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.
Best Answer
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.