What is the meaning of the derivative of the real and complex part ($\text{Re}'$, $\text{Im}'$)? Given a complex valued function $z(t) = a\exp((b+i \omega)t)$, with $a \in \mathbb{C}$ and $b, \omega, t \in \mathbb{R}$, what is the meaning of $\text{Re}'(z(t))$ and $\text{Im}'(z(t))$ in the partial derivatives
$$\frac{d}{da}\text{Re}(z(t)) = \text{Re}'(z(t))\frac{d}{da}z(t)$$
$$\frac{d}{da}\text{Im}(z(t)) = \text{Im}'(z(t))\frac{d}{da}z(t)$$
after applying the chain rule?
I have difficulties to understand what $\text{Re}'$ and $\text{Im}'$ means conceptually, since I understand $\text{Re}$ and $\text{Im}$ as helper functions to extract the real and imaginary part respectively and thus I don't understand how I can analytically calculate their derivatives?
 A: Treat $a,\,a^\ast$ as independent variables so$$\operatorname{Re}a=\frac{a+a^\ast}{2},\,\operatorname{Im}a=\frac{a-a^\ast}{2i}\implies\partial_a=\frac12\partial_{\operatorname{Re}a}+\frac{1}{2i}\partial_{\operatorname{Im}a}.$$Since $\operatorname{Re}z=e^{bt}(\operatorname{Re}a\cdot\cos\omega t-\operatorname{Im}a\cdot\sin\omega t)$, $$\partial_a\operatorname{Re}z=\frac12 e^{bt}\cos\omega t-\frac{1}{2i}e^{bt}\sin\omega t=\frac12 e^{(b+i\omega)t}.$$Dividing out $dz/da=e^{(b+i\omega)t}$, the first equation requires $$\operatorname{Re}^\prime(z)=\frac12,$$which makes sense because $\operatorname{Re}z=\frac{z+z^\ast}{2}$. Similarly, you can show $\partial_a\operatorname{Im}z=\frac{1}{2i}e^{(b+i\omega)t}$, whence$$\operatorname{Im}^\prime z=\frac{1}{2i},$$which makes sense because $\operatorname{Im}z=\frac{z-z^\ast}{2i}$.
A: The solution can be found by using the Cauchy-Riemann equations:
\begin{align*}
Re'(z(t)) &= \frac{\partial\text{Re}(\text{Re}(z(t)))}{\partial\text{Re}(a)} + i\frac{\partial\text{Im}(\text{Re}(z(t)))}{\partial\text{Re}(a)}\\
&= \frac{\partial\text{Im}(\text{Re}(z(t)))}{\partial\text{Im}(a)} + i\frac{\partial\text{Re}(\text{Re}(z(t)))}{\partial\text{Im}(a)}\\
\end{align*}
\begin{align*}
Im'(z(t)) &= \frac{\partial\text{Re}(\text{Im}(z(t)))}{\partial\text{Re}(a)} + i\frac{\partial\text{Im}(\text{Im}(z(t)))}{\partial\text{Re}(a)}\\
&= \frac{\partial\text{Im}(\text{Im}(z(t)))}{\partial\text{Im}(a)} + i\frac{\partial\text{Re}(\text{Im}(z(t)))}{\partial\text{Im}(a)}\\
\end{align*}
Where the real part of $z(t)$ is $$\text{Re}(z(t)) = \text{Re}(a)\exp(bt)\cos(\omega t) - \text{Im}(a)\exp(bt)\sin(\omega t)$$ and the imaginary part of $z(t)$ is $$\text{Im}(z(t)) = \text{Re}(a)\exp(bt)\sin(\omega t) + \text{Im}(a)\exp(bt)\cos(\omega t)$$
and therefore $$Re'(z(t)) = \exp(bt)\cos(\omega t)$$ and $$Im'(z(t)) = \exp(bt)\sin(\omega t)\text{.}$$
For the whole solution, we have to apply the chain rule, which then finally yields
\begin{align*}
\frac{\partial}{\partial a}Re(z(t)) &= \exp(bt)\cos(\omega t)\exp((b+i\omega)t)\\
&= \exp((2b+i\omega)t)\cos(\omega t)
\end{align*} and 
\begin{align*}
\frac{\partial}{\partial a}Im(z(t)) &= \exp(bt)\sin(\omega t)\exp((b+i\omega)t)\\
&= \exp((2b+i\omega)t)\sin(\omega t)\text{.}
\end{align*}
