Prove that the following function is holomorphic Let $g:B(0,R)\to\mathbb{C}$, $g(z)=a(\rho,\alpha)+ib(\rho,\alpha)$ an holomorphic function I have to prove that $h(z)=\overline{f\left(\frac{R^2}{\overline{z}}\right)}$ is holomorphic over $\mathbb{C}\setminus \overline{B(0,R)}$. I have troubles in proving the validity of Cauchy-Riemann equations for $h$.
C-R in polar coordinates are:
$$
\frac{\partial a}{\partial \rho}=
\frac{1}{\rho}\frac{\partial b}{\partial \alpha}\\
\frac{\partial b}{\partial \rho}=
-\frac{1}{\rho}\frac{\partial a}{\partial \alpha}\\
$$
My attemp so far:
$$
h(z)=a\left(\frac{R^2}{\rho},\alpha\right)-ib\left(\frac{R^2}{\rho},\alpha\right)
$$
$$
\frac{\partial }{\partial \rho}a\left(\frac{R^2}{\rho},\alpha\right)=\frac{\partial a}{\partial x_1}\frac{\partial x_1}{\partial \rho}=\frac{\partial a}{\partial x_1}\left(-\frac{R^2}{\rho^2}\right)\\
-\frac{\partial }{\partial \alpha}b\left(\frac{R^2}{\rho},\alpha\right)=
-\frac{\partial b}{\partial \alpha}=-\rho\frac{\partial a}{\partial \rho}=-\rho\frac{\partial a}{\partial x_1}\frac{\partial x_1}{\partial \rho}=-\rho\frac{\partial }{\partial \rho}a\left(\frac{R^2}{\rho},\alpha\right)
$$
but the minus sign is wrong...
Any help will be appreciated.
 A: For me, the best way to show this kind of things is to use the definition of holomorphic maps as conformal maps. Your function $h$ is the composition of three maps:
$$z\mapsto g(z) = \frac{R^2}{\overline{z}}$$
$$z \mapsto  f(z)$$
$$z \mapsto c(z) = \overline{z}$$
The maps $g$ and $c$ are easily shown to be antiholomorphic, that is, they preserve angles and switch orientations, and $f$ is holomorphic, that is, it preserves oriented angles. The composition $h = c\circ f\circ g$ then preserves the magnitude of angles (because the three factors do), and it switches orientation two times, so in the end it preserves oriented angles. That proves that $h$ is holomorphic.
In general, the composition of a holomorphic function with an even number of antiholomorphic functions is holomorphic. Antiholomorphic functions are recognized as functions of the conjugated variable $\overline{z}$.
EDIT: This is even more trivial if you think of $f$ as a power series of $z$. If you substitute $z$ by $R^2/\overline{z}$ in this power series, you get a power series of $\overline{z}$ (because $R^2/\overline{z}$ is itself a power series of $\overline{z}$). Applying conjugation, you get a power series of $z$, which is equal to $h$, hence $h$ is holomorphic.
A: You have not followed up on my suggestion to be careful about where partial derivatives are evaluated, so I'll do half of the proof here.  (I think the simultaneous use of the letters $a$ and $\alpha$ is bound to lead to disaster.)
Let $A(\rho,\alpha) = a(R^2/\rho,\alpha)$ and $B(\rho,\alpha) = -b(R^2/\rho,\alpha)$. You want to show, for example, that
$$\frac{\partial B}{\partial\alpha} = \rho\frac{\partial A}{\partial\rho}.$$
We have
$$
\frac{\partial B}{\partial\alpha} = -\color{red}{\frac{\partial b}{\partial\alpha}\big(R^2/\rho,\alpha\big) \overset{\text{C-R}}=} -\color{red}{\frac{R^2}{\rho}\frac{\partial a}{\partial\rho}\big(R^2/\rho,\alpha\big)}
= \rho\frac{\partial A}{\partial \rho},$$
because $\dfrac{\partial A}{\partial\rho} = \dfrac{\partial a}{\partial\rho}\big(R^2/\rho,\alpha\big)\cdot \left(-\frac{R^2}{\rho^2}\right)$. Note the portion in red is crucial, and this is why I told you to be careful about where partial derivatives are evaluated.
