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In short, is the derivative of the conjugate of a function, the conjugate of the derivative of that function?

We assume that the function is complex-valued.

I think this property is true when the derivation variable is real, but my question is: if this is the case, can't we always do a change of variable so that we change from a real variable to a complex one? It's a bit confusing to me.

Hans
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  • With the complex-analysis tag, I assume you mean you have a complex-differentiable function $f$ from a domain in $\Bbb C$ to $\Bbb C$. In that case, unless $f$ is constant, the conjugate function $\bar{f}$ is not differentiable, as follows from the Cauchy-Riemann equations. – Travis Willse May 11 '17 at 14:04
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    If both $f(z)$ and $\overline{f(z)}$ are differentiable with respect to the complex variable $z$, then $f$ must be constant. – Ben Grossmann May 11 '17 at 14:04
  • You're right about complex-valued functions of a real variable. – Ben Grossmann May 11 '17 at 14:05
  • You may want to experiment with polynomials in $z$ and $\bar z$. – Somos May 11 '17 at 22:07

1 Answers1

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In all non-trivial cases the conjugate of a complex-differentiable function is not complex-differentiable itself, so it makes no sense talking about its derivative. To see this, let $f(z)$ be a non-constant complex-differentiable function. Then $f(z)$ and $\overline{f(z)}$ agree on this uncountable set with a limit point:

$$\{z\in\Bbb C\mid \Im (f(z))=0\}.$$

For complex differentiable functions, this suffices to show that they are equal which is a contradiction because they are non-constant, hence cannot be real-valued. (I honestly forgot the name of the theorem stating the equality).

But if you only talk about complex valued functions $f:\Bbb R\to\Bbb C$, then yes, $\overline{f'(x)}=(\overline{f(x)})'$. You can see this via linearity. Differentiation commutes with $i$. Let $f(x)=a(x)+i b(x)$, then

$$(\overline{f(x)})'=(a(x)-i b(x))'=a'(x)-(i b(x))'=a'(x)-ib'(x)=\overline{f'(x)}.$$

M. Winter
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