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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?

Genius
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  • Firstly, be aware of what $\text{Re} \circ z$ is a function $\mathbb{R} \rightarrow \mathbb{R}^2 \rightarrow \mathbb{R}$ given by $t \mapsto z(t) \mapsto \text{Re}(z(t))$. – G. Chiusole Aug 27 '19 at 13:42

2 Answers2

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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 $\operatorname{Im}z=\frac{z-z^\ast}{2i}$.

J.G.
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    Thank you! The trick of writing $\text{Re}$ and $\text{Im}$ explicitly is great. However, I think there is a few mistakes: $\frac{d}{da} \text{Re}(z(t)) = \frac{z(t)+z^(t)}{2} = \frac{1}{2}(\frac{d}{da} z(t) + \frac{d}{da} z^(t)) = \frac{1}{2}(\exp(bt)(\cos(t\omega)+i\sin(t\omega))+\exp(bt)(\cos(t\omega)-i\sin(t\omega))) = \exp(bt)cos(t\omega) = \text{Re}(\frac{d}{da} z(t))$ and accordingly $\frac{d}{da} \text{Im}(z(t)) = -i\text{Im}(\frac{d}{da} z(t))$ – Genius Aug 27 '19 at 17:14
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    @Genius $\frac{d}{da}\left(z^\ast\right)\ne\left(\frac{dz}{da}\right)^\ast$ in general because $a$ is complex viz. $\frac{da^\ast}{da}=0$. For example, with $z=a$ this inequality becomes $\frac{d0}{da}\ne1^\ast$. – J.G. Aug 27 '19 at 17:22
  • According to what I have found out by now, $a$ and $a*$ can not be treated as independent variables, one has to follow the Cauchy-Riemann equations instead to find the solution – Genius Aug 30 '19 at 17:38
  • @Genius You read wrong, then. See also here. – J.G. Aug 30 '19 at 18:00
  • Posted my solution. Let me know what you think. For the gradient approach I am implementing, my solution actually works. – Genius Aug 30 '19 at 18:14
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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*}

Genius
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  • You might want to double-check how often $b$ appears in your final results. Your Cauchy-Riemann equations can be simplified with results such as $$\operatorname{Re}\operatorname{Re}w=\operatorname{Re}w,,\operatorname{Re}\operatorname{Im}w=\operatorname{Im}w,,\operatorname{Im}\operatorname{Im}w=\operatorname{Im}\operatorname{Re}w=0.$$ – J.G. Aug 30 '19 at 18:37
  • @J.G. Why are our solutions different? What is the correct solution and why? – Genius Sep 02 '19 at 09:04
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    You seem to be differentiating with respect to $a$ rather than $z$. The equations you asked about in the original question were clearly doing the latter. – J.G. Sep 02 '19 at 09:43