How to prove that $(z^{-1})^{-1} = z$ and $(zw)^{-1} = z^{-1}w^{-1}$? I need to prove that $(z^{-1})^{-1} = z$ and $(zw)^{-1} = z^{-1}w^{-1}$ but the only thing I can think about is to consider

$$z = a+bi, w = c+di$$

and then prove it algebraically using laws of multiplication for complex numbers. Any ideas to prove it structuraly?
You know that $\mathbb{C}\setminus\left\{0\right\},\cdot$ is a group. Hence the inverse of an element is unique. Since $zz^{-1}=1=z^{-1}z$, we conclude that $z$ is the inverse of $z^{-1}$, this element is denoted by $(z^{-1})^{-1}$. Hence $(z^{-1})^{-1}=z$.

In the same fashion we can conclude that $(zw)^{-1}=w^{-1}z^{-1}$.
Generally, $(z^a)^b=z^{ab}$, so $(z^{-1})^{-1}=z$.

We have $(zw)(zw)^{-1}=1.$ Pre-multiply by $z^{-1}$ and then $w^{-1}$ to get the result.