2

So I was trying to prove that for $x,y\in \mathbb{C}$ we have that: $4 \langle x,y \rangle=||x+y||^2-||x-y||^2+i||x+iy||^2-i||x-iy||^2$.

I got that $||x+y||^2-||x-y||^2=4\Re\langle x,y \rangle$ and $||x+iy||-||x-iy||=-4\Im \langle x,y \rangle$. Combining those two I get $4\overline{\langle x,y \rangle}$ and not $4\langle x,y \rangle$.

Imaginary part:

$||x\pm iy||^2=||x||^2+||y||^2 \pm 2\Re\langle x,iy \rangle$. We know that $\Re\langle x,iy \rangle =\Re \overline{\langle iy,x \rangle}=\Re (\overline i \cdot \overline{\langle y,x \rangle})=-\Re(i \cdot {\langle x,y \rangle})=-\Im \langle x,y \rangle.$ Now we have: $||x\pm iy||^2=||x||^2+||y||^2 \pm 2\Re\langle x,iy \rangle=||x||^2+||y||^2 \mp 2\Im\langle x,y \rangle$.

Can someone explain me where my error is?

CryoDrakon
  • 3,392
  • It will be easier to explain if you make your calculation of the imaginary part explicit. – Justpassingby Jan 10 '16 at 22:43
  • I added my calculation for the imaginary part. – CryoDrakon Jan 10 '16 at 22:51
  • The real part of $iz$ is not the imaginary part of $z.$ – Justpassingby Jan 10 '16 at 23:09
  • 1
    Well, I am goint with the logic: $\Re(iz)=\Re(i(a+ib))=\Re(ai-b)=-b=-\Im(z)$. – CryoDrakon Jan 10 '16 at 23:12
  • 1
    That is not what you have done in the line that ends in Now we have: – Justpassingby Jan 10 '16 at 23:24
  • Now I see it. The two minuses cancel out leaving $\Re\langle x,iy \rangle = \Im \langle x,y \rangle$. – CryoDrakon Jan 10 '16 at 23:31
  • 1
    Might be worth noting that the above uses the "mathematics" convention for the complex (Hermitian) inner product $\left\langle ax,by\right\rangle =ab^{}\left\langle x,y\right\rangle$; if one uses the "physics" convention $\left\langle ax,by\right\rangle =a^{}b\left\langle x,y\right\rangle$, the polarization identity is $$4\left\langle x,y\right\rangle=\left\Vert x+y\right\Vert {}^{2}-\left\Vert x-y\right\Vert {}^{2}+i\left\Vert x-iy\right\Vert {}^{2}-i\left\Vert x+iy\right\Vert {}^{2}$$ – Adam Marsh Mar 16 '18 at 17:39

1 Answers1

2

We have $\|x+y\|^{2}-\|x-y\|^{2}=4\Re\langle x,y\rangle$ and \begin{align}i\|x+iy\|^{2}-i\|x-iy\|^{2}&=i(\langle x+iy,x+iy\rangle-\langle x-iy,x-iy\rangle)\\ &=i(\langle x,x\rangle +\langle x,iy\rangle +\langle iy,x\rangle+\langle iy,iy\rangle\\ &\phantom{=I(}-\langle x,x\rangle +\langle x,iy\rangle +\langle iy,x\rangle -\langle iy,iy\rangle)\\ &=2i(\langle x,iy\rangle +\langle iy,x\rangle)\\ &=2i(-i\langle x,y\rangle+i\langle y,x\rangle)\\ &=-2(-\langle x,y\rangle +\overline{\langle x,y\rangle})\\ &=-2(-2\Im\langle x,y\rangle)=4\Im\langle x,y\rangle.\end{align} So $$\|x+y\|^{2}-\|x-y\|^{2}+i\|x+iy\|^{2}-i\|x-iy\|^{2}=4\Re\langle x,y\rangle+4\Im\langle x,y\rangle=4\langle x,y\rangle$$

hbghlyj
  • 2,115
Rafael
  • 3,789
  • If we have a complex number $z\in \mathbb{C}$ so that $z=a+ib$, shouldn't $\Re(z)=a$ and $\Im(z)=b$ and not $Im(z)=ib$? Maybe I am wrong, but then we would also have $\langle x,y \rangle =\Re \langle x,y \rangle + i \cdot \Im \langle x,y \rangle$ ? – CryoDrakon Jan 10 '16 at 22:59
  • 1
    Sorry. The correct is: $$i|x+iy|^{2}-i|x-iy|^{2}=-2(-\langle x,y\rangle +\overline{\langle x,y\rangle})=-2(-2i\Im\langle x,y\rangle)=4i\Im\langle x,y\rangle$$ Therefore $$|x+y|^{2}-|x-y|^{2}+i|x+iy|^{2}-i|x-iy|^{2}=4\Re\langle x,y\rangle+4i\Im\langle x,y\rangle=4\langle x,y\rangle$$ – Rafael Jan 10 '16 at 23:17