I'm working on a homework problem regarding the proof for the polarization identity for complex scalars. I've taken a look at another question on this community (Polarization Identity for Complex Scalars) and have tried working it out on my own, but am getting stuck towards the end, particularly towards dealing with the imaginary part. I'll elaborate on my approach.
Starting with the definition:
$$
\langle x, y \rangle = \frac{1}{4} \left( \Vert x + y \Vert^2 – \Vert x – y \Vert^2 – i\Vert x – iy \Vert^2 + i \Vert x + iy \Vert^2 \right)
$$
Focusing only on the imaginary part (i.e. $i \Vert x + iy \Vert^2 – i \Vert x – iy \Vert^2$):
$$
\begin{align}
i \Vert x + iy \Vert^2 – i \Vert x – iy \Vert^2 & = i \left( \Vert x + iy \Vert^2 – \Vert x – iy \Vert^2 \right) \\
& = i \left( \langle x + iy, x + iy \rangle – \langle x -iy, x- iy \rangle \right) \\
& = i \left[ (\langle x, x \rangle + \langle x, iy \rangle + \langle iy, x \rangle + \langle iy, iy \rangle ) – ( \langle x, x \rangle + \langle x, -iy \rangle + \langle -iy, x \rangle + \langle -iy, -iy \rangle ) \right] \\
& = 2i \left( \langle x, iy \rangle + \langle iy, x \rangle \right)
\end{align}
$$
Using $\langle x, iy \rangle = -i \langle x, y \rangle$ and $\langle iy, x \rangle = \overline{\langle x, iy \rangle} = \overline{-i\langle x, y \rangle} = i\overline{\langle x, y \rangle}$,
$$
\begin{align}
2i\left( \langle x, iy \rangle + \langle iy, x \rangle \right) & = 2i \left( -i\langle x, y \rangle + i \overline{\langle x, y \rangle} \right) \\
& = 2 \langle x, y \rangle – 2\overline{\langle x, y \rangle}
\end{align}
$$
I'm not sure how to proceed from here. I believe that I should end up with something like $-2(-2 \mathfrak{I} \langle x, y \rangle )$ but how does the last line become expressed as this? Thanks.
Best Answer
Hint: For $z \in \mathbb{C},$ $$z-\overline{z}=2i\,\Im{z},$$ where $\Im z$ denotes the imaginary part of $z.$
Btw, once you have shown that $$ \Vert x + y \Vert^2 - \Vert x - y \Vert^2=4\Re \langle x,y\rangle,$$ then replacing $y$ by $iy:$
$$ \Vert x + iy \Vert^2 - \Vert x - iy \Vert^2=4\Re\langle x,iy\rangle=4\Im \langle x,y\rangle.$$