Is the conjugate of a complex function holomorphic?

• Nov 18th 2010, 05:53 PM
aukie
Is the conjugate of a complex function holomorphic?
Hello

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

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

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

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 $\displaystyle k(z)=\overline{ f( z ) }$ is differentiable at $\displaystyle a \in D(0,1)$ 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.
• Nov 18th 2010, 06:00 PM
Prove It
You would need to check the Cauchy-Riemann equations.
• Nov 18th 2010, 06:24 PM
aukie
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.
• Nov 18th 2010, 06:57 PM
Prove It
First of all, when you're asking "Is the complex conjugate of a complex function holomorphic?" do you mean $\displaystyle \displaystyle \overline{f(z)}$?
• Nov 18th 2010, 08:22 PM
aukie
no i meant the function g(z) as defined above, didnt quite know how to word this briefly and accurately for the title.
• Nov 18th 2010, 09:07 PM
Jose27
In general if $\displaystyle U\subset \mathbb{C}$ is an open set and $\displaystyle f:U\rightarrow \mathbb{C}$ is holomorphic then $\displaystyle g(z)=\overline{f(\overline{z})}$ is holomorphic in $\displaystyle U'=\{ z\in \mathbb{C} : \overline{z}\in U\}$, and to see this just notice $\displaystyle \frac{g(z)-g(z_0)}{z-z_0}=\frac{\overline{f(\overline{z})-f(\overline{z_0})}}{z-z_0}=\overline{ \left( \frac{f(\overline{z})-f(\overline{z_0})}{\overline{z-z_0}} \right) } \rightarrow \overline{f'(\overline{z_0})}$.

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.
• Nov 19th 2010, 03:19 AM
aukie
Can you descirbe how the limit operation is performed please. I dont see how to take the limit of the lhs. Is this correct:

$\displaystyle \lim_{z \to z_0} \frac{g(z)-g(z_0)}{z-z_0}=\lim_{z \to z_0}\frac{\overline{f(\overline{z})-f(\overline{z_0})}}{z-z_0}=\lim_{z \to z_0}\overline{ \left( \frac{f(\overline{z})-f(\overline{z_0})}{\overline{z-z_0}} \right) } = \overline{ \left( \lim_{z \to z_0}\frac{f(\overline{z})-f(\overline{z_0})}{\overline{z-z_0}} \right) } = \overline{f'(\overline{z_0})}$

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