Show that $|z+w|^2$ + $|z-w|^2$ = $2|z|^2 + 2|w|^2$ 
Show that $|z+w|^2$ + $|z-w|^2$ = $2|z|^2 + 2|w|^2$.

Every time I work this problem out I have $2|z|^2$ but I can not get the other piece.
Here is what I got so far. Let $z = a+ bi$ and $w = c + di$, then
\begin{align*}
|z+w|^2 + |z-w|^2 &= a^2 + b^2 + c^2 + d^2 + a^2 + b^2 - c^2 - d^2\\ 
&= 2a^2 + 2b^2\\ 
&= 2(a^2 + b^2)\\ 
&=2|z|^2.
\end{align*}
I do not know if I have the wrong approach or not.  

You seem to be using $|z + w|^2 = a^2 + b^2 + c^2 + d^2$ and $|z - w|^2 = a^2 + b^2 - c^2 - d^2$. These are both false. Instead we have
$$|z + w|^2 = |(a+bi) + (c+di)|^2 = |(a + c) + (b + d)i|^2 = (a + c)^2 + (b + d)^2$$
and a similar calculation gives $|z - w|^2 = (a-c)^2 + (b - d)^2$. If you use these, you will find that their sum is $2(a^2 + b^2 + c^2 + d^2) = 2|z|^2 + 2|w|^2$.
