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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.

Sahiba Arora
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Sean
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2 Answers2

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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 $iy:$

$$ \Vert x + iy \Vert^2 - \Vert x - iy \Vert^2=4\Re\langle x,iy\rangle=4\Im \langle x,y\rangle.$$

Sahiba Arora
  • 10,847
  • Hi! Thanks for the answer. Sorry about the week-delayed comment, but when I put $z - \bar{z} = 2i \mathfrak{I}(z)$ into the last equation I obtained I get $2(\langle x, y \rangle - \overline{\langle x, y \rangle}) = 4i \mathfrak{I}(\langle x, y \rangle)$ rather than $4 \mathfrak{I} (\langle x, y \rangle)$. My assumption is that since we're dealing with the imaginary part it suffices to omit the $i$, but I'm not 100% sure. – Sean Jun 24 '20 at 06:34
  • @Seankala You should be getting $4i\Im\langle x,y\rangle$ and then $|x+y|^{2}-|x-y|^{2}+i|x+iy|^{2}-i|x-iy|^{2}=4 \Re\langle x,y\rangle+4i\Im \langle x,y\rangle=4\langle x,y \rangle.$ I don't understand why you want $4\Im \langle x,y\rangle.$ – Sahiba Arora Jun 24 '20 at 08:41
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It's perhaps easier to work out the coefficient across all norms of each type of term, viz.$$\begin{align}\sum_{n=0}^{3}i^{n}\left\langle x+i^{n}y,\,x+i^{n}y\right\rangle&=\underbrace{\sum_{n=0}^{3}i^{n}}_0{}\left(\left\langle x,\,x\right\rangle +\left\langle y,\,y\right\rangle \right)+\underbrace{\sum_{n=0}^{3}1}_{4}\left\langle x,\,y\right\rangle +\underbrace{\sum_{n=0}^{3}\left(-1\right)^{n}}_{0}\left\langle y,\,x\right\rangle \\&=4\left\langle x,\,y\right\rangle.\end{align}$$

J.G.
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