https://file.aiquickdraw.com/mathbot/file/1729237164164h2wz3v06.jpg

Let's solve the given problems one by one, using the provided complex numbers \( z_1 = 1 + 2i \), \( z_2 = \frac{1-i}{2} \), and \( z_3 = -2 + i \).

### Problem 1:
We need to calculate \( z_1^3 - 3z_1^2 + 2z_1 \).

#### Step 1: Find \( z_1^2 \)
\[
z_1^2 = (1 + 2i)^2 = 1^2 + 2(1)(2i) + (2i)^2 = 1 + 4i - 4 = -3 + 4i
\]

#### Step 2: Find \( z_1^3 \)
\[
z_1^3 = (1 + 2i)(-3 + 4i) = 1(-3 + 4i) + 2i(-3 + 4i)
= -3 + 4i - 6i - 8 = -11 - 2i
\]

#### Step 3: Compute \( z_1^3 - 3z_1^2 + 2z_1 \)
\[
3z_1^2 = 3(-3 + 4i) = -9 + 12i
\]
\[
2z_1 = 2(1 + 2i) = 2 + 4i
\]
\[
z_1^3 - 3z_1^2 + 2z_1 = (-11 - 2i) - (-9 + 12i) + (2 + 4i)
= -11 - 2i + 9 - 12i + 2 + 4i = 0 - 10i
\]
So, the result is \( -10i \).

### Problem 2:
We need to calculate \( \frac{z_1 - z_3}{4z_2 - z_3} \).

#### Step 1: Find \( z_1 - z_3 \)
\[
z_1 - z_3 = (1 + 2i) - (-2 + i) = 1 + 2i + 2 - i = 3 + i
\]

#### Step 2: Find \( 4z_2 - z_3 \)
\[
4z_2 = 4\left(\frac{1-i}{2}\right) = 2 - 2i
\]
\[
4z_2 - z_3 = (2 - 2i) - (-2 + i) = 2 - 2i + 2 - i = 4 - 3i
\]

#### Step 3: Compute \( \frac{z_1 - z_3}{4z_2 - z_3} \)
\[
\frac{3 + i}{4 - 3i}
\]
We multiply the numerator and the denominator by the conjugate of the denominator:
\[
\frac{(3 + i)(4 + 3i)}{(4 - 3i)(4 + 3i)} = \frac{12 + 9i + 4i - 3}{16 + 9} = \frac{9 + 13i}{25} = \frac{9}{25} + \frac{13i}{25}
\]
So, the result is \( \frac{9}{25} + \frac{13}{25}i \).

### Problem 3:
We need to calculate \( \frac{z_3^2}{2z_1 + 2z_2 + i} \).

#### Step 1: Find \( z_3^2 \)
\[
z_3^2 = (-2 + i)^2 = (-2)^2 + 2(-2)(i) + i^2 = 4 - 4i - 1 = 3 - 4i
\]

#### Step 2: Find \( 2z_1 + 2z_2 + i \)
\[
2z_1 = 2(1 + 2i) = 2 + 4i
\]
\[
2z_2 = 2\left(\frac{1-i}{2}\right) = 1 - i
\]
\[
2z_1 + 2z_2 + i = (2 + 4i) + (1 - i) + i = 3 + 4i - i + i = 3 + 4i
\]

#### Step 3: Compute \( \frac{z_3^2}{2z_1 + 2z_2 + i} \)
\[
\frac{3 - 4i}{3 + 4i}
\]
Again, we multiply the numerator and the denominator by the conjugate of the denominator:
\[
\frac{(3 - 4i)(3 - 4i)}{(3 + 4i)(3 - 4i)} = \frac{9 - 12i - 12i + 16}{9 + 16} = \frac{25 - 24i}{25} = 1 - \frac{24i}{25}
\]
So, the result is \( 1 - \frac{24}{25}i \).

---

Let me know if you need more details on any step or have further questions! Here are five related questions:

1. Can you verify the calculation of \( z_1^2 \) using another method?
2. How would you approach multiplying complex numbers in polar form?
3. What happens when you square a complex number with equal real and imaginary parts?
4. What role does the conjugate play when dividing complex numbers?
5. How can the modulus of a complex number be used to simplify division?

**Tip:** When dividing complex numbers, multiplying by the conjugate helps eliminate the imaginary part from the denominator.

z1 = 1 + 2i und z2 = (1 - i)/2 und z3 = -2 + i. Bestimme: 1) z1^3 - 3z1^2 + 2z1, 2) (z1 - z3) / (4z2 - z3), 3) z3^2 / (2z1 + 2z2 + i)

Solve complex number problems involving powers and division: z1 = 1 + 2i, z2 = (1 - i)/2, z3 = -2 + i. Learn to compute expressions such as z1^3 - 3z1^2 + 2z1, (z1 - z3) / (4z2 - z3), and z3^2 / (2z1 + 2z2 + i).

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