Let $\alpha,\beta:[-1,1]\to\Bbb{C}\simeq\Bbb{R^2}$ be two paths, s.t $\alpha(0)=\beta(0)=\zeta$. Let $A:\hat{\Bbb{C}}\to\hat{\Bbb{C}}$ be a mobius transform s.t $A\zeta\neq\infty, A\in SL_2(\Bbb{C})$. Prove: $$\frac{\langle\alpha'(0),\beta'(0)\rangle}{|\alpha'(0)||\beta'(0)|}=\frac{\langle(A\circ\alpha)'(0),(A\circ\beta)'(0)\rangle}{|(A\circ\alpha)'(0)||(A\circ\beta)'(0)|}$$.
It was given as a remark that we can think of everything as in $\Bbb{R^2}$ and the inner product is the regular inner product, or to think of everything as compelx and think of the inner product as the real part of the hermitian inner product. I got stuck calculating because I'm not even sure how to think of $(A\circ \alpha)'$ as a function in $\Bbb{R^2}$ (I have no background in complex analysis).
Any help would be appreciated.
Edit: My attempt:
$$\frac{\langle(A\circ\alpha)'(0),(A\circ\beta)'(0)\rangle}{|(A\circ\alpha)'(0)||(A\circ\beta)'(0)|}=\frac{\langle A'(\alpha(0))\alpha'(0),A'(\beta(0))\beta'(0)\rangle}{|(A'(\alpha(0))\alpha'(0)||(A'(\beta(0))\beta'(0)|}=\frac{\langle A'(\zeta)\alpha'(0),A'(\zeta)\beta'(0)\rangle}{|(A'(\zeta)\alpha'(0)||(A'(\zeta)\beta'(0)|}=\frac{\mathfrak{Re}(|A'(\zeta)|^2(\alpha'(0)\bar\beta'(0)}{|A'(\zeta)^2|\alpha'(0)||(A'(\zeta)\beta'(0)|}=\frac{\langle\alpha'(0),\beta'(0)\rangle}{|\alpha'(0)||\beta'(0)|}$$
The inner product of two complex numbers (identifying $\mathbb{C}\cong\mathbb{R}^2$) is $\langle w,z\rangle=\mathrm{Re}(\overline{w}z)$.
The chain rule says $(A\circ\alpha)'(0)=A'(\alpha(0))\alpha'(0)$ and similarly for $(A\circ\beta)'(0)$.
You should be able to start simplifying the right side now and show you end up with the same thing as the left side. Notice this only depends on $A$ being holomorphic in a neighborhood of $\zeta$, so this fact about preserving angles (conformality) is much more general than Mobius transformations.
Note that $\lambda\overline{\lambda}=|\lambda|^2$ and real scalars can be pulled out of $\mathrm{Re}(\,)$.