Find the value of x and the value of y in the following equation, give that 

𝑥(1 + 𝑖)

2 + 𝑦(2 − 𝑖)

2 = 3 + 10𝑖 where 𝑥 ∈ 𝑅 𝑎𝑛𝑑 𝑦 ∈ 𝑅 

B) Given that 𝑧 = −3 + 4𝑖 𝑎𝑛𝑑 𝑧𝑤 = −14 + 2𝑖, find 

 𝑖) 𝑤 in the form 𝑎 + 𝑖𝑏, where 𝑎 𝑎𝑛𝑑 𝑏 are real numbers. 

 𝑖𝑖) the modulus and the argument of 𝑤.

Let's solve each part of the problem step by step.

### Part A: Find the values of \( x \) and \( y \)
Given:
\[
x(1 + i)^2 + y(2 - i)^2 = 3 + 10i
\]
where \( x \) and \( y \) are real numbers.

1. **Expand the complex squares:**

   - \( (1 + i)^2 = 1^2 + 2i + i^2 = 1 + 2i - 1 = 2i \)
   - \( (2 - i)^2 = 2^2 - 2 \times 2 \times i + i^2 = 4 - 4i - 1 = 3 - 4i \)

   Thus, the equation becomes:
   \[
   x(2i) + y(3 - 4i) = 3 + 10i
   \]

2. **Distribute \( x \) and \( y \):**
   \[
   2xi + 3y - 4yi = 3 + 10i
   \]

   Group the real and imaginary parts:
   \[
   (3y) + (2x - 4y)i = 3 + 10i
   \]

3. **Set up the system of equations:**

   - For the real part: \( 3y = 3 \)
   - For the imaginary part: \( 2x - 4y = 10 \)

4. **Solve the system of equations:**

   From \( 3y = 3 \), we get:
   \[
   y = 1
   \]
   Substitute \( y = 1 \) into \( 2x - 4y = 10 \):
   \[
   2x - 4(1) = 10
   \]
   \[
   2x - 4 = 10
   \]
   \[
   2x = 14 \implies x = 7
   \]

So, the values are:
\[
x = 7, \quad y = 1
\]

### Part B: Given \( z = -3 + 4i \) and \( zw = -14 + 2i \)

1. **Find \( w \) in the form \( a + ib \):**

   Given \( zw = -14 + 2i \) and \( z = -3 + 4i \), find \( w \).

   Let \( w = a + ib \). Then:
   \[
   (-3 + 4i)(a + ib) = -14 + 2i
   \]

   Expand and equate real and imaginary parts:
   \[
   -3a - 3ib + 4ai + 4ib^2 = -14 + 2i
   \]
   \[
   (-3a - 4b) + (4a - 3b)i = -14 + 2i
   \]

   Set up the system of equations:
   \[
   -3a - 4b = -14 \quad \text{(Real part)}
   \]
   \[
   4a - 3b = 2 \quad \text{(Imaginary part)}
   \]

   Solve these equations:

   From \( -3a - 4b = -14 \):
   \[
   3a + 4b = 14
   \]

   Solve the system:
   \[
   \begin{cases}
   3a + 4b = 14 \\
   4a - 3b = 2
   \end{cases}
   \]

   Multiply the first equation by 3 and the second by 4:
   \[
   9a + 12b = 42
   \]
   \[
   16a - 12b = 8
   \]

   Add the equations:
   \[
   25a = 50 \implies a = 2
   \]

   Substitute \( a = 2 \) into \( 3a + 4b = 14 \):
   \[
   3(2) + 4b = 14 \implies 6 + 4b = 14 \implies 4b = 8 \implies b = 2
   \]

   Therefore, \( w = 2 + 2i \).

2. **Find the modulus and argument of \( w \):**

   - **Modulus:**
     \[
     |w| = \sqrt{a^2 + b^2} = \sqrt{2^2 + 2^2} = \sqrt{4 + 4} = \sqrt{8} = 2\sqrt{2}
     \]

   - **Argument:**
     \[
     \arg(w) = \tan^{-1}\left(\frac{b}{a}\right) = \tan^{-1}\left(\frac{2}{2}\right) = \tan^{-1}(1) = \frac{\pi}{4}
     \]

### Summary:
1. **Values of \( x \) and \( y \):**
   \[
   x = 7, \quad y = 1
   \]

2. **For \( w \):**
   - \( w = 2 + 2i \)
   - \( |w| = 2\sqrt{2} \)
   - \( \arg(w) = \frac{\pi}{4} \)

### Do you want details or have any questions? 
Here are 5 relative questions for you:
1. How do you multiply two complex numbers?
2. How is the modulus of a complex number calculated?
3. What are the standard forms of complex numbers?
4. How does the argument of a complex number relate to its position on the complex plane?
5. What does solving simultaneous equations involve?

**Tip:** When solving for real and imaginary parts in complex equations, separate the components carefully to avoid errors.

Learn how to solve complex equations involving both real and imaginary components. Find the values of x and y in a complex equation and determine the form, modulus, and argument of a complex number w. Suitable for high school students studying complex numbers and systems of equations.

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