# Let $f$ be an analytic function defined on $D = \{ z\in\mathbb{C}\colon|z| <1\}$. Then $g\colon D\to \mathbb C$ is analytic if

Let $f$ be an analytic function defined on $D = \{ z\in \mathbb{C}\colon |z| <1\}$. Then $g \colon D\to \mathbb{C}$ is analytic if

1. $g(z) = f( \bar{z})$ for all $z\in D$
2. $g(z) = \overline{(f (z))}$, for all $z\in D$
3. $g(z) = \overline{(f (\bar{z}))}$ for all $z\in D$
4. $g(z) =\bar{i} f (z )$ for all $z\in D$

-
Is this homework? If so, you should add the homework tag. Also, what are your thoughts on the problem? –  Antonio Vargas Dec 16 '12 at 10:06
Is there any particular reason not to write the complex conjugate of $z$ as $\overline z$ (and that of $i$ as $-i$)? This is not very readable on first sight (looks like there are derivatives all around, but there are none). –  Marc van Leeuwen Dec 16 '12 at 10:27
1 and 2 are clearly false if you take $f(z)=z$. The 4th is clearly true since $g(z)=-if(z)$. What can you say about 3? (Hint: Cauchy Riemann Equations)
Let me be more thorough about 3. Let $f(x,y)=u(x,y)+iv(x,y)$. Then $g(x,y)=u(x,-y)-iv(x,-y)$. Therefore, $$\frac{\partial u(x,-y)}{\partial x}=\frac{\partial u}{\partial x}(x,-y)$$ $$\frac{\partial u(x,-y)}{\partial y}=-\frac{\partial u}{\partial y}(x,-y)$$ $$\frac{\partial [-v(x,-y)]}{\partial x}=-\frac{\partial v}{\partial x}(x,-y)$$ $$\frac{\partial [-v(x,-y)]}{\partial y}=-\frac{-\partial v}{\partial y}(x,-y)=\frac{\partial v}{\partial y}(x,-y)$$ The CR equations are satisfied for $f$ and so...