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Prove the following.
If a + ib = (x+iy)/(x-iy), prove that a² + b² = 1 and b/a = (2xy)/(x²-y²).
If $a + ib = \frac{x+iy}{x-iy}$, prove that $a^2 + b^2 = 1$ and $\frac{b}{a} = \frac{2xy}{x^2-y^2}$.
Given: $a + ib = \frac{x+iy}{x-iy}$
To show: $a^2 + b^2 = 1 \text{ and } \frac{b}{a} = \frac{2xy}{x^2-y^2}$
Answer
Answer
Multiply numerator and denominator by the conjugate $x + iy$ and compare real and imaginary parts.
- Step 1: $a + ib = \frac{x+iy}{x-iy} = \frac{(x+iy)^2}{(x-iy)(x+iy)}$, multiplying by the conjugate of the denominator.
- Step 2: $(x+iy)^2 = x^2 - y^2 + 2ixy$ (using $i^2 = -1$) and $(x-iy)(x+iy) = x^2 + y^2$.
- Step 3: $a + ib = \frac{x^2-y^2}{x^2+y^2} + i\,\frac{2xy}{x^2+y^2}$.
- Step 4: Equating real and imaginary parts: $a = \frac{x^2-y^2}{x^2+y^2}$ and $b = \frac{2xy}{x^2+y^2}$.
- Step 5: $a^2 + b^2 = \frac{(x^2-y^2)^2 + 4x^2y^2}{(x^2+y^2)^2} = \frac{(x^2+y^2)^2}{(x^2+y^2)^2} = 1$, since $(x^2-y^2)^2 + 4x^2y^2 = (x^2+y^2)^2$.
- Step 6: $\frac{b}{a} = \frac{2xy}{x^2+y^2} \times \frac{x^2+y^2}{x^2-y^2} = \frac{2xy}{x^2-y^2}$. Hence $a^2 + b^2 = 1$ and $\frac{b}{a} = \frac{2xy}{x^2-y^2}$.