How to find the real or imaginary part of an equation involving complex numbers? I am currently using the Debye model and need to find the real and imaginary parts of the equation. The Debye equation is
$$ \epsilon_\text{r} = \epsilon_\infty + \frac{\epsilon_\text{s} - \epsilon_\infty}{1+i\frac{\omega\epsilon}{\sigma}} - i\frac{\sigma}{\omega\epsilon_0}, $$
where $i^2 = -1$.
How would I find $\mathfrak{R}\{\epsilon_\text{r}\}$ and $\mathfrak{I}\{\epsilon_\text{r}\}$ from this equation?
 A: Assuming all symbols are real except for the complex unit. Then the second term is fine; consider the first:
\begin{eqnarray}
\frac{\epsilon_{s}-\epsilon_{\infty}}{1+i\frac{\omega \epsilon}{\sigma}} &=& \frac{\epsilon_{s}-\epsilon_{\infty}}{1+i\frac{\omega \epsilon}{\sigma}}\frac{i}{i}\\
&=& \frac{i(\epsilon_{s}-\epsilon_{\infty})}{i-\frac{\omega \epsilon}{\sigma}}
\end{eqnarray}
Now multiply both numerator and denominator by the conjugate of the denominator:
\begin{eqnarray}
\frac{(-i-\frac{\omega \epsilon}{\sigma})}{(-i-\frac{\omega \epsilon}{\sigma})}\frac{i(\epsilon_{s}-\epsilon_{\infty})}{i-\frac{\omega \epsilon}{\sigma}}
&=& \frac{i(\epsilon_{s}-\epsilon_{\infty})(-i-\frac{\omega \epsilon}{\sigma})}{1+\frac{\omega^{2}}{\epsilon^{2}\sigma^{2}}}\\
\end{eqnarray}
Now multiply out the brackets on the numerator and include the second term I omitted above and then collect real and imaginary parts.
Good luck!
A: Outcome with Autolatry's help:
$$ \epsilon_r = \epsilon_\infty + \frac{ \epsilon_\text{s} + \left(\frac{\omega\epsilon}{\sigma}-1 \right)  \epsilon_\infty }{1+\frac{\omega^2\epsilon^2}{\sigma^2}} - i\left[\frac{\omega\epsilon\sigma}{\sigma^2+\omega^2\epsilon^2} + \frac{\sigma}{\omega\epsilon_0}\right],$$
and so
$$ \mathfrak{R}(\epsilon_r) = \epsilon_\infty + \frac{ \epsilon_\text{s} + \left(\frac{\omega\epsilon}{\sigma}-1 \right)  \epsilon_\infty }{1+\frac{\omega^2\epsilon^2}{\sigma^2}} $$
$$ \mathfrak{I}(\epsilon_r) = -\frac{\omega\epsilon\sigma}{\sigma^2+\omega^2\epsilon^2} - \frac{\sigma}{\omega\epsilon_0}. $$
I'm sure these could be simplified a little further but they suffice for now.
