点 A 表示一个固定的复数 $a$，满足 $-\dfrac{\pi}{2}<\arg a<0$。复数 $ia$ 和 $a+ia$ 分别由点 B 和 C 表示。在清晰标注的阿甘图（Argand diagram）上，画出点 A、B、C 以及表示满足以下条件的复数 $z$ 的点集： (i) $|z-i a|=|a|$, (ii) $|z|\ge|z-(a+ia)|$。 写出使 $|z|$ 达到最大值的复数 $z$，用 $a$ 表示你的答案。

1. Draw the square OACB to establish the geometric locations of the points.

Let $O$ be the origin. Point $A$ represents $a$, which is in the 4th quadrant.
Point $B$ represents $ia$, which is the rotation of $A$ by $90^\circ$ anti-clockwise about $O$. Thus, $OA \perp OB$ and $|OA|=|OB|=|a|$. $B$ is in the 1st quadrant.
Point $C$ represents $a+ia$. By vector addition, $\vec{OC} = \vec{OA} + \vec{OB}$. Hence, $OACB$ is a square of side length $|a|$.

2. Identify and sketch locus (i).

The equation $|z-ia| = |a|$ describes a circle centred at $B$ (which represents $ia$) with radius $|a|$.
Since the distance from $B$ to $O$ is $|ia| = |a|$ and the distance from $B$ to $C$ is $|(a+ia)-ia| = |a|$, this circle passes through the points $O$ and $C$.

3. Identify and sketch locus (ii).

The inequality $|z| \ge |z-(a+ia)|$ represents the set of points whose distance from $O$ is greater than or equal to their distance from $C$.
The boundary is the perpendicular bisector of the line segment $OC$. In the square $OACB$, the perpendicular bisector of the diagonal $OC$ is the line containing the other diagonal, $AB$.
The valid region is the half-plane bounded by the line $AB$ that contains the point $C$.

4. Find the point in the intersection set that maximizes $|z|$.

The set of points satisfying both conditions is the semicircle of locus (i) that lies on the $C$-side of line $AB$.
To find the greatest value of $|z|$, we look for the point on this semicircle furthest from $O$.
On the full circle centred at $B$, the point furthest from $O$ is diametrically opposite to $O$, located at $O + 2\vec{OB} = 2ia$.
Since $2ia$ lies on the valid semicircle (it is geometrically on the $C$-side of $AB$), it is the required complex number.