已知 $w$ 满足方程 $|2w - 1| \le |w + 1|$。

1. Square both sides of the modulus inequality to form a quadratic inequality.

$|2w - 1| \le |w + 1|$
$(2w - 1)^2 \le (w + 1)^2$
$4w^2 - 4w + 1 \le w^2 + 2w + 1$

2. Solve the resulting quadratic inequality to find the range of $w$.

$3w^2 - 6w \le 0$
$3w(w - 2) \le 0$
$0 \le w \le 2$

3. Deduce the inequality for the 'Hence' part by relating it to the earlier result.

$\left|\frac{2w - 1}{w + 1}\right| \ge 1$
$|2w - 1| \ge |w + 1|$ and $w + 1 \neq 0 \implies w \neq -1$
This inequality is the reverse of the earlier inequality, inclusive of the equality roots $w = 0$ and $w = 2$.

4. State the final range of values, omitting the restricted value.

The raw solution to $|2w - 1| \ge |w + 1|$ is $w \le 0$ or $w \ge 2$.
Applying the denominator restriction $w \neq -1$:
$w < -1 \text{ or } -1 < w \le 0 \text{ or } w \ge 2$

1. Substitute $w = x + \mathrm{i}y$ into the given inequality.

Let $w = x + \mathrm{i}y$, where $x, y \in \mathbb{R}$.
$|2(x + \mathrm{i}y) - 1| \le |x + \mathrm{i}y + 1|$
$|(2x - 1) + \mathrm{i}(2y)| \le |(x + 1) + \mathrm{i}y|$

2. Express the modulus in terms of $x$ and $y$ and square both sides to eliminate the square roots.

Using the definition $|a + \mathrm{i}b| = \sqrt{a^2 + b^2}$:
$(2x - 1)^2 + (2y)^2 \le (x + 1)^2 + y^2$

3. Expand, simplify, and complete the square to find the Cartesian inequality for the locus.

$4x^2 - 4x + 1 + 4y^2 \le x^2 + 2x + 1 + y^2$
$3x^2 - 6x + 3y^2 \le 0$
$x^2 - 2x + y^2 \le 0$
$(x - 1)^2 - 1 + y^2 \le 0$
$(x - 1)^2 + y^2 \le 1$

(Note: The cartesian equation for the boundary of this locus is $(x - 1)^2 + y^2 = 1$.)

4. Sketch the locus on an Argand diagram.

The locus is the region on and inside the circle with centre $(1, 0)$ and radius $1$.

• The boundary circle passes through the origin $(0, 0)$ and touches the imaginary axis.
• It crosses the real axis at $\operatorname{Re}(w) = 0$ and $\operatorname{Re}(w) = 2$.
• The interior of the circle should be shaded to accurately represent the inequality $\le$.