Given $f(z)$ is an analytic complex function, show $f^*(z^*)$ is also analytic This exercise is from Arfken mathematical methods for physicists): "The function $f(z)=u(x,y)+iv(x,y)$ is analytic. Show that $f^*(z^*)$ is also analytic."
There must be some simple proof (and not related to series), because there is little said about complex analysis in the book before this exercise (The only important thing said is Cauchy-Riemann conditions). Not sure, but I think if $f(z)=u(x,y)+iv(x,y)$ then $f^*(z^*)=u(x,-y)-iv(x,-y)=g(x,y)+ih(x,y)$. Now $g$ and $h$ must satisfy the Cauchy-Reimann conditions and their first partial derivatives with respect to $x$ and $y$ must be continuous. Yet I can't show any of them. Any suggestions for developing this idea?
 A: $f(z) = u(x,y) + iv(x,y)$
You want to show that the function
$f^*(z^*) = u(x,-y) - iv(x,-y)$
is also analytic.
Denote $f^*(z^*)$ by
$f^*(z^*) = u_1 + iv_1$
with
$u_1(x,y) = u(x,-y)$
$v_1(x,y) = -v(x,-y)$
The partial derivatives with respect to $x$ are related by
$$\frac{\partial{u_1}}{\partial{x}}(x,y)
=\frac{\partial{u}}{\partial{x}}(x,-y), \:\:\:\:
\frac{\partial{v_1}}{\partial{x}}(x,y)
=-\frac{\partial{v}}{\partial{x}}(x,-y)
$$
The partial derivatives with respect to $y$ are related by
$$\frac{\partial{u_1}}{\partial{y}}(x,y)
=-\frac{\partial{u}}{\partial{y}}(x,y), \:\:\:\:
\frac{\partial{v_1}}{\partial{y}}(x,y)
=\frac{\partial{v}}{\partial{y}}(x,y)
$$
You know that the Cauchy Riemann equations holds for $f(z) = u + iv$.
The above relations shows that they also hold for
$f^*(z^*) = u_1 + iv_1$
and thus the function $f^*(z^*)$
is also analytic.
A: JKnecht's answer is partially wrong, it is true that
$$\frac{\partial u_1}{\partial x}(x, y) =\frac{\partial u}{\partial x}(x, -y)$$
$$\frac{\partial v_1}{\partial x}(x, y) =-\frac{\partial v}{\partial x}(x, -y)$$
Let's prove the first one, for example. Observe that $u_1(x, y) = u(c(x, y)) $, with $c(x, y) = (c_1(x, y),c_2(x, y))= (x, -y) $ is the complex conjugate.
Applying chain rule, we get:
$$\frac{\partial u_1}{\partial x}(x, y) = \frac{\partial(u\circ c) }{\partial x}(x, y) =   \frac{\partial u}{\partial x}(c(x, y)) \frac{\partial c_1}{\partial x}(x, y) +  \frac{\partial u}{\partial y}(c(x, y)) \frac{\partial c_2}{\partial x}(x, y) =\frac{\partial u}{\partial x}(x, -y) .$$
We have used that $\frac{\partial c_1}{\partial x} = 1, \frac{\partial c_2}{\partial x} = 0$.
But the partial derivatives w. r. t. $y$ are related by another formula:
$$\frac{\partial u_1}{\partial y}(x, y) =\frac{\partial (u\circ c)}{\partial y}(x, y)=\frac{\partial u}{\partial x}(c(x, y)) \frac{\partial c_1}{\partial y}(x, y) +  \frac{\partial u}{\partial y}(c(x, y)) \frac{\partial c_2}{\partial y}(x, y) =-\frac{\partial u}{\partial y}(x, -y) . $$
We have used that $\frac{\partial c_1}{\partial y} = 0,\frac{\partial c_2}{\partial y} = -1$.
Similarly, one proves that:
$$\frac{\partial v_1}{\partial y}(x, y) =\frac{\partial v}{\partial y}(x, -y) . $$
In conclusion, the partial derivatives must be compared at different points. This still implies the CR equations.
$$  \frac{\partial u_1}{\partial x}(x, y) =\frac{\partial u}{\partial x}(x, -y)=  \frac{\partial v}{\partial y}(x, -y)= \frac{\partial v_1}{\partial y}(x, y). $$
We have used that $u, v$ verify Cauchy-Riemann at the point $(x, -y) $, and the previous equalities.
Same goes for the other equation:
$$  \frac{\partial u_1}{\partial y}(x, y) =-\frac{\partial u}{\partial y}(x, -y)=  \frac{\partial v}{\partial x}(x, -y)= -\frac{\partial v_1}{\partial x}(x, y). $$
EDIT: Another easy way of showing this goes by seeing that the complex conjugate preserves angles (but not orientations), so clearly $c(f(c(x, y))$ preserves angles (because $f$ also does) AND orientations (because $f$ preserves orientations and $c$ reverses them twice, so the composition preserves them). Every map preserving orientations and angles is holomorphic.
EDIT2: Proof of the equality of derivatives w. r. t. $y$. Recall that $v_1= -v\circ c$:
$$\frac{\partial v_1}{\partial y}(x, y) =-\frac{\partial (v\circ c)}{\partial y}(x, y)=-\frac{\partial v}{\partial x}(c(x, y)) \frac{\partial c_1}{\partial y}(x, y) -  \frac{\partial v}{\partial y}(c(x, y)) \frac{\partial c_2}{\partial y}(x, y) =\frac{\partial v}{\partial y}(x, -y) . $$
We have used that $\frac{\partial c_1}{\partial y} = 0,\frac{\partial c_2}{\partial y} = -1$.
A: Let's start with JKnecht's answer:
$f(z) = u(x,y) + iv(x,y)$
You want to show that the function
$f^*(z^*) = u(x,-y) - iv(x,-y)$
is also analytic
Denote $f^*(z^*)$ by
$f^*(z^*) = u_1 + iv_1$
with
$u_1(x,y) = u(x,-y)$
$v_1(x,y) = -v(x,-y)$
Let $-y = w$
Then we have
$$\frac{\partial{u_1}}{\partial{x}}(x,y)
=\frac{\partial{u}}{\partial{x}}(x,w), \:\:\:\:
\frac{\partial{v_1}}{\partial{x}}(x,y)
=-\frac{\partial{v}}{\partial{x}}(x,w)
$$
The partial derivatives with respect to $y$ are related by
$$\frac{\partial{u_1}}{\partial{y}}(x,y)
=\frac{\partial{u(x,-y)}}{\partial{y}} = -\frac{\partial{u}}{\partial{w}}(x,w), \:\:\:\:
\frac{\partial{v_1}}{\partial{y}}(x,y)
=-\frac{\partial{v(x,-y)}}{\partial{y}} = \frac{\partial{v}}{\partial{w}}(x,w)
$$
the chain rules that produce these results can be intuitively seen if we treat u and v as multiple different single variable functions.
The key here is that we are assuming that f is holomorphic at the complex conjugate of every point that f is holomorphic on. we are assuming that the domain is symmetrical about the real axis
