The image you've uploaded seems to contain a mathematical expression involving complex numbers. Let me analyze it.

The equation appears to be:

$\text{If} \quad z = x + yi = \frac{a - bi}{c + bi} \quad \text{and} \quad z = x - yi$

You are asked to show that:

$x^2 + y^2 = 1$

Let me proceed with solving this step by step.

1. First, express $z = x + yi$ as a complex number.
2. Next, express the right-hand side $\frac{a - bi}{c + bi}$ in a simpler form by multiplying both numerator and denominator by the conjugate of the denominator.
3. Then, equate the real and imaginary parts of both sides of the equation to extract values for $x$ and $y$.

Let me begin this process.### Step 1: Express $z = \frac{a - bi}{c + bi}$

To simplify this complex fraction, we multiply both the numerator and the denominator by the conjugate of the denominator, which is $c - bi$.

$z = \frac{a - bi}{c + bi} \times \frac{c - bi}{c - bi}$

This gives:

$z = \frac{(a - bi)(c - bi)}{(c + bi)(c - bi)}$

Step 2: Simplify the denominator

The denominator is a product of a complex number and its conjugate, which simplifies as follows:

$(c + bi)(c - bi) = c^2 + b^2$

Step 3: Expand the numerator

Now expand the numerator:

$(a - bi)(c - bi) = ac - abi - bci + b^2i^2$

Since $i^2 = -1$, we have:

$ac - abi - bci - b^2$

This simplifies to:

$(ac - b^2) - i(ab + bc)$

Step 4: Combine everything

Thus, we have:

$z = \frac{(ac - b^2) - i(ab + bc)}{c^2 + b^2}$

Now, write $z = x + yi$, so that:

$x + yi = \frac{(ac - b^2)}{c^2 + b^2} - i \cdot \frac{(ab + bc)}{c^2 + b^2}$

By comparing the real and imaginary parts, we obtain:

• $x = \frac{ac - b^2}{c^2 + b^2}$
• $y = \frac{ab + bc}{c^2 + b^2}$

Step 5: Show that $x^2 + y^2 = 1$

Now, let's compute $x^2 + y^2$:

$x^2 + y^2 = \left( \frac{ac - b^2}{c^2 + b^2} \right)^2 + \left( \frac{ab + bc}{c^2 + b^2} \right)^2$

Combine the terms:

$x^2 + y^2 = \frac{(ac - b^2)^2 + (ab + bc)^2}{(c^2 + b^2)^2}$

Expanding the numerator:

$(ac - b^2)^2 = a^2c^2 - 2ab^2c + b^4$ $(ab + bc)^2 = a^2b^2 + 2ab^2c + b^2c^2$

Add them together:

$a^2c^2 - 2ab^2c + b^4 + a^2b^2 + 2ab^2c + b^2c^2$

Simplifying:

$a^2c^2 + b^2c^2 + a^2b^2 + b^4$

This is the same as the expanded form of:

$(c^2 + b^2)^2$

Thus, we get:

$x^2 + y^2 = \frac{(c^2 + b^2)^2}{(c^2 + b^2)^2} = 1$

Conclusion

We have shown that $x^2 + y^2 = 1$, as required.

Let me know if you would like more details or have any further questions. Here are some related questions you might be interested in:

1. How do you find the modulus of a complex number?
2. What is the geometric interpretation of the expression $x^2 + y^2 = 1$ in the complex plane?
3. Can this method be applied to other complex numbers in a similar form?
4. How do you use complex conjugates to simplify expressions involving complex numbers?
5. What is the significance of complex numbers in solving real-world problems?

Tip: When simplifying complex fractions, multiplying by the conjugate of the denominator is often a helpful technique.