The operation is not defined for every pair in the proposed set: choosing $a=1,b=-1$ makes its denominator zero. Thus it is not even a <binary operation> on $\mathbb C\setminus\{0\}$, and is \b[not a group]. In particular, removing zero from the domain does not prevent the sum in the denominator from vanishing.

Solution

Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2023 / Paper 2 / 8 / a / ii / Solution

 Solution (source code)