I want to prove this : $$H^2(G,{\bf C}^\times ) = {\bf Z}_2 $$ where $$ G:={\rm Gal}\ ({\bf C}/{\bf R})={\bf Z}_2=\langle\alpha\rangle$$
Step 1 : normalized $2$-cocycle condition is $$ f(g,h)+ f(gh,k)=g\cdot f(h,k)+ f(g,hk)$$
So $$ f(\alpha,1)=f(1,\alpha)=0$$
Hence $$ f(\alpha,\alpha )=\alpha f( \alpha ,\alpha) $$
So $f(\alpha,\alpha)\in {\bf R}$.
Step 2 : Consider a $1$-coboundary :
$$ F(g,h):=gf_1(h)-f_1(gh) + f_1(g) $$
$$F(1,\alpha)= f_1(1):=0,\ F(\alpha,1)=\alpha f_1(1) =0 $$ $$ F(\alpha,\alpha)=\alpha f_1(\alpha ) +f_1(\alpha) $$
So let $$ f_1(\alpha ):=\frac{1}{2}f(\alpha,\alpha ) + yi $$ for any $y$. Hence $f=F$.
So we have $H^2(G,{\bf C}^\times )=0$. Am I right ?
Hint : Step 1 : Let $G=\langle\tau \rangle$. Let $f$ be a normalized $2$-cocycle : $$ f(1,\tau)=f(\tau,1)=1$$
And $$ f(\tau,\tau)f(\tau,\tau x)=\tau f(\tau ,x)f(\tau, \tau x) $$
If $x=\tau$ then we have $$ \tau f(\tau,\tau)=f(\tau,\tau)\Rightarrow f(\tau,\tau)\in {\bf R} $$
Step 2 : Consider coboundary $F$ : $$ F(\tau,1)=\tau f_1(1)f(\tau)^{-1} f(\tau )=1 \Rightarrow f_1(1):=1$$
$$ F(1,\tau)=f_1(\tau )f_1(\tau)^{-1} f_1(1)=1$$
That is $F$ is normalized.
And $$ F(\tau,\tau)=\tau f_1(\tau) f_1(\tau )=a^2+b^2 \geq 0\ (f_1(\tau):=a+b_i) $$