Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 $z\in D$

How can I solve this problem? Please help me anyone.

share|cite|improve this question

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...

share|cite|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.