac{\frac{\pi}{4}}{2} = \frac{\pi}{8}$

3. Evaluate $\tan(\arg(z+w))$ using the Cartesian form to complete the proof.

For a complex number in the first quadrant, $\tan(\arg(z+w)) = \frac{\text{Im}(z+w)}{\text{Re}(z+w)}$.

$\tan\left(\frac{\pi}{8}\right) = \frac{1 - \frac{\sqrt{2}}{2}}{\frac{\sqrt{2}}{2}}$

$\tan\left(\frac{\pi}{8}\right) = \frac{\frac{2 - \sqrt{2}}{2}}{\frac{\sqrt{2}}{2}} = \frac{2 - \sqrt{2}}{\sqrt{2}}$

$\tan\left(\frac{\pi}{8}\right) = \frac{2}{\sqrt{2}} - 1 = \sqrt{2} - 1$ (shown)