下方阿冈德图上的点 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 $w_2 = -w_1$ 且 $w_3 = i w_2$。 设 B、C 和 D 分别表示复数 $w_2$、$w_3$ 和 $w_2 + w_3$。

1. Determine the position, modulus, and argument for point B representing $w_2$.

Given $w_2 = -w_1$.
Multiplying a complex number by $-1$ is equivalent to a $180^\circ$ ($\pi$ radians) rotation about the origin.
$|w_2| = |-w_1| = |-1| |w_1| = r$
$\arg(w_2) = \arg(-1) + \arg(w_1) = \pi + \theta$
Plot point B such that O is the midpoint of AB. Label the length of OB as $r$ and the angle from the positive real axis as $\theta + \pi$ (or $\theta - \pi$).

2. Determine the position, modulus, and argument for point C representing $w_3$.

Given $w_3 = iw_2$.
Multiplying by $i$ rotates a complex number by $90^\circ$ ($\frac{\pi}{2}$ radians) anti-clockwise about the origin.
$|w_3| = |i||w_2| = (1)(r) = r$
$\arg(w_3) = \arg(i) + \arg(w_2) = \frac{\pi}{2} + (\theta + \pi) = \theta + \frac{3\pi}{2} \text{ (or } \theta - \frac{\pi}{2})$
Plot point C by rotating OB anti-clockwise by $90^\circ$. Label the length of OC as $r$ and the angle from the positive real axis as $\theta + \frac{3\pi}{2}$ (or $\theta - \frac{\pi}{2}$).

3. Determine the position of point D representing the sum $w_2 + w_3$.

Addition of complex numbers follows the parallelogram law of vector addition.
Since $\overrightarrow{OD} = \overrightarrow{OB} + \overrightarrow{OC}$, point D forms a parallelogram $OBDC$ with OB and OC.
As $|w_2| = |w_3| = r$ and they are perpendicular, $OBDC$ is specifically a square of side $r$. Plot D at the fourth vertex of this square.

1. State the angle BOC based on the geometrical effect of multiplying by $i$

Since $w_3 = i w_2$, the position vector $\overrightarrow{OC}$ is obtained by rotating the position vector $\overrightarrow{OB}$ counter-clockwise by $\frac{\pi}{2}$ radians about the origin. Thus, the angle between the two vectors is $\frac{\pi}{2}$.

$\therefore \angle BOC = \frac{\pi}{2}$

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

Since $w_3 = iw_2$, multiplying by $i$ represents a $\frac{\pi}{2}$ anticlockwise rotation. Thus, $OC = OB = r$ and $\angle BOC = \frac{\pi}{2}$.
Since $D$ represents $w_2 + w_3$, $\overrightarrow{OD} = \overrightarrow{OB} + \overrightarrow{OC}$.
Therefore, the quadrilateral OBDC is a parallelogram. Given adjacent equal sides and a $90^\circ$ included angle, OBDC is a square.
Using Pythagoras' theorem on $\triangle OBD$:
$|w_2 + w_3| = OD = \sqrt{r^2 + r^2} = r\sqrt{2}$

2. Find the argument of $w_2 + w_3$ by considering the angle of the diagonal OD.

The diagonal OD bisects $\angle BOC$, so the angle between OB and OD is $\frac{\pi}{4}$ radians anticlockwise.
The point B represents $w_2 = -w_1$, so its argument is diametrically opposite to $w_1$:
$\arg(w_2) = \theta + \pi$ (or $\theta - \pi$)
The argument of $w_2 + w_3$ is the angle of OD from the positive real axis:
$\arg(w_2 + w_3) = \arg(w_2) + \frac{\pi}{4} = (\theta + \pi) + \frac{\pi}{4} = \theta + \frac{5\pi}{4}$
Subtracting $2\pi$ to give a standard principal value form, this can also be written as $\theta - \frac{3\pi}{4}$.