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It looks like you've uploaded an image with a mathematical problem. The problem appears to involve showing that some conditions are satisfied in both cases.

Let me summarize what I can see in the image:

You are given the following conditions:

- \( x + iy = \frac{a - bi}{a + bi} \)
- \( x - iy = \frac{a + bi}{a - bi} \)

The goal is to show that \( x^2 + y^2 = 1 \) in both cases.

To begin, we will need to simplify the complex fractions and use algebra to prove that \( x^2 + y^2 = 1 \).

Let's break this down:

### Step 1: Simplify the expression for \( x + iy \)
Starting with \( x + iy = \frac{a - bi}{a + bi} \), we multiply both the numerator and denominator by the conjugate of the denominator \( a - bi \) to simplify:

\[
x + iy = \frac{(a - bi)(a - bi)}{(a + bi)(a - bi)}
\]

We know that \( (a + bi)(a - bi) = a^2 + b^2 \). Now, expand the numerator:

\[
(a - bi)(a - bi) = a^2 - 2abi + b^2i^2 = a^2 - 2abi - b^2
\]

This gives:

\[
x + iy = \frac{a^2 - b^2 - 2abi}{a^2 + b^2}
\]

Now, separate the real and imaginary parts:

\[
x = \frac{a^2 - b^2}{a^2 + b^2}, \quad y = \frac{-2ab}{a^2 + b^2}
\]

### Step 2: Simplify the expression for \( x - iy \)
Next, simplify the second equation \( x - iy = \frac{a + bi}{a - bi} \) by following a similar process:

\[
x - iy = \frac{(a + bi)(a + bi)}{(a - bi)(a + bi)} = \frac{a^2 - b^2 + 2abi}{a^2 + b^2}
\]

Again, separating the real and imaginary parts:

\[
x = \frac{a^2 - b^2}{a^2 + b^2}, \quad y = \frac{2ab}{a^2 + b^2}
\]

### Step 3: Show that \( x^2 + y^2 = 1 \)
Now, calculate \( x^2 + y^2 \):

For both cases, we have the same expressions for \( x \) and \( y \), except for the sign of \( y \). In either case:

\[
x^2 + y^2 = \left( \frac{a^2 - b^2}{a^2 + b^2} \right)^2 + \left( \frac{2ab}{a^2 + b^2} \right)^2
\]

Simplifying:

\[
x^2 + y^2 = \frac{(a^2 - b^2)^2}{(a^2 + b^2)^2} + \frac{4a^2b^2}{(a^2 + b^2)^2}
\]

\[
x^2 + y^2 = \frac{(a^2 - b^2)^2 + 4a^2b^2}{(a^2 + b^2)^2}
\]

Expanding the numerator:

\[
(a^2 - b^2)^2 + 4a^2b^2 = a^4 - 2a^2b^2 + b^4 + 4a^2b^2 = a^4 + 2a^2b^2 + b^4
\]

\[
x^2 + y^2 = \frac{(a^2 + b^2)^2}{(a^2 + b^2)^2} = 1
\]

Thus, we have shown that \( x^2 + y^2 = 1 \) in both cases.

Let me know if you need any further explanation or clarification!

---

Would you like more details on any specific part? Here are some related questions:

1. What is the geometric interpretation of the equation \( x^2 + y^2 = 1 \)?
2. How do complex conjugates help simplify complex expressions?
3. What other algebraic techniques can be used to simplify complex fractions?
4. How do the real and imaginary parts of complex numbers relate to their polar form?
5. What are some other common problems where complex conjugates play a role?

Tip: Remember that multiplying a complex number by its conjugate eliminates the imaginary part, which is useful when simplifying complex fractions!

If x + iy = (a - bi) / (a + bi) and x - iy = (a + bi) / (a - bi), show that x² + y² = 1 in both cases.

Learn how to prove x² + y² = 1 given x + iy = (a - bi) / (a + bi) and x - iy = (a + bi) / (a - bi). This step-by-step solution covers complex numbers, algebra, and conjugates for Grades 10-12.

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