下方阿甘图（Argand diagram）上的点 $A$ 表示模为 $r$、辐角为 $\theta$ 的复数 $w_1$。已知复数 $w_2$ 和 $w_3$ 满足方程
$w_2 = -w_1 \text{ and } w_3 = \text{i}w_2.$
令 $B$、$C$ 和 $D$ 分别表示复数 $w_2$、$w_3$ 和 $w_2 + w_3$。

1. Determine the modulus and argument of $w_2$ and plot point $B$.

Given $w_2 = -w_1$, the point $B$ is a rotation of $A$ by $\pi$ radians about the origin (equivalent to a reflection through the origin).
$|w_2| = |-w_1| = |w_1| = r$
$\arg(w_2) = \arg(w_1) - \pi = \theta - \pi$
Plot $B$ in the third quadrant such that $AOB$ is a straight line. On the diagram, label the length $OB$ as $r$ and draw an angle arc from the positive real axis to $OB$ labelled $\theta - \pi$.

2. Determine the modulus and argument of $w_3$ and plot point $C$.

Given $w_3 = \text{i}w_2$, multiplying by $\text{i}$ scales the modulus by $|\text{i}|=1$ and adds $\frac{\pi}{2}$ to the argument (an anti-clockwise rotation by $90^\circ$).
$|w_3| = |\text{i}| |w_2| = 1 \times r = r$
$\arg(w_3) = \arg(w_2) + \frac{\pi}{2} = \theta - \pi + \frac{\pi}{2} = \theta - \frac{\pi}{2}$
Plot $C$ in the fourth quadrant such that $\angle BOC = 90^\circ$ anti-clockwise. Label the length $OC$ as $r$ and draw an angle arc from the positive real axis to $OC$ labelled $\theta - \frac{\pi}{2}$.

3. Plot point $D$ representing $w_2 + w_3$.

Point $D$ represents the sum $w_2 + w_3$. By the parallelogram law of vector addition on the Argand diagram, $D$ is the fourth vertex of the parallelogram $OBDC$. Since adjacent sides $OB$ and $OC$ are equal in length ($r$) and perpendicular ($\angle BOC = 90^\circ$), complete the construction such that $OBDC$ forms a square.

1. Determine the angle $BOC$ using the geometric effect of multiplying by $\text{i}$

Since $w_3 = \text{i}w_2$, the complex number $w_3$ is obtained by rotating $w_2$ by $\frac{\pi}{2}$ radians anti-clockwise about the origin.
Therefore, the angle between the line segments $OB$ and $OC$ is $\frac{\pi}{2}$.
Angle $BOC = \frac{\pi}{2}$

1. Identify the shape of the quadrilateral $OBDC$ to find the modulus $|w_2 + w_3|$.

The length $OB = |w_2| = |-w_1| = r$.
The length $OC = |w_3| = |\text{i}w_2| = r$.
Since $w_3 = \text{i}w_2$, $C$ is obtained by rotating $B$ counter-clockwise by $\frac{\pi}{2}$ radians about the origin, so $\angle BOC = \frac{\pi}{2}$.
The point $D$ represents $w_2 + w_3$, making $OBDC$ a parallelogram. Since adjacent sides $OB$ and $OC$ are equal and perpendicular, $OBDC$ is a square of side length $r$.
The distance $OD = |w_2 + w_3|$ is the diagonal of the square:
$|w_2 + w_3| = \sqrt{r^2 + r^2} = r\sqrt{2}$

2. Use the geometric properties of the square to find $\arg(w_2 + w_3)$.

Since $w_2 = -w_1$ and $w_1$ is in the first quadrant with argument $\theta$, $w_2$ is in the third quadrant. Its principal argument is:
$\arg(w_2) = \theta - \pi$
Since $OBDC$ is a square, the diagonal $OD$ bisects $\angle BOC$.
Thus, $OD$ is rotated counter-clockwise from $OB$ by exactly $\frac{\pi}{4}$ radians.
$\arg(w_2 + w_3) = \arg(w_2) + \frac{\pi}{4}$
$\arg(w_2 + w_3) = (\theta - \pi) + \frac{\pi}{4} = \theta - \frac{3\pi}{4}$