点 $A$ 表示复数 $a$，满足 $|a|=1$ 且 $\theta=\arg(a),\; 0<\theta<\dfrac{\pi}{4}$。点 $B$ 表示复数 $ia$。 复数 $z$ 满足关系式 $|z-ia|\le\sqrt{2}\quad\text{and}\quad|z-ia|=|z-a|.$

1. Identify the geometrical meaning of the equations $|z-ia| = |z-a|$ and $|z-ia| \le \sqrt{2}$.

The equation $|z-ia| = |z-a|$ represents the perpendicular bisector of the line segment $AB$. Since $|a| = 1$ and $|ia| = 1$, the origin $O$ is equidistant from $A$ and $B$, meaning this bisector passes through $O$.

The inequality $|z-ia| \le \sqrt{2}$ represents the region on and inside the circle centered at $B$ with radius $\sqrt{2}$.
Note that the distance $AB = |ia - a| = |a||i - 1| = (1)\sqrt{1^2 + (-1)^2} = \sqrt{2}$. Thus, the boundary of this circle passes through $A$.

2. Find the angle the locus makes with the positive real axis.

The point $A$ has argument $\theta$ and $B(ia)$ has argument $\theta + \frac{\pi}{2}$.
The perpendicular bisector of $AB$ passes through the origin and bisects $\angle AOB$.
The angle the locus makes with the positive real axis is the average of their arguments:
$\text{Angle} = \theta + \frac{1}{2}\left(\frac{\pi}{2}\right) = \theta + \frac{\pi}{4}$

3. Describe the required sketch.

• Draw the real and imaginary axes.
• Plot point $A$ in the first quadrant with angle $\theta$ ($0 < \theta < \frac{\pi}{4}$).
• Plot point $B$ in the second quadrant such that $\angle AOB = \frac{\pi}{2}$ and $OA = OB$.
• Draw an arc of a circle centered at $B$, passing through $A$ (radius $\sqrt{2}$).
• Draw the perpendicular bisector of $AB$ passing through $O$ at an angle of $\theta + \frac{\pi}{4}$.
• The locus of $z$ is the straight line segment on this bisector, bounded by its intersections with the circular arc.

1. Relate the given expression for the complex number at $C$ to the points $A$ and $B$ using displacement vectors.

Let the complex numbers represented by $A$, $B$, and $C$ be $z_A = a$, $z_B = ia$, and $z_C = 2ia-a$.
Using the hint provided, we express $z_C$ as:
$z_C = ia + (ia - a)$
Substituting the respective complex numbers for $A$ and $B$:
$z_C = z_B + (z_B - z_A)$
$z_C - z_B = z_B - z_A$
This corresponds to the vector relationship:
$\vec{BC} = \vec{AB}$
(On the Argand diagram, point $C$ is indicated by extending the line segment from $A$ through $B$ by a distance equal to the length of $AB$.)

2. State the geometrical relationship between the three points.

The vector relation $\vec{BC} = \vec{AB}$ implies two things:

1. The vectors have the same direction, meaning $A$, $B$, and $C$ are collinear.
2. The vectors have the same magnitude ($|\vec{BC}| = |\vec{AB}|$), meaning the distance from $B$ to $C$ is equal to the distance from $A$ to $B$.

Therefore, the point $B$ is the midpoint of the line segment $AC$.

1. Interpret the given expression as a distance on the Argand diagram.

The expression $|z+a-2ia|$ can be rewritten as $|z - (2ia - a)|$, which represents the distance between a point $z$ on the locus and the point $C$.

2. Identify the point on the locus that minimises this distance geometrically.

From earlier parts, $z$ lies on the perpendicular bisector of $AB$, passing through the midpoint $M$ of $AB$.
Since $A, B,$ and $C$ are collinear (forming the line segment $AC$), the line $AC$ is perpendicular to the locus of $z$. The shortest distance from $C$ to the perpendicular bisector is the straight line distance $CM$.
Checking the constraint $|z-ia| \le \sqrt{2}$, we see $|M-B| = \frac{\sqrt{2}}{2} \le \sqrt{2}$, so $M$ is a valid point on the restricted locus segment.

3. Calculate the least distance $CM$.

We know $AB = |ia - a| = |a||i - 1| = (1)\sqrt{1^2 + (-1)^2} = \sqrt{2}$.
Since $B$ is the midpoint of $AC$, the length $CB = AB = \sqrt{2}$.
$M$ is the midpoint of $AB$, so the length $BM = \frac{1}{2}AB = \frac{\sqrt{2}}{2}$.
Thus, the least value is $CM = CB + BM = \sqrt{2} + \frac{\sqrt{2}}{2} = \frac{3\sqrt{2}}{2}$.

1. Recognize the geometric meaning of $\arg(z+a-2ia)$ and identify the points giving the maximum argument.

Let $C$ represent $c = 2ia - a$. We need the largest value of $\arg(z - c)$, which geometrically is the maximum angle the vector $\vec{Cz}$ makes with the positive real axis.
From earlier parts, $C, B, A$ form a straight line with $B$ being the midpoint of $AC$. The locus of $z$ is the line segment $PQ$ on the perpendicular bisector of $AB$.
Let $M$ be the midpoint of $AB$. The segment $PQ$ is perpendicular to the line $CMA$ at $M$. As $z$ moves along $PQ$, the maximum argument of $\vec{Cz}$ occurs at one of the endpoints, $P$ or $Q$.

2. Calculate the perpendicular distance $CM$ and the half-length of the locus segment $MP$.

The length $AB = |a - ia| = |a(1 - i)| = 1 \times \sqrt{2} = \sqrt{2}$.
Since $B$ is the midpoint of $AC$, $CB = AB = \sqrt{2}$. Since $M$ is the midpoint of $AB$, $BM = \frac{\sqrt{2}}{2}$.
Thus, the distance $CM = CB + BM = \sqrt{2} + \frac{\sqrt{2}}{2} = \frac{3\sqrt{2}}{2}$.
At the endpoints $P$ and $Q$ of the locus, $z$ lies on the boundary circle $|z - ia| = \sqrt{2}$, meaning $BP = \sqrt{2}$.
In the right-angled $\triangle BMP$, applying Pythagoras' theorem:
$MP = \sqrt{BP^2 - BM^2} = \sqrt{2 - \left(\frac{\sqrt{2}}{2}\right)^2} = \sqrt{\frac{3}{2}} = \frac{\sqrt{6}}{2}$

3. Determine the angle $\angle MCP$ and the baseline argument of the vector $\vec{CM}$.

In the right-angled $\triangle CMP$:
$\tan(\angle MCP) = \frac{MP}{CM} = \frac{\frac{\sqrt{6}}{2}}{\frac{3\sqrt{2}}{2}} = \frac{\sqrt{3}}{3} \implies \angle MCP = \frac{\pi}{6}$
The vector $\vec{CM}$ points from $C$ to $M$, which is identical to the direction from $B$ to $A$ ($\vec{BA}$). The complex number representing $\vec{BA}$ is $a - ia$:
$\arg(a - ia) = \arg(a(1 - i)) = \arg(a) + \arg(1 - i) = \theta - \frac{\pi}{4}$

4. Evaluate the largest argument of $z+a-2ia$.

The maximum argument is obtained by adding the angle subtended by $MP$ at $C$, $\angle MCP$, to the baseline argument of $\vec{CM}$ (representing a positive, counter-clockwise rotation to the upper endpoint $P$):
$\text{Largest } \arg(z+a-2ia) = \arg(M-C) + \angle MCP = \left(\theta - \frac{\pi}{4}\right) + \frac{\pi}{6} = \theta - \frac{\pi}{12}$