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Please do not vote to close it as I want to find errors in my proof, which cannot be rectified on previously answered question.

I want a different proof using Cauchy Riemann equation.

Let $D\subset \mathbb C$ be a domain and suppose f is holomorphic on $D$.

Show that $\overline{f(\overline{z})}$ is holomorphic on the domain $D^*:=\{\overline z: z\in D\}$.

Attempt:

let $z= x+i y$ and $f(z)=u(x,y)+iv(x,y)$

$f$ is holomorphic on $D \Rightarrow u_x=v_y$ and $u_y=-v_x$

To show: $\overline{f(\overline{z})}$ is holomorphic on the domain $D^*$

Let $w\in D^* \Rightarrow w=\overline z$ for some $z \in D$

To show: $\overline{f(\overline{w})}$ satisfy Cauchy Riemann equation.

i.e. To Show: $\overline{f({z})}$ satisfy Cauchy Riemann equation.

$\overline{f({z})}= u(x,y)-iv(x,y)$

Let $v_1=-v$

$\overline{f({z})}= u(x,y)+iv_1(x,y)$

i.e. To show: $u_x={v_1}_y$ and $u_y=-{v_1}_x$

But $-v_y={v_1}_y$ and $-v_x=-{v_1}_x$

$\Rightarrow u_x=-v_y$ and $u_y=v_x$

which is not what I want.

Where I go wrong ?

User
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  • @NajibIdrissi, Can you please help me in finding error in this. – User Mar 09 '16 at 09:02
  • Please do not vote to close it as I want to find errors in my proof, which cannot be rectified on previously answered question. – User Mar 09 '16 at 09:21
  • @Najibldrissi, I have doubt in that solution too, can you help me. – User Mar 09 '16 at 09:54
  • It seems the discussion of the domain $D$ or $D^* = \overline D$ is a bit confused. As the original Question would seem to suggest, the point is that if $f(z)$ is holomorphic on $D$, then $\overline{f(\overline z)}$ is also holomorphic on $D$. – hardmath Mar 09 '16 at 14:35

2 Answers2

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Your problem: $f(\bar z) = u(x,-y) + i v(x,-y)$.

Alternative solution: use that $f$ is locally a power series.

EDIT: $$f(z) = f(x+iy) = u(x,y) + i v(x,y),$$ $$ \tilde u(x,y) + i\tilde v(x,y) = \tilde f(z) = \overline{f(\overline z)} = u(x,-y) - i v(x,-y), $$ $$\eqalign{ \frac{\partial\tilde u}{\partial x}(x,y) = \frac{\partial u}{\partial x}(x,-y)\frac{\partial x}{\partial x} = \frac{\partial u}{\partial x}(x,-y) &= \frac{\partial v}{\partial y}(x,-y) = \frac{\partial(-v)}{\partial y}(x,-y)\frac{\partial(-y)}{\partial y} = \frac{\partial\tilde v}{\partial y}(x,y),\cr \frac{\partial\tilde u}{\partial y}(x,y) = \frac{\partial u}{\partial y}(x,-y)\frac{\partial(-y)}{\partial y} = -\frac{\partial u}{\partial y}(x,-y) &= \frac{\partial v}{\partial x}(x,-y) = -\frac{\partial(-v)}{\partial x}(x,-y)\frac{\partial x}{\partial x} = -\frac{\partial\tilde v}{\partial x}(x,y). }$$

Martín-Blas Pérez Pinilla
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We have to prove that the function $$g(w):=\overline{f(\bar w)}$$ is holomorphic on $D^*$. To this end fix a point $w\in D^*$ and consider a variable complex increment vector $W$ attached at $w$. Then $$g(w+W)-g(w)=\overline{f(\bar w+\bar W)-f(\bar w)}=\overline{f'(\bar w)\bar W+o(|\bar W|)}\qquad(W\to0)\ .$$ It follows that $$g(w+W)-g(w)=\overline{f'(\bar w)}\>W+o(|W|)\qquad(W\to0)\ ,$$ and this shows that $g$ is complex differentiable at $w$ with $g'(w)=\overline{f'(\bar w)}$.

Christian Blatter
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