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Let $f: \mathbb{C} \rightarrow \mathbb{C}$ a $C^1$ function (i.e. real and imaginary part $f_1, f_2$ are continuously differentiable, where $f=f_1 + i \cdot f_2$) and let $\gamma: \mathbb{R} \rightarrow \mathbb{C}, t \mapsto \gamma(t)\, \, $ $C^1$. Then we have that \begin{align} \frac{d}{dt} \, \, f(\gamma(t)) = \frac{\partial f(\gamma(t))}{\partial z} \cdot \gamma'(t) + \frac{\partial f}{\partial \overline{z}} \cdot \overline{\gamma'(t)} \end{align}

I have some trouble to do the calculation to get the formula (I tried to use the Cauchy-Riemann equations but it doesn't work, because the function $f$ is not holomorphic). Any suggestion? Thanks in advance!

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  • $\begingroup$ I think real and imaginary parts of $f$ will not help you. Perhaps you need $z = \gamma(t)$ and $\overline{z} = \overline{\gamma(t))$. $\endgroup$ – GEdgar Sep 25 '18 at 19:01
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What you are saying is that you have the path in the complex plane that is specified by a parametrization $\gamma(t)$. So you have that $z(t) = \gamma(t) \Rightarrow \bar{z} = \bar{\gamma}$. Then from the multivariable chain rule you have that

$$ \frac{df(z,\bar{z})}{dt} = \frac{\partial f}{\partial z}\frac{d z}{d t} + \frac{\partial f}{\partial \bar{z}}\frac{d \bar{z} }{d t}$$

Hence,

$$ \frac{df(z,\bar{z})}{dt} = \frac{\partial f}{\partial z} \gamma'(t) + \frac{\partial f}{\partial \bar{z}} \bar \gamma'(t)$$

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  • $\begingroup$ I think you mean $$\frac{df(z, \bar{z})}{dt} = \frac{\color{red} \partial f}{\color{red} \partial z} \frac{\color{blue} dz}{\color{blue} dt} + \frac{\color{red} \partial f}{\color{red} \partial \bar{z}} \frac{\color{blue} d \bar{z}}{\color{blue} dt}$$ $\endgroup$ – Mattos Sep 26 '18 at 4:54
  • $\begingroup$ Yes, thank you. My mistake. You mind making an edit? I’m on my phone. $\endgroup$ – InertialObserver Sep 26 '18 at 4:55

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