复数 $a$ 的模为 $r$，辐角为 $\theta$，其中 $r \in \mathbb{R}^+$ 且 $0 < \theta < \frac{\pi}{2}$。复数 $b$ 满足 $b = -2ia$，复数 $c$ 为纯虚数且虚部为负。已知 $|b| = |c|$。 设点 $A, B, C$ 和 $D$ 分别表示复数 $a, b, c$ 和 $b+c$，$O$ 为原点。

1. Determine the position, modulus, and argument of point A.

Given $a$ has modulus $r$ and argument $\theta$ where $0 < \theta < \frac{\pi}{2}$.
Point $A$ is in the first quadrant. It is drawn at a distance $r$ from the origin $O$, and the line segment $OA$ makes an angle of $\theta$ with the positive real axis.

2. Calculate the modulus and argument of $b$ and locate point B.

$b = -2ia$
Modulus: $|b| = |-2i| \cdot |a| = 2r$
Argument: $\arg(b) = \arg(-2i) + \arg(a) = -\frac{\pi}{2} + \theta$
Since $0 < \theta < \frac{\pi}{2}$, the argument $\theta - \frac{\pi}{2}$ is negative and lies strictly between $-\frac{\pi}{2}$ and $0$. Therefore, point $B$ is in the fourth quadrant, with distance $2r$ from the origin.

3. Determine the modulus and argument of $c$ and locate point C.

We are given that $|c| = |b| = 2r$.
Since $c$ is purely imaginary with a negative imaginary part, it lies on the negative imaginary axis.
Thus, the argument of $c$ is $-\frac{\pi}{2}$.
Point $C$ is drawn on the negative imaginary axis at a distance of $2r$ from $O$.

4. Locate point D on the Argand diagram.

Point $D$ represents the complex number $b+c$.
Geometrically, complex addition corresponds to vector addition. Point $D$ is found by completing the parallelogram $OBDC$.
Since adjacent sides are equal in length ($OB = OC = 2r$), the quadrilateral $OBDC$ is specifically a rhombus in the fourth quadrant.

1. Determine $\angle AOB$ using the relationship between $a$ and $b$.

We are given that $b = -2ia$.
The complex number $b$ is obtained by multiplying $a$ by $-2i$.
Since $\arg(-2i) = -\frac{\pi}{2}$, multiplying by $-2i$ geometrically corresponds to scaling the modulus by $2$ and rotating the complex number clockwise about the origin by $\frac{\pi}{2}$ radians.
Therefore, the geometric angle between $OA$ and $OB$ is simply the magnitude of this rotation:
$\angle AOB = \frac{\pi}{2}$

2. Determine $\angle BOC$ by finding the arguments of $b$ and $c$.

We are given that $c$ is purely imaginary with a negative imaginary part. Thus, it lies on the negative imaginary axis, so:
$\arg(c) = -\frac{\pi}{2}$
Using the property of arguments for a product, the argument of $b$ is:
$\arg(b) = \arg(-2ia) = \arg(-2i) + \arg(a) = -\frac{\pi}{2} + \theta$
The angle $\angle BOC$ is the positive difference between the arguments of $b$ and $c$ (since both points lie in the fourth quadrant / negative imaginary axis region):
$\angle BOC = \arg(b) - \arg(c)$
$\angle BOC = \left(-\frac{\pi}{2} + \theta\right) - \left(-\frac{\pi}{2}\right) = \theta$

1. State the shape of the quadrilateral $OBDC$.

Since $D$ represents the sum $b+c$, $OBDC$ is a parallelogram.
Given $|b| = |c| = 2r$, $OBDC$ is a parallelogram with equal adjacent sides.
Therefore, $OBDC$ is a rhombus.

2. Determine the argument of $b+c$ using the properties of a rhombus.

In a rhombus, the diagonal $OD$ bisects the interior angle $\angle BOC$.
From earlier parts, $c$ lies on the negative imaginary axis, so $\arg(c) = -\frac{\pi}{2}$, and $\angle BOC = \theta$.
Since $\arg(b) > \arg(c)$, the diagonal $OD$ is formed by a counter-clockwise rotation of $\frac{\theta}{2}$ from $OC$.
$\arg(b+c) = \arg(c) + \frac{\theta}{2} = -\frac{\pi}{2} + \frac{\theta}{2}$

3. Determine the modulus of $b+c$.

The diagonals of a rhombus bisect each other at right angles.
Using right-angled trigonometry on the half-rhombus, the distance from $O$ to the midpoint of $BC$ is $|c| \cos\left(\frac{\theta}{2}\right) = 2r \cos\left(\frac{\theta}{2}\right)$.
The total length of the diagonal $OD$ gives the modulus:
$|b+c| = 2 \times 2r \cos\left(\frac{\theta}{2}\right) = 4r \cos\left(\frac{\theta}{2}\right)$

4. Express $b+c$ in exponential form.

Since $0 < \theta < \frac{\pi}{2}$, $\cos\left(\frac{\theta}{2}\right) > 0$, confirming the modulus is positive.
Combining the modulus and argument:
$b+c = 4r \cos\left(\frac{\theta}{2}\right) e^{i\left(\frac{\theta}{2} - \frac{\pi}{2}\right)}$