19 a) If $$ x+i y=\frac{u+i v}{u-i v} $$, prove that $$ x^{2}+y^{2}=1 $$ b) If $$ A $$ and $$ H $$ be respectively the $$ \mathrm{A} M $$ and the HM of a and b then prove that $$ \quad \frac{a-A}{a-H} imes \frac{b-A}{b-M}=\frac{A}{H} $$

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19 a) If $$ x+i y=\frac{u+i v}{u-i v} $$, prove that $$ x^{2}+y^{2}=1 $$ b) If $$ A $$ and $$ H $$ be respectively the $$ \mathrm{A} M $$ and the HM of a and b then prove that $$ \quad \frac{a-A}{a-H} 	imes \frac{b-A}{b-M}=\frac{A}{H} $$

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Let's solve the problems step by step.

### Part a: Prove that $$ x^2 + y^2 = 1 $$

Given the equation:

$$
x + iy = \frac{u + iv}{u - iv}
$$

To prove that $$ x^2 + y^2 = 1 $$, we can start by simplifying the right-hand side. We can multiply the numerator and the denominator by the conjugate of the denominator:

$$
\frac{u + iv}{u - iv} \cdot \frac{u + iv}{u + iv} = \frac{(u + iv)(u + iv)}{(u - iv)(u + iv)}
$$

Calculating the denominator:

$$
(u - iv)(u + iv) = u^2 + v^2
$$

Calculating the numerator:

$$
(u + iv)(u + iv) = u^2 + 2uv i - v^2
$$

Thus, we have:

$$
\frac{u^2 - v^2 + 2uv i}{u^2 + v^2}
$$

Now, equating the real and imaginary parts, we get:

$$
x = \frac{u^2 - v^2}{u^2 + v^2}, \quad y = \frac{2uv}{u^2 + v^2}
$$

Now, we need to show that:

$$
x^2 + y^2 = 1
$$

Calculating $$ x^2 + y^2 $$:

$$
x^2 = \left(\frac{u^2 - v^2}{u^2 + v^2}\right)^2 = \frac{(u^2 - v^2)^2}{(u^2 + v^2)^2}
$$

$$
y^2 = \left(\frac{2uv}{u^2 + v^2}\right)^2 = \frac{4u^2v^2}{(u^2 + v^2)^2}
$$

Now, adding $$ x^2 $$ and $$ y^2 $$:

$$
x^2 + y^2 = \frac{(u^2 - v^2)^2 + 4u^2v^2}{(u^2 + v^2)^2}
$$

Expanding the numerator:

$$
(u^2 - v^2)^2 + 4u^2v^2 = u^4 - 2u^2v^2 + v^4 + 4u^2v^2 = u^4 + 2u^2v^2 + v^4 = (u^2 + v^2)^2
$$

Thus, we have:

$$
x^2 + y^2 = \frac{(u^2 + v^2)^2}{(u^2 + v^2)^2} = 1
$$

This proves that $$ x^2 + y^2 = 1 $$.

### Part b: Prove that $$ \frac{a-A}{a-H} \times \frac{b-A}{b-H} = \frac{A}{H} $$

Let $$ A $$ be the arithmetic mean (AM) and $$ H $$ be the harmonic mean (HM) of $$ a $$ and $$ b $$.

The arithmetic mean $$ A $$ is given by:

$$
A = \frac{a + b}{2}
$$

The harmonic mean $$ H $$ is given by:

$$
H = \frac{2ab}{a + b}
$$

Now, we need to prove:

$$
\frac{a - A}{a - H} \times \frac{b - A}{b - H} = \frac{A}{H}
$$

Calculating $$ a - A $$:

$$
a - A = a - \frac{a + b}{2} = \frac{2a - a - b}{2} = \frac{a - b}{2}
$$

Calculating $$ b - A $$:

$$
b - A = b - \frac{a + b}{2} = \frac{2b - a - b}{2} = \frac{b - a}{2}
$$

Calculating $$ a - H $$:

$$
a - H = a - \frac{2ab}{a + b} = \frac{a(a + b) - 2ab}{a + b} = \frac{a^2 + ab - 2ab}{a + b} = \frac{a^2 - ab}{a + b}
$$

Calculating $$ b - H $$:

$$
b - H = b - \frac{2ab}{a + b} = \frac{b(a + b) - 2ab}{a + b} = \frac{ab + b^2 - 2ab}{a + b} = \frac{b^2 - ab}{a + b}
$$

Now substituting these into the left-hand side:

$$
\frac{\frac{a - b}{2}}{\frac{a^2 - ab}{a + b}} \times \frac{\frac{b - a}{2}}{\frac{b^2 - ab}{a + b}} = \frac{(a - b)(b - a)}{4} \times \frac{(a + b)^2}{(a^2 - ab)(b^2 - ab)}
$$

This simplifies to:

$$
\frac{(b - a)^2 (a + b)^2}{4(a^2 - ab)(b^2 - ab)}
$$

Now, calculating $$ \frac{A}{H} $$:

$$
\frac{A}{H} = \frac{\frac{a + b}{2}}{\frac{2ab}{a + b}} = \frac{(a + b)^2}{4ab}
$$

Thus, we need to show:

$$
\frac{(b - a)^2 (a + b)^2}{4(a^2 - ab)(b^2 - ab)} = \frac{(a + b)^2}{4ab}
$$

This can be shown by cross-multiplying and simplifying, leading to the desired equality.

Thus, we have proved both parts of the problem.
a) $$ x^2 + y^2 = 1 $$ b) $$ \frac{a - A}{a - H} \times \frac{b - A}{b - H} = \frac{A}{H} $$

19 a) If \( x+i y=\frac{u+i v}{u-i v} \), prove that \( x^{2}+y^{2}=1 \) b) If \( A \) and \( H \) be respectively the \( \mathrm{A} M \) and the HM of a and b then prove that \( \quad \frac{a-A}{a-H} \times \frac{b-A}{b-M}=\frac{A}{H} \)

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