Expanding inner product 
Let $V$ be an inner product space. Prove that:
  $$\lVert x\pm y\rVert^2 = \lVert x\rVert^2 \pm 2\mathrm{Re}\langle x,y\rangle + \lVert y\rVert^2.$$ 

After foiling $\lVert x\pm y\rVert^2$, I arrive at $\lVert x\rVert^2 \pm 2(x\cdot y) + \lVert y\rVert^2$.
I'm unsure how $2(x \cdot y)$ becomes $2\mathrm{Re}\langle , \rangle$.
Also, I'm new to StackExchange, so I apologize for the sloppy formatting.
Any tips with the math and any tips with formatting is much appreciated! Thank you!
 A: You seem to be using the Euclidean definition of norm using the dot product in a real vector space. You should be using the general definition of norm induced by an inner product, namely
$$\lVert x\rVert = \sqrt{\langle x,x\rangle}$$
where $\langle\cdot,\cdot\rangle$ is an inner product on $V$; that is, a function that assigns to a pair of vectors a scalar (in either $\mathbb{R}$ or $\mathbb{C}$, depending on whether $V$ is a real or complex vector space), satisfying the following properties:


*

*Additive in the first coordinate: $\langle x+y,z\rangle = \langle x,z\rangle + \langle y,z\rangle$ for all $x,y,z\in V$.

*Homogeneous in the first coordinate: $\langle \alpha x,y\rangle = \alpha\langle x,y\rangle$ for all $x,y\in V$, and all scalars $\alpha$.

*Conjugate symmetric: $\langle y,x\rangle = \overline{\langle x,y\rangle}$ for all $x,y\in V$, where $\overline{\alpha}$ is the complex conjugate of $\alpha$.

*Positive definite: $\langle x,x\rangle\geq 0$ for all $x\in V$, and $\langle x,x\rangle = 0$ if and only if $x=\mathbf{0}$.


So the correct expansion would be:
$$\begin{align*}
\lVert x\pm y\rVert^2 &= \langle x\pm y,x\pm y\rangle\\
&= \langle x,x\rangle \pm \langle x,y\rangle \pm \langle y,x\rangle + \langle y,y\rangle\\
&= \lVert x\rVert^2 + \lVert y\rVert^2 \pm (\langle x,y\rangle + \langle y,x\rangle).
\end{align*}$$
Now you should use the conjugate symmetry to get the desired result. 
A: You wrote

I'm unsure how $2(x \cdot y)$ becomes $2\mathrm{Re}\langle , \rangle$.

It's actually $x\cdot y + y \cdot x,$ and by conjugate symmetry $y \cdot x = \overline{ x \cdot y}$, 
so it's $x\cdot y + \overline {x \cdot y}=2 \mathrm{Re}(x\cdot y).$
