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

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.

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