Answer
See below
Work Step by Step
Let $w,v \in V$
$||v+w||^2\\
=(\sqrt )^2\\
=\\
=+\\
=\bar{}+\bar{}\\
=\bar{}+\bar{}+\bar{}+\bar{}\\
=||v||^2+++||w||^2$
We have $z=a+bi\\
\rightarrow z+\bar z=a+bi+a-bi=2a=2Re(z)$
Hence, $+\bar{}=2Re()$
Consequently, $||v+w||^2=||v||^2+2Re()+||w||^2$