Question

If x + iy = a+ib/a-ib , prove that x² + y² = 1.

Answer 1

To prove that \( x^2 + y^2 = 1 \) given that

\[
x + iy = \frac{a + ib}{a - ib},
\]

we start by simplifying the right-hand side. We can multiply the numerator and denominator by the conjugate of the denominator:

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

Calculating the denominator:

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

Now calculating the numerator:

\[
(a + ib)(a + ib) = a^2 + 2a(ib) + (ib)^2 = a^2 + 2abi - b^2 = a^2 - b^2 + 2abi.
\]

Thus, we have:

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

This can be separated into real and imaginary parts:

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

Now, we want to compute \( x^2 + y^2 \):

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

Calculating \( x^2 \):

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

and calculating \( y^2 \):

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

Combine \( x^2 + y^2 \):

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

Next, we simplify the numerator:

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

Thus,

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

Therefore, we have proved that

\[
x^2 + y^2 = 1.
\]

Hence,

\[
\boxed{1}.
\]