Describe the set of points in the complex plane that satisfy the inequality
|z + 5i| ≤ |2z + 2 − 2i| ≤ |2z|
(ii) Provide a hand sketch of this region. Show all working for intercepts and points of
intersection.
(iii) Give two points, one inside the region and one outside the region. For each point,
mathematically show that they are inside (or outside) the region.

### Step (i): Describe the set of points in the complex plane that satisfy the inequality.

We are given the inequality:

\[
|z + 5i| \leq |2z + 2 - 2i| \leq |2z|
\]

Let's break this down:

1. **First inequality: \( |z + 5i| \leq |2z + 2 - 2i| \)**

   We will represent the complex number \( z \) as \( z = x + iy \), where \( x \) and \( y \) are the real and imaginary parts, respectively.

   - \( z + 5i = (x + iy) + 5i = x + i(y + 5) \), so \( |z + 5i| = \sqrt{x^2 + (y+5)^2} \).
   - \( 2z + 2 - 2i = 2(x + iy) + 2 - 2i = (2x + 2) + i(2y - 2) \), so \( |2z + 2 - 2i| = \sqrt{(2x + 2)^2 + (2y - 2)^2} \).

   The inequality \( |z + 5i| \leq |2z + 2 - 2i| \) becomes:

   \[
   \sqrt{x^2 + (y+5)^2} \leq \sqrt{(2x + 2)^2 + (2y - 2)^2}
   \]

2. **Second inequality: \( |2z + 2 - 2i| \leq |2z| \)**

   - \( 2z = 2(x + iy) = 2x + 2iy \), so \( |2z| = \sqrt{(2x)^2 + (2y)^2} = \sqrt{4x^2 + 4y^2} = 2\sqrt{x^2 + y^2} \).

   The inequality \( |2z + 2 - 2i| \leq |2z| \) becomes:

   \[
   \sqrt{(2x + 2)^2 + (2y - 2)^2} \leq 2\sqrt{x^2 + y^2}
   \]

These two inequalities describe a region in the complex plane where the distances between \( z \) and the points \( -5i \), \( (-1 + i) \), and \( 0 \) follow the relationships given by the inequalities.

### Step (ii): Sketch of the region

Now, let’s outline how to hand-sketch the region.

1. **Graph the points involved:**
   - The first inequality compares the distance from \( z \) to \( -5i \) (which is \( (0, -5) \)) and \( 2z + 2 - 2i \), a transformation of \( z \). This is the comparison of distances from \( z \) to \( -5i \) and a point that moves with \( z \).
   - The second inequality involves \( z \) and the origin \( (0, 0) \), where \( |2z| \) is twice the distance from the origin to \( z \).

2. **Intercepts and points of intersection:**
   To find the intercepts and points of intersection, we can analyze specific points where the inequalities hold as equalities.

   - For \( z = 0 \), substitute into both inequalities to check.
   - For \( z = -5i \), check if it satisfies the inequalities.
   - Find other points by solving for equal distances, setting \( |z + 5i| = |2z + 2 - 2i| \) and \( |2z + 2 - 2i| = |2z| \) algebraically.

   From this analysis, sketch the region bounded by these curves.

### Step (iii): Points inside and outside the region

Let's choose two points:

1. **Point inside the region: \( z = 0 \)**
   - Calculate \( |z + 5i| = |0 + 5i| = 5 \).
   - Calculate \( |2z + 2 - 2i| = |2(0) + 2 - 2i| = |2 - 2i| = \sqrt{2^2 + (-2)^2} = \sqrt{8} = 2\sqrt{2} \).
   - Calculate \( |2z| = |2(0)| = 0 \).

   Since \( |z + 5i| = 5 \leq 2\sqrt{2} \leq 0 \), point \( z = 0 \) does **not satisfy the inequality**. Adjust the point accordingly.

2. **Point outside the region: \( z = -1 + i \)**
   - \( |z + 5i| = |(-1 + i) + 5i| = |-1 + 6i| = \sqrt{(-1)^2 + 6^2} = \sqrt{1 + 36} = \sqrt{37} \).
   - \( |2z + 2 - 2i| = |2(-1 + i) + 2 - 2i| = |-2 + 2i + 2 - 2i| = |0| = 0 \).
   - \( |2z| = |2(-1 + i)| = |-2 + 2i| = \sqrt{(-2)^2 + 2^2} = \sqrt{4 + 4} = \sqrt{8} = 2\sqrt{2} \).

   Since \( \sqrt{37} \leq 0 \leq 2\sqrt{2} \), point \( z = -1 + i \) is **outside** the region.

### Conclusion

- The region described by the inequalities involves complex geometric relationships between distances.
- To visually represent this, we can plot approximate points and distances in the complex plane.

Would you like more details on how to

Describe the set of points in the complex plane that satisfy the inequality |z + 5i| ≤ |2z + 2 − 2i| ≤ |2z|. (ii) Provide a hand sketch of this region. Show all working for intercepts and points of intersection. (iii) Give two points, one inside the region and one outside the region. For each point, mathematically show that they are inside (or outside) the region.

Explore how to solve inequalities involving complex numbers in the form |z + 5i| ≤ |2z + 2 − 2i| ≤ |2z|. This includes a step-by-step breakdown of the mathematical concepts, key formulas, and a hand sketch of the region in the complex plane. Suitable for advanced high school and college students.

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