Prove an identity for an inner product

Question

Prove that identity $\langle u,v\rangle = \text{Re}\langle v,u \rangle - i\text{Re}\langle v,iu\rangle $ Where $V$ is an inner product over $\mathbb{C}$.

I tried to use the common properties of inner product, but couldn't get much farther.

Answer 1

Note that $$\mathrm{Re}\langle v,u\rangle= \mathrm{Re}\langle u,v\rangle$$ Also, $$\langle v,iu\rangle=-i \langle v,u\rangle$$ And $$\mathrm{Im}\langle v,u\rangle=- \mathrm{Im}\langle u,v\rangle$$ Hence when we multiply by $-i$ we are squaring the $-i$ that appears from conjugating by switching the order, obtaining the negative of the imaginary part of the original as the real part. Hence $$\mathrm{Re}\langle v,iu\rangle=- \mathrm{Im}\langle u,v\rangle$$ This proves the identity.

For a less wordy derivation, note that for any complex number $z$, $$\mathrm{Re}\ -iz=\mathrm{Im}\ z$$ So $$\mathrm{Re}\langle v,iu\rangle= \mathrm{Re}-i\langle v,u\rangle= \mathrm{Im}\langle v,u\rangle=-\mathrm{Im}\langle u,v\rangle$$

Answer 2

Hint:

Start from the right side and use $Re(z) = \frac 12 (z+\overline z)$.