We are given the equation:

$\tan(i x) = x + i y$

We are asked to prove that:

$x^2 + y^2 + \frac{2}{\sqrt{3}} = 1$

Step 1: Recall the definition of the tangent function for complex arguments

For any complex number $z$, we have:

$\tan(z) = \frac{\sin(z)}{\cos(z)}$

For $z = i x$, we need to compute $\tan(i x)$.

Step 2: Express $\sin(i x)$ and $\cos(i x)$

Using the known results for trigonometric functions of imaginary numbers:

$\sin(i x) = i \sinh(x)$ $\cos(i x) = \cosh(x)$

Thus:

$\tan(i x) = \frac{\sin(i x)}{\cos(i x)} = \frac{i \sinh(x)}{\cosh(x)}$

Step 3: Separate real and imaginary parts

Now, simplify $\frac{i \sinh(x)}{\cosh(x)}$:

$\tan(i x) = i \cdot \frac{\sinh(x)}{\cosh(x)} = i \cdot \tanh(x)$

Since we are given that $\tan(i x) = x + i y$, we conclude:

$i \tanh(x) = x + i y$

Step 4: Equate real and imaginary parts

From the above equation, equating the real and imaginary parts:

• The real part is $0$ on the left-hand side, and $x$ on the right-hand side. Thus, we have $x = 0$.
• The imaginary part is $\tanh(x)$ on the left-hand side, and $y$ on the right-hand side. Thus, we have $y = \tanh(x)$.

Step 5: Prove the given equation

Substituting $x = 0$ and $y = \tanh(0) = 0$ into the expression $x^2 + y^2 + \frac{2}{\sqrt{3}} = 1$:

$0^2 + 0^2 + \frac{2}{\sqrt{3}} = 1$

$\frac{2}{\sqrt{3}} = 1$

However, this does not satisfy the equation as it stands. Let's reconsider the approach for any mistake or different interpretations of the given problem.

Would you like further clarification or details on the approach?

Here are 5 related questions for better understanding:

1. What are the trigonometric functions for imaginary numbers?
2. How does one derive the hyperbolic tangent function?
3. What does equating real and imaginary parts in a complex equation imply?
4. What are common mistakes when dealing with complex numbers?
5. How are hyperbolic functions related to exponential functions?

Tip: When working with complex functions, separating real and imaginary parts is crucial to solving equations.