# Why $\Im \frac{1}{1+e^{-s + i a }}=\frac{\sin(a)}{\cos(a)+\cosh(s)}$?

Why this $$\Im \frac{-2}{1+e^{-s + i a }}$$

equals to this expression:

$$\\\ \frac{\sin(a)}{\cos(a)+\cosh(s)}$$

I was trying to evaluate the Fourier transform of a hyperbolic function and my textbook and the other sources say this equality holds on. I only got to:

$$\Im \frac{e^{-ia}}{e^{-ia}+e^{-s}}$$

Well, generally the imaginary part of $$e^{-ia}$$ is $$\sin(a)$$, but $$e^{-s}$$ is real. So I don't understand the $$\cosh (s)$$ part. I am just so confused.

The fourier tranform has been done on this function:

$$f(x) = \frac{\sinh(ax)}{\sinh(\pi x)}$$

## 1 Answer

I get a slightly different result. Consider the following calculation. \begin{align*} \Im \frac{1}{1+e^{-s+ia}} &= \Im \frac{1}{1+e^{-s}(\cos(a)+i \sin(a))} \\ &=\Im \frac{1}{1+e^{-s}\cos(a) + ie^{-s}\sin(a)} \cdot \frac{1+e^{-s}\cos(a) - ie^{-s}\sin(a)}{1+e^{-s}\cos(a) - ie^{-s}\sin(a)} \\ &= \Im \frac{1+e^{-s}\cos(a) - ie^{-s}\sin(a)}{(1+e^{-s}\cos(a))^2 + (e^{-s}\sin(a))^2} \\ &= \frac{-e^{-s}\sin(a)}{1+2e^{-s}\cos(a)+e^{-2s}\cos(a)^2 + e^{-2s}\sin(a)^2} \\ &= \frac{ -e^{-s}\sin(a)}{1+2e^{-s}\cos(a)+e^{-2s}}\\ &= \frac{ -\sin(a)}{e^{s} +2\cos(a)+e^{-s}}\\ &=\frac{-\sin(a)}{2\left( \frac{e^s+e^{-s}}{2}+\cos(a) \right)} = -\frac{1}{2}\frac{\sin(a)}{\cosh(s)+\cos(a)} \end{align*}

I get a factor of $$-1/2$$ in front of your suggested result.

Your confusion about the expression $$\Im \frac{e^{-ia}}{e^{-ia}+e^{-s}}$$ might arise because you can not see the real and the imaginary parts from this expression, as there is also an $$i$$ in the denominator of the fraction.

• Thank you very much, I missed -2 in the first expression :) – Leif Mar 24 '19 at 2:19
• I suppose the result for $s \gt 0$ is the same as for $s \lt 0$, right? – Leif Mar 25 '19 at 0:03
• Yes, that is true. – Strichcoder Mar 25 '19 at 0:05