复数 $z$ 满足 $\arg(z-1-2i)=\theta$，其中 $\theta$ 是区间 $-\pi < \theta \le \pi$ 内的一个固定角。

1. State the standard geometrical interpretation of the given argument equation.

The equation $\arg(z - (1 + 2i)) = \theta$ describes a half-line (or ray) starting from the point representing $1 + 2i$, which corresponds to the coordinates $(1, 2)$ on the Argand diagram.
It makes an angle $\theta$ with the positive real axis, and the starting point $(1, 2)$ is strictly excluded since the argument of $0$ is undefined.

1. Translate the algebraic minimum value problem into a geometrical one on the Argand diagram.

Let $A$ be the point representing $1+6i$, which is $(1, 6)$.
Let $B$ be the starting point of the locus of $z$, representing $1+2i$, which is $(1, 2)$.
The locus of $z$ is a half-line $L$ starting from $B$ making an angle of $\frac{\pi}{6}$ with the positive real axis.
We seek the shortest distance from $A$ to $L$, which corresponds to the minimum value of $|z-(1+6i)|$.

2. Determine the distance between the two reference points and the angle between their connecting segment and the locus.

The distance $AB = |(1+6i) - (1+2i)| = |4i| = 4$.
The vector $\vec{BA}$ is parallel to the positive imaginary axis, so its argument is $\frac{\pi}{2}$.
The angle between $AB$ and the half-line $L$ is $\alpha = \frac{\pi}{2} - \frac{\pi}{6} = \frac{\pi}{3}$.

3. Calculate the exact minimum distance.

Since $\frac{\pi}{3} < \frac{\pi}{2}$, the foot of the perpendicular from $A$ to the line containing $L$ lies on the half-line $L$ itself.
The minimum distance $d$ is the perpendicular distance from $A$ to $L$:
$d = AB \sin\left(\frac{\pi}{3}\right) = 4 \left(\frac{\sqrt{3}}{2}\right) = 2\sqrt{3}$

1. Evaluate the modulus on the right hand side of the equation.

$|2(\cos\alpha + i\sin\alpha)| = 2\sqrt{\cos^2\alpha + \sin^2\alpha} = 2$

2. Identify and sketch the locus represented by the simplified equation.

The equation simplifies to $|w - (6+2i)| = 2$, which represents a circle on the Argand diagram. The center of the circle is at $(6, 2)$ and its radius is $2$.

Since the imaginary part of the center is equal to the radius, sketch the circle such that it is tangent to the real axis at $(6, 0)$.

1. Recall the loci of $P$ and $Q$ and state the geometrical condition for multiple intersections.

The locus of $P$ is a half-line starting from $A(1,2)$.
The locus of $Q$ is given by $|w-(6+2i)| = |2(\cos\alpha + i\sin\alpha)| = 2$, which is a circle with centre $C(6,2)$ and radius $r = 2$.

The distance between the starting point $A$ and the centre $C$ is $5$. Since $A$ lies strictly outside the circle ($5 > 2$), the half-line $P$ will meet the circle $Q$ at exactly two distinct points if and only if it lies strictly between the two tangents from $A$ to the circle.

2. Calculate the critical angles of the tangents to determine the range of $\theta$.

Let the angle between the line segment $AC$ and a tangent from $A$ be $\phi$. In the right-angled triangle formed by $A$, $C$, and the point of tangency, the opposite side is the radius $r$ and the hypotenuse is $AC$.
$\sin \phi = \frac{r}{AC} = \frac{2}{5}$
$\phi = \sin^{-1}\left(\frac{2}{5}\right) \approx 0.4115 \text{ rad}$

Since $A(1,2)$ and $C(6,2)$ lie on the same horizontal line ($y=2$), the direction of $AC$ corresponds to an angle of $0$.
Therefore, for the locus of $P$ to intersect the circle twice, $\theta$ must fall strictly within the tangent bounds:
$-\sin^{-1}\left(\frac{2}{5}\right) < \theta < \sin^{-1}\left(\frac{2}{5}\right)$
$-0.412 < \theta < 0.412 \text{ (to 3 s.f.)}$