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.