In ahlfors text it states that(formally)

If the rules of calculus were applicable, we would obtain $$\frac{\partial{f}}{\partial{z}}=\frac12\left(\frac{\partial{f}}{\partial{x}}-i\frac{\partial{f}}{\partial{y}}\right), \frac{\partial{f}}{\partial{\overline{z}}}=\frac12\left(\frac{\partial{f}}{\partial{x}}+i\frac{\partial{f}}{\partial{y}}\right)$$

And we see that $f$ being analytic,

$$\frac{\partial{f}}{\partial{x}} = -i\frac{\partial{f}}{\partial{y}}$$ holds, which gives that, $$\frac{\partial{f}}{\partial{\overline{z}}}=0$$

Now , in text following he states that,

the conjugate function $\overline{f(z)}$ has derivative $0$ w.r.t $z$.

That is , $$\frac{\partial{\overline{f}}}{\partial{z}}=0$$

I didn't get this.

Won't this mean , $\frac{\partial{\overline{f}}}{\partial{z}}=\frac12\left(\frac{\partial{\overline{f}}}{\partial{x}}-i\frac{\partial{\overline{f}}}{\partial{y}}\right)$ , and from $\frac{\partial{\overline{f}}}{\partial{x}} = -i\frac{\partial{\overline{f}}}{\partial{y}}$

I didn't see how it's coming, i suppose their a mistake of mine considering $\frac{\partial{\overline{f}}}{\partial{z}} = \overline {\frac{\partial{f}}{\partial{z}}}$

Please explain, where's the mistake , and what does actually is the difference between differentiating a analytic function and its conjugate.

  • $\begingroup$ The conjugate function is not analytic, so by the chain rule it breaks analycity. $\endgroup$
    – user65203
    Mar 19 '21 at 17:06
  • $\begingroup$ $\frac{\partial{\overline{f}}}{\partial{z}} = \overline {({\frac{\partial{f}}{\partial{\bar z}}})}=0$ (just check definitions of the partials in $z, \bar z$ with $f=u+iv, \bar f=u-iv$) $\endgroup$
    – Conrad
    Mar 19 '21 at 17:15
  • $\begingroup$ @YvesDaoust , if it's not analytic, then, what does applying chain rule means, sorry for my ignorance, please explain a bit more. $\endgroup$
    – Rkb
    Mar 20 '21 at 3:44
  • $\begingroup$ @Conrad i understand the definition of partials in $z$ but please could you restate or provide a source, about the same, thing in $\overline{z}$ $\endgroup$
    – Rkb
    Mar 20 '21 at 3:46
  • 1
    $\begingroup$ with the motivation as noted above, we define $\frac{\partial{f}}{\partial{\overline{z}}}=\frac12\left(\frac{\partial{f}}{\partial{x}}+i\frac{\partial{f}}{\partial{y}}\right)$; from here it is straightforward that $\frac{\partial{\overline{f}}}{\partial{z}} = \overline {({\frac{\partial{f}}{\partial{\bar z}}})}$ $\endgroup$
    – Conrad
    Mar 20 '21 at 13:55

The conjugate function is analytic as a function of $\overline{z}.$ So, letting $u =\overline{z},$ the computation you give implies that $\frac{\partial \overline{f}}{\partial \overline{u}} = 0.$ But $\overline{u} = z,$ so what your professor says is true.

  • $\begingroup$ Thanks i understand, and does the equations, $\frac{\partial{\overline{f}}}{\partial{z}}=\frac12\left(\frac{\partial{\overline{f}}}{\partial{x}}-i\frac{\partial{\overline{f}}}{\partial{y}}\right)$ and $\frac{\partial{\overline{f}}}{\partial{x}} = -i\frac{\partial{\overline{f}}}{\partial{y}}$ holds? Please explain a bit. $\endgroup$
    – Rkb
    Mar 20 '21 at 3:41

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