I have managed to prove chain rule property which says that:

$\large \frac{\partial (f \circ g)}{\partial z} = \frac{\partial f}{\partial z} \frac{\partial g}{\partial z} + \frac{\partial f}{\partial \bar{z}} \frac{\partial \bar{g}}{\partial z} $


$\large \frac{\partial (f \circ g)}{\partial \bar{z}} = \frac{\partial f}{\partial z} \frac{\partial g}{\partial \bar{z}} + \frac{\partial f}{\partial \bar{z}} \frac{\partial \bar{g}}{\partial \bar{z}} $

But I am supposed to show from the above the conjugate properties of the devivatives, which are:

$\large \overline{\frac{\partial f}{\partial z}} = \frac{\partial \bar{f}}{\partial \bar{z}}$ and $\large \overline{\frac{\partial f}{\partial \bar{z}}} = \frac{\partial \bar{f}}{\partial {z}}$

I'm not getting anywhere trying to prove this using what I proved before. Any help would be awesome.


You can think of $f$ as a function of $x$ and $y$ and those as functions of $z$ and $\overline{z}$ given the relations $$ x = \frac{1}{2}(z+\overline{z}) \hspace{0.5cm} \text{and} \hspace{0.5cm} y=\frac{1}{2i}(z-\overline{z}).$$ Note that $$ \frac{\partial x}{\partial z}=\frac{1}{2}, \hspace{0.5cm} \frac{\partial x}{\partial \overline{z}}=\frac{1}{2}, \hspace{0.5cm} \frac{\partial y}{\partial z}=\frac{1}{2i}=-\frac{i}{2}, \hspace{0.5cm} \text{and} \hspace{0.5cm} \frac{\partial y}{\partial \overline{z}}=-\frac{1}{2i}=\frac{i}{2}. \hspace{0.4cm} (\star)$$ Then $$ \overline{\frac{\partial f}{\partial z}}=\overline{\frac{\partial f}{\partial x}\frac{\partial x}{\partial z}+\frac{\partial f}{\partial y}\frac{\partial y}{\partial z}} =\overline{\frac{1}{2}\big(\frac{\partial f}{\partial x}-i\frac{\partial f}{\partial y}\big)} =\frac{1}{2}\big(\overline{\frac{\partial f}{\partial x}}+i\overline{\frac{\partial f}{\partial y}}\big) =\overline{\frac{\partial f}{\partial x}}\frac{\partial x}{\partial \overline{z}}+\overline{\frac{\partial f}{\partial y}}\frac{\partial y}{\partial \overline{z}} =\frac{\partial \overline{f}}{\partial \overline{z}}. \hspace{0.4cm} (1)$$ Similarly $$ \overline{\frac{\partial f}{\partial \overline{z}}}=\overline{\frac{\partial f}{\partial x}\frac{\partial x}{\partial \overline{z}}+\frac{\partial f}{\partial y}\frac{\partial y}{\partial \overline{z}}} =\overline{\frac{1}{2}\big(\frac{\partial f}{\partial x}+i\frac{\partial f}{\partial y}\big)} =\frac{1}{2}\big(\overline{\frac{\partial f}{\partial x}}-i\overline{\frac{\partial f}{\partial y}}\big) =\overline{\frac{\partial f}{\partial x}}\frac{\partial x}{\partial z}+\overline{\frac{\partial f}{\partial y}}\frac{\partial y}{\partial z} =\frac{\partial \overline{f}}{\partial z}. \hspace{0.4cm} (2)$$

Edit: To clarify (1).

Let $f=u+iv$ we $u= \Re f$ and $v = \Im f$ are the real and imaginary parts of $f$, which are real functions. Then $\overline{f}=u-iv.$ Note that the complex conjugate satisfies $\overline{a+b}=\overline{a}+\overline{b}$ and $\overline{ab}=\overline{a}\overline{b}.$ Also note that $\overline{i}=-i.$ Thus $$\overline{\frac{\partial f}{\partial z}}= \overline{\frac{1}{2}\big(\frac{\partial f}{\partial x}-i\frac{\partial f}{\partial y}\big)} =\frac{1}{2}\big(\overline{\frac{\partial f}{\partial x}}-\overline{i}\overline{\frac{\partial f}{\partial y}}\big) =\frac{1}{2}\big(\overline{\frac{\partial f}{\partial x}}+i\overline{\frac{\partial f}{\partial y}}\big). \hspace{0.4cm} (3)$$ Since $u$ and $v$ are real valuead functions, we also have $$ \frac{\partial f}{\partial x}=\frac{\partial u}{\partial x}+i\frac{\partial v}{\partial x} \hspace{0.4cm} \text{and} \hspace{0.4cm} \frac{\partial f}{\partial y}=\frac{\partial u}{\partial y}+i\frac{\partial v}{\partial y},$$ so $$ \overline{\frac{\partial f}{\partial x}}=\frac{\partial u}{\partial x}-i\frac{\partial v}{\partial x}={\frac{\partial \overline{f}}{\partial x}} \hspace{0.4cm} \text{and} \hspace{0.4cm} \overline{\frac{\partial f}{\partial y}}=\frac{\partial u}{\partial y}-i\frac{\partial v}{\partial y}={\frac{\partial \overline{f}}{\partial y}}.$$ Therefore, using $(\star)$ we have $$ \frac{1}{2}\big(\overline{\frac{\partial f}{\partial x}}+i\overline{\frac{\partial f}{\partial y}}\big) ={\frac{\partial \overline{f}}{\partial x}}\frac{\partial x}{\partial \overline{z}}+{\frac{\partial \overline{f}}{\partial y}}\frac{\partial y}{\partial \overline{z}}=\frac{\partial \overline{f}}{\partial \overline{z}}. \hspace{0.4cm} (4)$$ Hence $(3)$ and $(4)$ gives $(1)$. Similarly one justifies $(2)$.

  • $\begingroup$ the third line would prove that $\bar{\frac{\partial f}{\partial z}} = \frac{\partial f}{\partial \bar{z}}$ $\endgroup$ – user368063 Feb 11 '17 at 17:00
  • $\begingroup$ which is not true...am I wrong? $\endgroup$ – user368063 Feb 11 '17 at 17:09
  • $\begingroup$ @HexaFlexagon I corrected my answer. $\endgroup$ – positrón0802 Feb 11 '17 at 18:02
  • $\begingroup$ I'm not sure I follow the logic.. $\endgroup$ – user368063 Feb 11 '17 at 19:15
  • $\begingroup$ @HexaFlexagon See my edit. Feel free to ask if you don't understand something. $\endgroup$ – positrón0802 Feb 11 '17 at 19:40

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