Solution: Since $\vert z \vert = 1$, so $z \bar{z} =1$. Note that \begin{align*} \left\vert 1 + z \right\vert ^2 & = (1 + z) \overline{(1+z)} \\ & = (1 + z) \left( 1 + \bar{z} \right) \\ & = 1 + \bar{z} + z + z \bar{z} \\ & = 1 + \bar{z} + z + 1 \\ & = 2 + \bar{z} + z. \end{align*} Similarly, \[ \left\vert 1 - z \right\vert ^2 = (1 - z) \overline{(1 - z)} = 1 - \bar{z} - z + z \bar{z} = 2 - \bar{z} - z. \] Therefore, \[ \vert 1 + z \vert ^2 + \vert 1-z \vert ^2 = 4. \]