Is the conjugate of a complex function holomorphic?

**aukie** · November 18th 2010, 05:53 PM

Hello

Suppose $f(z)$ is holomorphic on the open disc $D(0,1)$ of the complex plane.

How would one go about proving $g(z) = \overline{ f( \overline{z} ) }$ is holomorphic in $D(0,1)$ ?

I attepted this by writing out the definition of the derivative, substituting $\overline{ f( \overline{z} ) } = u(Re(z), - Im(z)) - i v(Re(z), - Im(z))$ doing the algebra and using the facts that the real and imaginery parts of $f$ are differentiable and if $z \in D(0,1)$ then $\overline{z} \in D(0,1)$

Is this the way this question would be tackled?? i'm not sure if my proof is a hundred percent correct.

Also I want to prove $k(z)=\overline{ f( z ) }$ is differentiable at $a \in D(0,1)$ iff f'(a) = 0.

I'm a bit stuck on this part. Any ideas?

Incidently this is question 5.11 from Introduction to Complex Analysis by Priestly.

**Prove It** · November 18th 2010, 06:00 PM

You would need to check the Cauchy-Riemann equations.

**aukie** · November 18th 2010, 06:24 PM

how can i check the cauchy conditions? i have no information about the continuity of the partial derivatives of the real and imaginary parts of f. I only know that f is holomorphic on the open disc and thus continuous but the cauchy equations only provide a sufficient condition for differentiability when the partial derivatives are continuous.

**Prove It** · November 18th 2010, 06:57 PM

First of all, when you're asking "Is the complex conjugate of a complex function holomorphic?" do you mean $\displaystyle \overline{f(z)}$ ?

**aukie** · November 18th 2010, 08:22 PM

no i meant the function g(z) as defined above, didnt quite know how to word this briefly and accurately for the title.

**Jose27** · November 18th 2010, 09:07 PM

In general if $U\subset \mathbb{C}$ is an open set and $f:U\rightarrow \mathbb{C}$ is holomorphic then $g(z)=\overline{f(\overline{z})}$ is holomorphic in $U'=\{ z\in \mathbb{C} : \overline{z}\in U\}$ , and to see this just notice $\frac{g(z)-g(z_0)}{z-z_0}=\frac{\overline{f(\overline{z})-f(\overline{z_0})}}{z-z_0}=\overline{ \left( \frac{f(\overline{z})-f(\overline{z_0})}{\overline{z-z_0}} \right) } \rightarrow \overline{f'(\overline{z_0})}$ .

For the second one use the Cauchy-Riemann equations together with the definition of the conjugation operation to conclude that the partials must be zero.

**aukie** · November 19th 2010, 03:19 AM

Can you descirbe how the limit operation is performed please. I dont see how to take the limit of the lhs. Is this correct:

$\lim_{z \to z_0}<br /> \frac{g(z)-g(z_0)}{z-z_0}=\lim_{z \to z_0}\frac{\overline{f(\overline{z})-f(\overline{z_0})}}{z-z_0}=\lim_{z \to z_0}\overline{ \left( \frac{f(\overline{z})-f(\overline{z_0})}{\overline{z-z_0}} \right) } = \overline{ \left( \lim_{z \to z_0}\frac{f(\overline{z})-f(\overline{z_0})}{\overline{z-z_0}} \right) } = \overline{f'(\overline{z_0})}$

I've used the fact $z \mapsto \overline{z}$ is a continuous function so I can bring the limit inside the bracket. Is that right?

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