Prove $|z+iw|^2 + |w+iz|^2 = 2 (|z|^2 + |w|^2)$ Let $z,w \in \Bbb C$. 
Prove $|z+iw|^2 + |w+iz|^2 = 2 (|z|^2 + |w|^2)$
When i solve the left side of the above equation i get stuck with $2i(z\bar w + z\bar w)$.
if you could please help me with this. Thanks in advance.
V.
 A: Use $|a+b| = |a|^2 + \bar a b + a \bar b + |b|^2$. Then $$|z + iw|^2 = |z|^2 + \bar z (iw) + z(\overline{iw}) + |iw|^2 = |z|^2 + i \bar z w - i z\bar w + |w|^2$$ and $$|w + iz |^2 = |w|^2 + \bar w (iz) + w(\overline{iz}) + |iz|^2 = |w|^2 + i \bar w z - i w\bar z + |z|^2.$$  Now add.
A: Hint
$$|z+iw|^2=(z+iw)(\overline{z}-i\overline{w})=z \overline{z}+iw\overline{z}-iz\overline{w}+w\overline{w}=|z|^2+|w|^2+i(w\overline{z}-z\overline{w})$$
Do the same for 
$$|w+iz|^2=(w+iz)(\overline{w}-i\overline{z})$$
And get what you want.
Ps.: I'm using that $|k|^2=k\cdot \overline{k}$
A: Just calculate straight forward
$|z+iw|^2 + |w+iz|^2=(z+iw)(\overline{z} - i\overline{w})+(w+iz)(\overline{w}-i\overline{z})= |z|^2 -i\overline{w}z+iw\overline{z}+ |w|^2 +|w|^2 -i\overline{z}w +i\overline{w}z + |z|^2 = 2 (|w|^2 + |z|^2) $
A: For a geometric interpretation of the formula, note that the equality is homogeneous in $z,w$ so it can be assumed WLOG that $w=1$. Also, $|1+iz|=|i||z-i|=|z-i|$ so the equality becomes:
$$|z+i|^2 + |z-i|^2 = 2 |z|^2 + 2 \quad \iff \quad |z|^2 = \cfrac{|z+i|^2+|z-i|^2}{2} - \cfrac{2^2}{4}$$
But the latter is simply the median length formula in the triangle defined by $z,i,-i$ with side lengths $|z-i|,|z+i|,2$ and median length $|z|\,$.
