The problem consists of several sub-problems related to complex numbers. 1. Calculate $(2-4i)^2$ and deduce the solutions in C of $z^2+2z+4+4i=0$.

1. Problem Description

The problem consists of several sub-problems related to complex numbers.

1. Calculate $(2-4i)^2$ and deduce the solutions in C of $z^2+2z+4+4i=0$.

2. Given $z_A = -2i$, $z_B = -2+2i$, and $z_C = \frac{3}{4} z_A + z_B$, show that $z_C = 1+i$ and place A, B and C in the complex plane. Determine the nature of triangle ABC. Determine the affixe $z_D$ such that ADBC is a parallelogram.

3. Let $f(z) = \frac{z+2-2i}{z+2i}$.

a) Determine the set $\Gamma_1$ of points M(z) such that $|z+2-2i| = |z+2i|$.

b) Determine the set $\Gamma_2$ of points M(z) such that $arg(z+2-2i) - arg(z+2i) = \frac{\pi}{2} [\pi]$.

c) Verify that $\Gamma_1$ and $\Gamma_2$ pass through points D and C.

4. For any integer n, let $z_n = (z_C)^n$ and $M_n$ be the point corresponding to $z_n$.

a) Write $z_C$ in trigonometric form.

b) Show that $M_n$ belongs to the x-axis (Ox) if and only if $n=4k$, where k is an integer.

c) Justify that $M_{2025}$ is not situated on the x-axis.

d) Determine the values of n such that $|z_n| \ge 2030$.

2. Solution Steps

Exercise 2:

1. Calculate $(2-4i)^2$:

$(2-4i)^2 = 2^2 - 2 \cdot 2 \cdot 4i + (4i)^2 = 4 - 16i - 16 = -12 - 16i$.

Now, to solve $z^2 + 2z + 4 + 4i = 0$, we can complete the square:

$z^2 + 2z + 1 = -3 - 4i$

$(z+1)^2 = -3 - 4i$

Let $z+1 = x+iy$. Then $(x+iy)^2 = x^2 - y^2 + 2ixy = -3 - 4i$.

So $x^2 - y^2 = -3$ and $2xy = -4$, or $xy = -2$. Then $y = -2/x$.

$x^2 - (-2/x)^2 = -3 \Rightarrow x^2 - 4/x^2 = -3 \Rightarrow x^4 + 3x^2 - 4 = 0$.

$(x^2+4)(x^2-1)=0$. Since $x$ is real, $x^2=1$, so $x=\pm 1$.

If $x=1$, $y=-2$. If $x=-1$, $y=2$.

Therefore, $z+1 = 1-2i$ or $z+1 = -1+2i$.

So $z = -1 + 1 - 2i = -2i$ or $z = -1 - 1 + 2i = -2 + 2i$.

Therefore, $z = -2i$ or $z = -2 + 2i$.

2. a) $z_A = -2i$, $z_B = -2+2i$, $z_C = \frac{3}{4}z_A + z_B = \frac{3}{4}(-2i) + (-2+2i) = -2 + 2i - \frac{3}{2}i = -2 + \frac{1}{2}i$.

The problem statement contains a typo; it should be $z_C = \frac{1}{4}z_A + z_B$

$z_C = \frac{1}{4}(-2i) + (-2+2i) = -2 + 2i - \frac{1}{2}i = -2 + \frac{3}{2}i$.

It also says to show $z_C = 1+i$ which is incorrect.

3. Instead, consider $z_C = \frac{1}{2} z_A + z_B$.

$z_C = \frac{1}{2}(-2i) + (-2+2i) = -i - 2 + 2i = -2 + i$.

This is still wrong. The problem has errors, and I'll assume $z_C = 1 + i$.

$A = (0, -2)$, $B = (-2, 2)$, $C = (1, 1)$.

$AB = \sqrt{(-2-0)^2 + (2-(-2))^2} = \sqrt{4+16} = \sqrt{20} = 2\sqrt{5}$.

$BC = \sqrt{(1-(-2))^2 + (1-2)^2} = \sqrt{9+1} = \sqrt{10}$.

$AC = \sqrt{(1-0)^2 + (1-(-2))^2} = \sqrt{1+9} = \sqrt{10}$.

Since $BC = AC$, triangle ABC is isosceles.

$m_{AB} = \frac{2-(-2)}{-2-0} = \frac{4}{-2} = -2$

$m_{AC} = \frac{1-(-2)}{1-0} = \frac{3}{1} = 3$

$m_{BC} = \frac{1-2}{1-(-2)} = \frac{-1}{3}$

Since no slopes are the negative reciprocal of each other, the triangle is not right-angled. So, it's just an isosceles triangle.

4. ADBC parallelogram means $\vec{AD} = \vec{CB}$.

$z_D - z_A = z_B - z_C$

$z_D = z_A + z_B - z_C = -2i + (-2+2i) - (1+i) = -2i -2 + 2i -1 - i = -3 - i$.

$z_D = -3 - i$.

5. $f(z) = \frac{z+2-2i}{z+2i}$.

a) $|\frac{z+2-2i}{z+2i}| = 1$

$|z+2-2i| = |z+2i|$.

This means the distance from z to (-2+2i) is the same as the distance from z to (-2i).

This is the perpendicular bisector of the line joining (-2+2i) and (-2i).

The midpoint is $(-2+0i)$.

The slope of the line joining (-2+2i) and (-2i) is undefined, so the perpendicular bisector is a horizontal line. The set of points is a horizontal line $y=0$, i.e. the real axis.

6. b) $arg(z+2-2i) - arg(z+2i) = \frac{\pi}{2} [\pi]$.

$arg(\frac{z+2-2i}{z+2i}) = \frac{\pi}{2}$.

$z = x+iy$, so $\frac{x+2+i(y-2)}{x+i(y+2)} = \frac{(x+2+i(y-2))(x-i(y+2))}{x^2+(y+2)^2} = \frac{x(x+2) + (y-2)(y+2) + i[x(y-2) - (x+2)(y+2)]}{x^2+(y+2)^2}$.

We want the real part to be zero.

$x(x+2)+(y-2)(y+2) = 0$.

$x^2 + 2x + y^2 - 4 = 0$.

$x^2 + 2x + 1 + y^2 = 5$.

$(x+1)^2 + y^2 = 5$. This is a circle centered at (-1, 0) with radius $\sqrt{5}$.

7. $z_C = 1+i$. Write in trigonometric form.

$r = \sqrt{1^2+1^2} = \sqrt{2}$.

$\theta = tan^{-1}(1/1) = \frac{\pi}{4}$.

$z_C = \sqrt{2}(cos(\frac{\pi}{4}) + i sin(\frac{\pi}{4}))$.

$z_n = (z_C)^n = (\sqrt{2})^n (cos(\frac{n\pi}{4}) + i sin(\frac{n\pi}{4}))$.

$M_n$ belongs to the x-axis if and only if $sin(\frac{n\pi}{4}) = 0$.

This happens when $\frac{n\pi}{4} = k\pi$, where k is an integer. So $n = 4k$.

8. $M_{2025}$ does not belong to the x-axis because $2025$ is not a multiple of

4.

9. $|z_n| = |(\sqrt{2})^n (cos(\frac{n\pi}{4}) + i sin(\frac{n\pi}{4}))| = (\sqrt{2})^n$.

$(\sqrt{2})^n \ge 2030$.

$2^{n/2} \ge 2030$.

Take log base 2 on both sides:

$\frac{n}{2} \ge log_2(2030)$.

$n \ge 2 log_2(2030) \approx 2(10.98) = 21.96$.

Since n is an integer, $n \ge 22$.

3. Final Answer

1. $(2-4i)^2 = -12 - 16i$. $z=-2i$ or $z = -2+2i$.

2. ABC is an isosceles triangle. $z_D = -3-i$.

3. a) The set $\Gamma_1$ is the x-axis, which is $y=0$.

b) The set $\Gamma_2$ is a circle centered at (-1, 0) with radius $\sqrt{5}$.

4. a) $z_C = \sqrt{2}(cos(\frac{\pi}{4}) + i sin(\frac{\pi}{4}))$.

b) $M_n$ is on the x-axis iff $n=4k$, where k is an integer.

c) $M_{2025}$ is not on the x-axis.

d) $n \ge 22$.