We need to solve the given problem. The equation provided is:

$\tan\left(i \log \frac{x - iy}{x + iy}\right) = 2,$ and we are tasked with proving that $x^2 - y^2 = xy$.

Step-by-step solution:

1. Simplify the complex logarithm: The argument of the logarithm is a fraction: $z = \frac{x - iy}{x + iy}.$ Write $z$ in polar form: $x - iy = \sqrt{x^2 + y^2} e^{-i\theta}, \quad x + iy = \sqrt{x^2 + y^2} e^{i\theta}.$ So: $\frac{x - iy}{x + iy} = e^{-2i\theta},$ where $\theta = \tan^{-1}\left(\frac{y}{x}\right)$.

2. Apply the logarithm: Using the logarithmic property of exponential expressions: $\log\left(\frac{x - iy}{x + iy}\right) = \log\left(e^{-2i\theta}\right) = -2i\theta.$

3. Simplify the given expression: Substituting this result into the original equation: $\tan\left(i \log \frac{x - iy}{x + iy}\right) = \tan\left(i(-2i\theta)\right) = \tan(2\theta).$ Hence: $\tan(2\theta) = 2.$

4. Use the tangent double-angle identity: The identity for $\tan(2\theta)$ is: $\tan(2\theta) = \frac{2\tan\theta}{1 - \tan^2\theta}.$ Let $\tan\theta = \frac{y}{x}$, so: $\tan(2\theta) = \frac{2\frac{y}{x}}{1 - \left(\frac{y}{x}\right)^2}.$ Simplifying: $\tan(2\theta) = \frac{\frac{2y}{x}}{1 - \frac{y^2}{x^2}} = \frac{2y}{x - \frac{y^2}{x}} = \frac{2xy}{x^2 - y^2}.$

5. Set $\tan(2\theta) = 2$: From the equation: $\frac{2xy}{x^2 - y^2} = 2.$ Multiply through by $x^2 - y^2$: $2xy = 2(x^2 - y^2).$ Dividing by 2: $xy = x^2 - y^2.$

6. Rearrange to the required form: $x^2 - y^2 = xy.$

Conclusion:

We have successfully proven that: $x^2 - y^2 = xy.$

Let me know if you'd like detailed clarifications or further assistance!

5 Related Questions:

1. What is the geometric interpretation of $\tan^{-1}\left(\frac{y}{x}\right)$?
2. How does the complex logarithm work in polar coordinates?
3. Can you derive $\tan(2\theta)$ using the sum of angles formula?
4. What happens if $\tan(2\theta) = -2$ instead?
5. How do Euler’s formulas simplify complex expressions?

1 Tip:

Always represent complex fractions in polar form to simplify logarithms and trigonometric identities effectively.