Sei z=a+bi und wegen |z| = 1 gilt a^2 + b^2 = 1.
\( \frac{1+z}{|1+z|} = \frac{1+a+bi }{ \sqrt {(1+a)^2 + b^2} } = \frac{1+a+bi }{ \sqrt {1+2a +a^2 + b^2} } \)
Wegen a^2 + b^2 = 1. also
\( \frac{1+z}{|1+z|} = \frac{1+a+bi }{ \sqrt {1+2a +1} } \)
\( = \frac{1+a+bi }{ \sqrt {2+2a} } \)
==> \( (\frac{1+z}{|1+z|})^2 = ( \frac{1+a+bi }{ \sqrt {2+2a} } )^2 = \frac{(1+a)^2 +2(1+a)bi - b^2 }{ 2+2a} \)
\( = \frac{1+2a+a^2 }{ 2+2a} + bi - \frac{ b^2 }{ 2+2a} = \frac{1+2a+a^2 - b^2 }{ 2+2a} + bi \)
Wegen a^2 + b^2 = 1 gilt b^2 = 1-a^2 also weiter mit
\( = \frac{1+2a+a^2 - (1-a^2)}{ 2+2a} + bi = \frac{2a+2a^2}{ 2+2a} + bi = \frac{a(2+2a}{ 2+2a} + bi =a+bi=z\). q.e.d.
