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