Why "Re" when squaring norms? Why does the "Re"-operator pop up in equations when we take a norm and square it?
Take, for example, $$\| h - u \|^2 + \|h - w \|^2 = Re\langle h-w, w - u \rangle + \|u - w\|^2 $$
which is taken from a book. Could somebody explain this equation?
 A: It's simply what happens when we expand the inner product:
$$\begin{align*}\|a-b\|^2&=\langle a-b,a-b\rangle\\&=\langle a, a\rangle -\langle a,b\rangle -\langle b,a\rangle +\langle b,b\rangle\\&=\|a\|^2 + \|b\|^2 - \langle a,b\rangle -\langle b,a\rangle.\end{align*}$$
What happens next is that we use the property that $\langle a,b\rangle$ is the complex conjugate of $\langle b,a\rangle$ as part of the definition of an inner product. Since $z+z^*=2\text{Re}(z)$ we get
$$\|a-b\|^2=\|a\|^2+\|b\|^2-2\text{Re}\langle a,b\rangle$$
which is what we wanted. Moreover, notice that the norms of each vector is definitely real, so whatever remains definitely has to be real as well, which inner products might not always be.
One might also note that $\text{Re}\langle a+bi,c+di\rangle = ac+bd$ coincides with the dot product of $(a,b)$ and $(c,d)$; this essentially tells us that the appearance of the real part means that, when we're dealing with norms only, we can treat this as a real inner product space, splitting each dimension into real and complex components. More precisely, $\langle a,b\rangle$ induces the same norm as $\text{Re}\langle a,b\rangle$ and both are bonafide inner products. This makes sense, since $\mathbb C$ under the norm $\|z\|^2=zz^*$ is isomorphic to $\mathbb R^2$ under the Euclidean norm.
