Polarization identity if operator Let $\mathcal{H}$ be a Hilbert space over $\mathbb{C}$. I know (and proved) the following polarization identity:
\begin{eqnarray}
\langle x, y \rangle = \frac{1}{4}\sum_{k=0}^{3}i^{k}\langle x+i^{k}y, x+i^{k}y \rangle \tag{1}\label{1}
\end{eqnarray}
for $x,y \in \mathcal{H}$. Now, I'd like to prove the following result as a consequence of (\ref{1}):
\begin{eqnarray}
\langle Ax, y \rangle = \frac{1}{4}\sum_{k=0}^{3}i^{k}\langle A(x+iy), x+iy \rangle \tag{2}\label{2}
\end{eqnarray}
where $A$ is any given bounded linear operator on $\mathcal{H}$. If you change $x$ to $Ax$ in (\ref{1}) you get:
\begin{eqnarray}
\langle Ax, y \rangle = \frac{1}{4}\sum_{k=0}^{3}i^{k}\langle Ax+iy, Ax+iy \rangle \tag{3}\label{3}
\end{eqnarray}
but not (\ref{2}) and I'm stuck at this point. How does (\ref{1}) imply (\ref{2})?
 A: Thanks to the comments, I believe I got it. First, we must show that the polarization identity also holds in a more general context, namely if $T:\mathcal{H}\times \mathcal{H}\to \mathbb{C}$ satisfies the following properties:
(a) $T(x,\alpha y + \beta z) = \bar{\alpha}T(x,y)+\bar{\beta}T(x,z)$
(b) $T(\alpha x + \beta z, y) = \alpha T(x,y) + \beta T(z,y)$
then, it follows that:
\begin{eqnarray}
T(x,y) = \frac{1}{4}\sum_{k=0}^{3}i^{k}T(x+i^{k}y,x+i^{k}y) \tag{1'}\label{1.1}
\end{eqnarray}
Proof of (\ref{1.1}): Write:
$$T(x,y) = T\bigg{(}\frac{1}{2}(x+iy+x-iy), \frac{1}{2i}(x+iy - (x-iy))\bigg{)}$$
Using properties (a) and (b), we get:
$$T(x,y) = -\frac{1}{4i}[T(x+iy,x+iy)-T(x+iy,x-iy)+T(x-iy,x+iy)-T(x-iy,x-iy)]$$
Now, note that:
$$-\frac{1}{4i}[T(x+iy,x+iy)-T(x-iy,x-iy)] = \frac{1}{4}i[T(x+iy,x+iy)-T(x-iy,x-iy)]$$
and, using (a) and (b) again, we also have:
$$\frac{1}{4}i [-T(x+iy,x-iy)+T(x-iy,x+iy)] = \frac{1}{4}i[2T(x,iy)-2T(iy,x)] = \frac{1}{4}i[-2iT(x,y)-2iT(y,x)] = \frac{1}{2}[T(x,y)+T(y,x)] = \frac{1}{4}[T(x+y,x+y)-T(x-y,x-y)]$$
and (\ref{1.1}) follows.
Now, the result follows by setting $T(x,y) := \langle Ax,y\rangle$ for all $x,y \in \mathcal{H}$.
