Why ℑ 1 1 + e − s + i a = sin ⁡ ( a ) cos ⁡ ( a ) + cosh ⁡ ( s ) ?

Question

Why   ℑ      1          1      +              e                  −          s          +          i          a                      =            sin      ⁡      (      a      )              cos      ⁡      (      a      )      +      cosh      ⁡      (      s      )      ?

Solomon Ferguson · Accepted Answer

I get a slightly different result. Consider the following calculation.

\begin{aligned} ℑ \frac{1}{1 + e^{- s + i a}} & = ℑ \frac{1}{1 + e^{- s} (\cos (a) + i \sin (a))} \\ = ℑ \frac{1}{1 + e^{- s} \cos (a) + i e^{- s} \sin (a)} \cdot \frac{1 + e^{- s} \cos (a) - i e^{- s} \sin (a)}{1 + e^{- s} \cos (a) - i e^{- s} \sin (a)} \\ = ℑ \frac{1 + e^{- s} \cos (a) - i e^{- s} \sin (a)}{(1 + e^{- s} \cos (a))^{2} + (e^{- s} \sin (a))^{2}} \\ = \frac{- e^{- s} \sin (a)}{1 + 2 e^{- s} \cos (a) + e^{- 2 s} \cos (a)^{2} + e^{- 2 s} \sin (a)^{2}} \\ = \frac{- e^{- s} \sin (a)}{1 + 2 e^{- s} \cos (a) + e^{- 2 s}} \\ = \frac{- \sin (a)}{e^{s} + 2 \cos (a) + e^{- s}} \\ = \frac{- \sin (a)}{2 (\frac{e^{s} + e^{- s}}{2} + \cos (a))} = - \frac{1}{2} \frac{\sin (a)}{\cosh (s) + \cos (a)} \end{aligned}

I get a factor of -1/2 in front of your suggested result.
Your confusion about the expression

ℑ \frac{e^{- i a}}{e^{- i a} + 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.

Why <mi mathvariant="normal">ℑ 1 1 +

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