Understanding the Tanh Function in Complex Variables

We need to find the imaginary part of the complex hyperbolic tangent function, $\tanh(x+iy)$, where $x$ and $y$ are real numbers and $i$ is the imaginary unit.

Derivation of Tanh(x+iy)

We start with the definition of the hyperbolic tangent:

$ \tanh(z) = \frac{\sinh(z)}{\cosh(z)} $

Let $z = x+iy$. We use the identities for $\sinh(a+ib)$ and $\cosh(a+ib)$:

• $ \sinh(a+ib) = \sinh a \cos b + i \cosh a \sin b $
• $ \cosh(a+ib) = \cosh a \cos b + i \sinh a \sin b $

Applying these to $z = x+iy$:

• $ \sinh(x+iy) = \sinh x \cos y + i \cosh x \sin y $
• $ \cosh(x+iy) = \cosh x \cos y + i \sinh x \sin y $

Substitute these into the tanh formula:

$ \tanh(x+iy) = \frac{\sinh x \cos y + i \cosh x \sin y}{\cosh x \cos y + i \sinh x \sin y} $

To simplify, multiply the numerator and the denominator by the conjugate of the denominator, which is $ (\cosh x \cos y - i \sinh x \sin y) $:

Numerator calculation:

$ (\sinh x \cos y + i \cosh x \sin y)(\cosh x \cos y - i \sinh x \sin y) $ $ = \sinh x \cosh x \cos^2 y - i \sinh^2 x \sin y \cos y + i \cosh^2 x \sin y \cos y + \cosh x \sinh x \sin^2 y $ $ = \sinh x \cosh x (\cos^2 y + \sin^2 y) + i \sin y \cos y (\cosh^2 x - \sinh^2 x) $

Using the identities $ \cos^2 y + \sin^2 y = 1 $ and $ \cosh^2 x - \sinh^2 x = 1 $:

$ = \sinh x \cosh x + i \sin y \cos y $

Using the double angle identities $ 2\sinh x \cosh x = \sinh(2x) $ and $ 2\sin y \cos y = \sin(2y) $:

$ = \frac{1}{2}\sinh(2x) + i \frac{1}{2}\sin(2y) $

Denominator calculation:

$ (\cosh x \cos y + i \sinh x \sin y)(\cosh x \cos y - i \sinh x \sin y) $ $ = (\cosh x \cos y)^2 + (\sinh x \sin y)^2 $ $ = \cosh^2 x \cos^2 y + \sinh^2 x \sin^2 y $

Using the identities $ \cosh^2 x = \frac{\cosh(2x)+1}{2} $, $ \sinh^2 x = \frac{\cosh(2x)-1}{2} $, $ \cos^2 y = \frac{1+\cos(2y)}{2} $, and $ \sin^2 y = \frac{1-\cos(2y)}{2} $:

$ = \left(\frac{\cosh(2x)+1}{2}\right)\left(\frac{1+\cos(2y)}{2}\right) + \left(\frac{\cosh(2x)-1}{2}\right)\left(\frac{1-\cos(2y)}{2}\right) $ $ = \frac{1}{4} \left[ (\cosh(2x)+1)(1+\cos(2y)) + (\cosh(2x)-1)(1-\cos(2y)) \right] $ $ = \frac{1}{4} \left[ \cosh(2x) + \cosh(2x)\cos(2y) + 1 + \cos(2y) + \cosh(2x) - \cosh(2x)\cos(2y) - 1 + \cos(2y) \right] $ $ = \frac{1}{4} \left[ 2\cosh(2x) + 2\cos(2y) \right] $ $ = \frac{1}{2} (\cosh(2x) + \cos(2y)) $

Now, combine the simplified numerator and denominator:

$ \tanh(x+iy) = \frac{\frac{1}{2}\sinh(2x) + i \frac{1}{2}\sin(2y)}{\frac{1}{2}(\cosh(2x) + \cos(2y))} $ $ \tanh(x+iy) = \frac{\sinh(2x) + i \sin(2y)}{\cosh(2x) + \cos(2y)} $

Separate the real and imaginary parts:

$ \tanh(x+iy) = \frac{\sinh(2x)}{\cosh(2x) + \cos(2y)} + i \frac{\sin(2y)}{\cosh(2x) + \cos(2y)} $

Identifying the Imaginary Part

The imaginary part of $\tanh(x+iy)$ is the term multiplied by $i$.

$ \text{Im}(\tanh(x+iy)) = \frac{\sin(2y)}{\cosh(2x) + \cos(2y)} $