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I must verify if the real part of the following expression

$$z = \frac{1 + i}{\sigma \delta \left[ 1 - e^{-(1 + i)t/\delta} \right] }$$

is

$$\Re(z) = \frac{1}{\sigma \delta} \frac{1}{1 - e^{-t/\delta}}$$

where $t, \sigma, \delta$ are real, positive quantities and $i$ is the imaginary unit.

1) First of all, I think that

$$\frac{i}{\sigma \delta \left[ 1 - e^{-(1 + i)t/\delta} \right] }$$

is not purely imaginary, due to the complex exponential in the denominator, so it is anyway to be considered when trying to obtain $\Re(z)$.

2) Moreover $\Re(z)$ does not contain any $\cos$ or $\sin$ terms and this is confusing to me, because in the denominator I must consider $e^{iw} = \cos(w) + i\sin(w)$ and even Wolfram gives a result with those quantities.

Is there a way to simplify the $\cos$ and $\sin$ terms from $z$ and obtain the $\Re(z)$ I wrote? If yes, what can I observe in order to cancel them?

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I've checked my answer with Mathematica and it gives me the same, it's much more complicated as you thought, so your answer isn't good!


$$z = \frac{1 + i}{\sigma \delta \left(1 - e^{-(1 + i)t/\delta} \right)}= \frac{1 + i}{\sigma \delta \left(1 - e^{-\frac{(1+i)t}{\delta}} \right)}$$

Assuming $\sigma,\delta,t\in \mathbb{R}$:

$$\Re\left(\frac{1 + i}{\sigma \delta \left(1 - e^{-\frac{(1+i)t}{\delta}} \right)}\right)=$$ $$\Re\left(\frac{1 + i}{\sigma \delta - \sigma \delta e^{-\frac{(1+i)t}{\delta}} }\right)=$$

$$\Re\left(\frac{1+i}{\sigma \delta}\right)+\Re\left(\frac{1+i}{\sigma \delta\left(-1+e^{-\frac{(1+i)t}{\delta}}\right)}\right)=$$

$$\frac{1}{\sigma \delta}+\Re\left(\frac{1+i}{\sigma \delta\left(-1+e^{-\frac{(1+i)t}{\delta}}\right)}\right)=$$

$$\frac{1}{\sigma \delta}+\frac{\Re\left(\frac{1+i}{-1+e^{-\frac{(1+i)t}{\delta}}}\right)}{\sigma \delta}=$$

$$\frac{1}{\sigma \delta}+\frac{\Re\left(\frac{1}{-1+e^{-\frac{(1+i)t}{\delta}}}\right)-\Im\left(\frac{1}{-1+e^{-\frac{(1+i)t}{\delta}}}\right)}{\sigma \delta}=$$

$$\frac{1}{\sigma \delta}+\frac{\frac{1-e^{\frac{t}{\delta}}\cos\left(\frac{t}{\delta}\right)}{-e^{\frac{2t}{\delta}}+2e^{\frac{t}{b}}\cos\left(\frac{t}{\delta}\right)-1}-\left(-\frac{e^{\frac{t}{\delta}}\sin\left(\frac{t}{\delta}\right)}{e^{\frac{2t}{\delta}}\sin^2\left(\frac{t}{\delta}\right)+\left(e^{\frac{t}{\delta}}\cos\left(\frac{t}{\delta}\right)-1\right)^2}\right)}{\sigma \delta}=$$

$$\frac{1}{\sigma \delta}+\frac{\frac{1-e^{\frac{t}{\delta}}\cos\left(\frac{t}{\delta}\right)}{-e^{\frac{2t}{\delta}}+2e^{\frac{t}{\delta}}\cos\left(\frac{t}{\delta}\right)-1}+\frac{e^{\frac{t}{\delta}}\sin\left(\frac{t}{\delta}\right)}{e^{\frac{2t}{\delta}}\sin^2\left(\frac{t}{\delta}\right)+\left(e^{\frac{t}{\delta}}\cos\left(\frac{t}{\delta}\right)-1\right)^2}}{\sigma \delta}=$$

$$\frac{1}{\sigma \delta}+\frac{e^{\frac{t}{\delta}}\sin\left(\frac{t}{\delta}\right)+e^{\frac{t}{\delta}}\cos\left(\frac{t}{\delta}\right)-1}{\sigma \delta\left(e^{\frac{2t}{\delta}}-2e^{\frac{t}{\delta}}\cos\left(\frac{t}{\delta}\right)+1\right)}$$

So:

$$\Re\left(\frac{1 + i}{\sigma \delta \left(1 - e^{-\frac{(1+i)t}{\delta}} \right)}\right)=\frac{1}{\sigma \delta}+\frac{e^{\frac{t}{\delta}}\sin\left(\frac{t}{\delta}\right)+e^{\frac{t}{\delta}}\cos\left(\frac{t}{\delta}\right)-1}{\sigma \delta\left(e^{\frac{2t}{\delta}}-2e^{\frac{t}{\delta}}\cos\left(\frac{t}{\delta}\right)+1\right)}$$

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$$\mathrm{Re} \left[\frac{1 + i}{\sigma \delta \left( 1 - e^{-(1 + i)t/\delta} \right) }\right] = \mathrm{Re}\left[\frac{(1 + i)\left( 1 - e^{-(1 - i)t/\delta} \right)}{\sigma \delta \left( 1 - e^{-(1 + i)t/\delta} \right) \left( 1 - e^{-(1 - i)t/\delta} \right)} \right] = \frac{1 - \mathrm{Re}\left[e^{-(1 - i)t/\delta} \right] - \mathrm{Re}\left[i e^{-(1 - i)t/\delta}\right]}{2 \sigma \delta e^{-t/\delta} (\cosh(t/\delta) - \cos(t/\delta))} = \frac{1 - \mathrm{Re}\left[e^{-(1 - i)t/\delta} \right] + \mathrm{Im}\left[ e^{-(1 - i)t/\delta}\right]}{2 \sigma \delta e^{-t/\delta} (\cosh(t/\delta) - \cos(t/\delta))} = \frac{1 - e^{-t/\delta} (\cos (t/\delta) - \sin(t/\delta))}{2 \sigma \delta e^{-t/\delta} (\cosh(t/\delta) - \cos(t/\delta))} = \frac{e^{t/\delta} - \cos (t/\delta) + \sin(t/\delta)}{2 \sigma \delta (\cosh(t/\delta) - \cos(t/\delta))}.$$

It doesn't seem that this expression can be reduced to the supposed real part in the answer.

Furthermore, plotting the real part and its supposed value for $\delta = \sigma = 1$ shows they are quite different functions.

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  • $\begingroup$ A quickly check in Mathematica gives me that you're wrong! $\endgroup$ – Jan Jul 15 '15 at 14:10
  • $\begingroup$ try it for some values and you will see the difference, your $=$ signs are not allowed because it isn't equal to each other $\endgroup$ – Jan Jul 15 '15 at 14:18
  • $\begingroup$ @JanEerland Odd, Mathematica simplifies the difference between your answer and mine to $0$. $\endgroup$ – Budenn Jul 15 '15 at 14:45
  • $\begingroup$ But there is a difference so your $=$ (equal) signs are not good you're not allowed to use them! $\endgroup$ – Jan Jul 15 '15 at 14:47
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For $t/\delta=\pi$ and ignoring the inessential $\sigma\delta$,

$$z=\frac{1+i}{1+e^{-\pi}}$$

and

$$\Re(z)=\frac1{1+e^{-\pi}}.$$

This doesn't match

$$\frac1{1-e^{-\pi}}.$$

So, no.

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Without computing:

The denominator of $z$ has non-trivial roots (when the argument of the exponential is $i2k\pi$); not the denominator of $\Re(z)$.

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