Linear dependence of two complex vectors and Euler's formula. Let $x,y\in \mathbb{C}^n$ and $\vert\vert x\vert\vert_2=\vert\vert 
 y\vert\vert_2.$ Then $x$ and $y$ are linear independent if and only if
$y\neq x e^{i\theta}$ for some real $\theta$. Is this claim true?
I know $y=x e^{i\theta}\implies x\perp y$ because for example
$$
det
\begin{pmatrix}
x_1 & e^{i\theta}x_1\\
x_2 & e^{i\theta}x_2
\end{pmatrix}=0
$$
For the other direction, I tried using the definition of $x^*y=0$ and didn't get anywhere.
 A: If $x = e^{i\theta}y$, then
$$
x+(-e^{i\theta})y = e^{i\theta}y +(-e^{i\theta})y = (e^{i\theta} - e^{i\theta})y = 0y = \vec{0}, 
$$
i.e., $x$ and $y$ are linearly dependent.
Conversely, if $x$ and $y$ are linearly dependent, then there are complex numbers $\lambda$ and $\mu$, at least one of which is nonzero, such that $\lambda x + \mu y =0$.
If $\mu =0$, then $\lambda x = 0 \implies x = 0$. But $\vert\vert x\vert\vert_2=\vert\vert y\vert\vert_2$, so $y = 0$. Thus, $x = e^{i\theta}y$ holds for any $\theta$. A similar argument applies when $\lambda =0$.
If $\lambda, \mu \ne 0$, then $x = \frac{\mu}{\lambda}y$. Can you take it from here?
A: First direction
Lemma:
$x$ is lin. ind. of $y \implies y \neq xe^{i\theta}, \forall \theta \in \Bbb R$
Proof:
By definition of linear independence, $\not\exists \lambda \in \Bbb C$ such that $y = \lambda x$ where $\lambda \neq 0$.  Let $\lambda = e^{i\theta}$.  If they are linearly independent, this cannot be.
Other direction (by contrapositive)
Lemma:
$y \neq xe^{i\theta}, \forall \theta \in \Bbb R \implies x$ is lin. ind. of $y$
Proof:
Suppose $x$ and $y$ are not lin. ind.
$\implies \exists \lambda \in \Bbb C$  s.t. $y = \lambda x$ and $\lambda \neq 0$
$\implies \left\lVert \lambda x \right\rVert = \left\lVert y  \right\rVert$
$\implies$
$ |\lambda| \cdot \left\lVert x \right\rVert = \left\lVert y \right\rVert$
If $\left\lVert x \right\rVert = \left\lVert y \right\rVert$, then $|\lambda| = 1 \implies \lambda = e^{i\theta}$ for some $\theta \in \Bbb R$
Therefore, $\exists \theta \in \Bbb R$ s.t. $y = xe^{i\theta}$
