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Prove: if $f(z)$ differentiable in $|z|<R$ So Does $\overline {f(\overline{z})}$

I have went over previous similar questions but did not find an answer to the following:

We say that $f(z)$ differentiable in $|z|<R$ therefore

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

Looking at $$\overline {f(\overline{z})}=U(x,-y)-iV(x,-y)$$

We can say that it is defined in $|z|<R$ as $|z|=|\overline{z}|<R$

But looking at its $C-R$ equations we get

$$U_x=--V_y\iff U_x=V_y\iff u_x=v_y$$

$$-U_y=-V_x\neq u_y=-v_x$$

How can we conclude from $C-R$ that $\overline {f(\overline{z})}$ is differentiable?

gbox
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    Aside from your abuse of notation, you should have $U_y = - V_x$, since we are looking at $- \frac{\partial U(x,-y)}{\partial y} = - - \frac{\partial U}{\partial y} = U_y$ (I'm also abusing notation; I hope you understand what I wrote) –  Jun 11 '17 at 17:05
  • @OpenBall Why do you have $--U_y$ and not $-U_y$? – gbox Jun 11 '17 at 17:11
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    Well I'm abusing notation, so it will be quite difficult for me to explain stuff. If you can write an intelligible outline of what you have to prove, I can point out to you what you should do. –  Jun 11 '17 at 17:13
  • @OpenBall Got it, Thanks!! – gbox Jun 11 '17 at 17:21

2 Answers2

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If $f(z)$ is differentiable in $|z|<R$ (I guess, in the complex sense) then $f(z)$ can be written as a power series $$ f(z) = a_0 + a_1 z + a_2 z^2 +\ldots $$ uniformly converging to $f(z)$ over any compact subset of $|z|<R$. If you replace $z$ with $\overline{z}$, then conjugate both sides, you get $$ \overline{f(\overline{z})} = \overline{a_0} + \overline{a_1}z+\overline{a_2} z^2+\ldots $$ and that still is a power series in $z$ that is uniformly converging over any compact subset of $|z|<R$, since the radius of convergence is the same as the previous power series. In particular, $\overline{f(\overline{z})}$ is analytic, hence differentiable, in $|z|<R$.

Jack D'Aurizio
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  • Yes, this is answering it easily, I am trying to understand the answer via $C-R$ – gbox Jun 11 '17 at 17:09
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    @gbox: the Cauchy-Riemann equations imply that any differentiable function in the complex sense is analytic by Cauchy's integral formula. You may simply prove this classical result, then apply the power series approach. – Jack D'Aurizio Jun 11 '17 at 17:11
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    As a simple alternative, you may just show that "$f(z)$ fulfills the CR equations" implies "$\overline{f(\overline{z})}$ fulfills the CR equations as well" by simple algebraic manipulations. – Jack D'Aurizio Jun 11 '17 at 17:13
  • Yes, this is exactly what I am trying to do – gbox Jun 11 '17 at 17:15
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    @gbox: then do it. $f(z)=f(a+bi)= R(a,b)+i I(a,b)$ with $R,I$ being real functions over $\mathbb{R}^2$. The CR equations tell you how $R$ and $I$ are connected, and $\overline{f(\overline{z})} = \overline{f(a-bi)} = R(a,-b)-i I(a,-b)$. To check the previous claim is now straightforward. – Jack D'Aurizio Jun 11 '17 at 17:19
  • Got it, Thanks! – gbox Jun 11 '17 at 17:21
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Suppose $f(z)=f(x+yi)=u(x,y)+iv(x,y)$ is differentiable, so that all first-order partial derivatives of $f$ exist and satisfy $\frac{\partial u}{\partial x}=\frac{\partial v}{\partial y}$ and $\frac{\partial u}{\partial y}=-\frac{\partial v}{\partial x}$. Let $g(z) = \overline{f(\overline z)}$. Then $f(\overline z) = f(x-yi) = u(x,-y)+iv(x,-y)$, and so $g(z)=u(x,-y)-iv(x,-y)$ $=U(x,y)+iV(x,y)$. Since the map $z\mapsto \overline z$ is real-differentiable, $g$ is real differentiable, and further

$$U_x(x,y) = \frac{\partial}{\partial x} u(x,-y) = u_x(x,-y) = v_y(x,-y) = \frac{\partial}{\partial y}(-v(x,-y)) = V_y(x,y)$$ $$U_y(x,y) = \frac{\partial}{\partial y}u(x,-y) = -u_y(x,-y) = v_x(x,-y) = -\frac{\partial}{\partial x}(-v(x,-y)) = V_x(x,y)$$

florence
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