Question

The imaginary part of $tanh(x+iy)$ is:

• A. $\frac{sin(2x)}{cos(2x)+cosh(2y)}$
• B. $\frac{sinh(2y)}{cos(2x)+cosh(2y)}$
• C. $\frac{sinh(2x)}{cosh(2x)+cos(2y)}$
• D. $\frac{sin(2y)}{cosh(2x)+cos(2y)}$

Hint:

Remember the conversion $sinh(iy) = i \cdot sin(y)$ and $cosh(iy) = cos(y)$ when dealing with complex arguments in hyperbolic functions.

Answer 1

The Correct Option is D

Concept: To find the real and imaginary parts of a complex hyperbolic function like $tanh(z)$, we use the definition $tanh(z) = \frac{sinh(z)}{cosh(z)}$. We then multiply the numerator and denominator by the complex conjugate of the denominator to separate the terms.

Step 1: Express $tanh(x+iy)$ in terms of sine and cosine.
By definition: \[ tanh(x+iy) = \frac{sinh(x+iy)}{cosh(x+iy)} \] To rationalize, multiply the numerator and denominator by $2 \cdot cosh(x-iy)$: \[ tanh(x+iy) = \frac{2 \cdot sinh(x+iy)cosh(x-iy)}{2 \cdot cosh(x+iy)cosh(x-iy)} \]

Step 2: Apply product-to-sum identities.
Using the identities $2sinhA \cdot coshB = sinh(A+B) + sinh(A-B)$ and $2coshA \cdot coshB = cosh(A+B) + cosh(A-B)$: Numerator: $sinh(2x) + sinh(2iy) = sinh(2x) + i \cdot sin(2y)$ Denominator: $cosh(2x) + cosh(2iy) = cosh(2x) + cos(2y)$

Step 3: Identify the imaginary part.
The expression becomes: \[ tanh(x+iy) = \frac{sinh(2x) + i \cdot sin(2y)}{cosh(2x) + cos(2y)} \] The imaginary part is the coefficient of $i$: \[ Im[tanh(x+iy)] = \frac{sin(2y)}{cosh(2x)+cos(2y)} \]