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:

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


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.

  • $\begingroup$ Thank you very much, I missed -2 in the first expression :) $\endgroup$ – Leif Mar 24 at 2:19
  • $\begingroup$ I suppose the result for $s \gt 0$ is the same as for $s \lt 0$, right? $\endgroup$ – Leif Mar 25 at 0:03
  • 1
    $\begingroup$ Yes, that is true. $\endgroup$ – Strichcoder Mar 25 at 0:05

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.