Complex inner product identity

Question

Complex inner product

In $\mathbb{C}^d, $$v\not = 0$ and $w=\lambda v$ for some $\lambda \in \mathbb{C}$.

Let $\langle v, w \rangle = \rho e^{i\theta}$ with $\rho >0$. If $\tilde{v} = e^{-i\theta}v$, then $\langle \tilde{v}, \tilde{v} \rangle = \langle v, v \rangle$.

I tried that $$\langle \tilde{v}, \tilde{v} \rangle = \langle e^{-i \theta}v, e^{-i \theta}v \rangle = e^{-2i \theta} \rho e^{i \theta} = \rho e^{-i \theta}.$$

Is the second equality correct? How can we prove the identity $\langle \tilde{v}, \tilde{v} \rangle = \langle v, v \rangle$? Is it true that for any $v$ and $w$ in $\mathbb{C}^d$, $\langle v, w \rangle = \rho e^{i\theta}$? What decides the value of $\theta$?

Answer 1

$\langle \tilde{v}, \tilde{v} \rangle = \langle e^{-i \theta}v, e^{-i \theta}v \rangle= e^{-i \theta} \overline{e^{-i \theta}}\langle v,v \rangle=e^{-i \theta} e^{i \theta}\langle v,v \rangle=\langle v,v \rangle$.