Prove that such an inverse is unique Given $z$ is a non zero complex number, we define a new complex number $z^{-1}$ , called $z$ inverse to have the property that $z\cdot z^{-1} = 1$ 
$z^{-1}$  is also often written as $1/z$

Show that the identity element in a group $G$ is unique; that is there exists only one element $e\in G$ such that $eg=ge=g$ for all $g\in G$.
