to show that $\frac{df}{dz} = \bar{(\frac{d\bar{f}}{d\bar{z}})}$
I need to show to show that $\frac{df}{dz} = \bar{(\frac{d\bar{f}}{d\bar{z}})}$ given that $f : \omega$ ---> $ \mathbb{C} $ and all the partial derivatives are continuous.
I tried using $f=u+iv$ and then using the standard formula for differentiation by $\bar{z} $ but was not getting it.
$\endgroup$1 Answer
$\begingroup$We have
$$2\frac{\partial f}{\partial z} = \frac{\partial f}{\partial x} - i\frac{\partial f}{\partial y}$$
and
$$2\frac{\partial \bar{f}}{\partial \bar{z}} = \frac{\partial \bar{f}}{\partial x} + i\frac{\partial \bar{f}}{\partial y} = \overline{\frac{\partial f}{\partial x} - i\frac{\partial f}{\partial y}} = \overline{2\frac{\partial f}{\partial z}} = 2\overline{\frac{\partial f}{\partial z}},$$
therefore
$$\frac{\partial \bar{f}}{\partial \bar{z}} = \overline{\frac{\partial f}{\partial z}}.$$
By conjugation,
$$\frac{\partial f}{\partial z} = \overline{\frac{\partial \bar{f}}{\partial \bar{z}}}.$$
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