Math Problem Statement

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)

Solution

Let's solve the given problems one by one, using the provided complex numbers z1=1+2iz_1 = 1 + 2i, z2=1−i2z_2 = \frac{1-i}{2}, and z3=−2+iz_3 = -2 + i.

Problem 1:

We need to calculate z13−3z12+2z1z_1^3 - 3z_1^2 + 2z_1.

Step 1: Find z12z_1^2

z12=(1+2i)2=12+2(1)(2i)+(2i)2=1+4i−4=−3+4iz_1^2 = (1 + 2i)^2 = 1^2 + 2(1)(2i) + (2i)^2 = 1 + 4i - 4 = -3 + 4i

Step 2: Find z13z_1^3

= -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.

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Math Problem Analysis

Mathematical Concepts

Complex Numbers
Algebraic Manipulations
Division of Complex Numbers

Formulas

(a + bi)(c + di) = (ac - bd) + (ad + bc)i
Multiplying by the conjugate to divide complex numbers

Theorems

Complex Number Operations

Suitable Grade Level

Grades 10-12

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