Mathematics of phasor representations of sinusoidal variations of fields In physics, we often use phasor representations of sinusoidal variations of fields, such as
$$\mathbf{e}(\mathbf{r}, t) = Re[\mathbf{E}(\mathbf{r})e^{j \omega t}]$$
I have some questions about how to deal with the mathematics of this type of equation. Using Euler's formula, we have that $e^{j \omega t} = \cos(\omega t) + j \sin(\omega t)$. So would the real part just be $\mathbf{e}(\mathbf{r}, t) = \mathbf{E}(\mathbf{r})$? I'm confused, because if we use Euler's formula, we have that $e^{j \omega t} = \cos(\omega t) + j \sin(\omega t)$, so I'm unsure if $\cos(\omega t)$ would also be part of the real part? 
Also, how do we deal with the partial derivative $\dfrac{\partial{\mathbf{e}}}{\partial{t}}$? I ask because we have that 
$$\dfrac{\partial{\mathbf{e}}}{\partial{t}} = \dfrac{\partial{\left\{ Re[\mathbf{E}(\mathbf{r})e^{j \omega t}] \right\}}}{\partial{t}},$$
so it is natural to wonder if this is linear so that we can take the partial derivative first and then the real part, or whether we must take the real part first and then the partial derivative?
 A: The real part of ${\mathbf e}({\mathbf r})={\mathbf E}({\mathbf r})e^{j\omega t}$ is ${\mathbf E}({\mathbf r})\cos (\omega t)$, assuming, as is conventional, that ${\mathbf E}({\mathbf r})$ is real.
The partial time derivative
$$\frac{\partial {\mathbf e}}{\partial t} =  \omega{\mathbf E}({\mathbf r})\left[ -\sin(\omega t) + j\cos(\omega t)\right]   $$
So whether you differentiate the real part of ${\bf e}$ or take the real part of $\partial {\bf e}/\partial t$, makes no difference.
A: 
So would the real part just be $\mathbf{e}(\mathbf{r}, t) = \mathbf{E}(\mathbf{r})$? 



*

*$\mathbf{e}(\mathbf{r}, t) = Re[\mathbf{E}(\mathbf{r})e^{j \omega t}] = \mathbf{E}(\mathbf{r}) \cos(\omega t)$ is already real.



How do we deal with the partial derivative $\dfrac{\partial{\mathbf{e}}}{\partial{t}}$?



*

*$\dfrac{\partial \mathbf e}{\partial t} = \dfrac{\partial \ (\mathbf{E}(\mathbf{r}) \cos(\omega t))}{\partial t} = - \omega  \mathbf{E}(\mathbf{r}) \sin(\omega t)$
If you have not taken the real part yet, 
$$\dfrac{\partial \mathbf e}{\partial t} = \dfrac{\partial \ (\mathbf{E}(\mathbf{r}) \exp(j \omega t))}{\partial t} = j \omega \mathbf{E}(\mathbf{r}) \exp(j \omega t) = \omega  \mathbf{E}(\mathbf{r})[-\sin(\omega t) + j \cos(\omega t)]$$ whose real part will still be $-\omega \mathbf E(\mathbf r) \sin(\omega t)$. The order really does not matter.


*

*In fact, this trick is used many times while solving coupled oscillators. It is much easier to deal with $\exp(j\omega t)$ than $\sin$ and $\cos$ while differentiating.

