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let $f$ is analytic in domain $\Omega$ in complex plane $\mathbb{C}$

let $g(z)=f(\overline z) \ \forall \ z \in \ \Omega $

is $g$ analytic in $\Omega$ ??

if not why??

i was trying it by C-R equations

let $$f(z)=u(x,y)+iv(x,y) $$ then$$g(z)=U(x,y)+iV(x,y)=u(x,-y)+iv(x,-y)$$ now does $$U_x=V_y,U_y=-V_x \ \ ??$$

Eklavya
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  • See https://math.stackexchange.com/questions/1689734/show-that-overlinef-overlinez-is-holomorphic-on-the-domain-d-ove and https://math.stackexchange.com/questions/474754/fz-and-overlinef-overlinez-simultaneously-holomorphic. – Martín-Blas Pérez Pinilla Oct 23 '17 at 14:33

1 Answers1

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You're working too generally to see what's going on. Try an example instead! What happens if $f(z) = z$? Is $g$ analytic then?

If you want to be general, use the chain rule to express $U_x, V_x, U_y$ and $V_y$ in terms of $u_x,u_y, v_x$ and $v_y$. Then use $u_x = v_y, u_y = -v_x$ to find out what relationship holds between $U_x, V_x, U_y$ and $V_y$.

Arthur
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  • i tried this ,but have problems in finding the relation between $U_x,U_y.,V_x,V_y$ in terms $u_x,u_y.v_x,v_y$. – Eklavya Oct 23 '17 at 14:12
  • I can do $U$, and then perhaps you can do $V$. We have $U(x,y)=u(x,-y)$. Differentiating this with respect to $x$ gives $U_x(x,y)=u_x(x,-y)$ while differentiating with respect to $y$ gives $U_y(x,y)=-u_y(x,-y)$. – Arthur Oct 23 '17 at 14:20
  • why $U_y(x,y)= -u_y(x, -y)$? i think it is because of chain rule.but how ?? – Eklavya Oct 23 '17 at 14:28
  • It's exactly because of chain rule! If we set $t(y)=-y$, then the chain rule states$$\frac{d}{dy}u(x,-y)=\frac{dt}{dy}\cdot \frac{d}{dt}u(x,t)\=-1\cdot u_y(x,t)\=-u_y(x,-y)$$where I use $u_y$ for the partial derivative of $u$ with respect to its second argument, even though we call that argument $t$, because that's what it is. – Arthur Oct 23 '17 at 14:37
  • should not it be $-u_{-y}(x, -y)$ instead of $-u_y(x,-y)$ as $t= -y$?? – Eklavya Oct 23 '17 at 14:51
  • No. As I said, the $y$ in $u_y$ has nothing to do with the variable $y$. Instead, $u_y$ simply means "The partial derivative of $u$ with respect to the second input", which is exactly what we do in $\frac{d}{dt}u(x,t)$. I'm aware that this is a somewhat unfortunate notation, but that is how it is, sadly. – Arthur Oct 23 '17 at 15:08