Showing if scalar product of vectors is 1 then they are the same quantum state

Question

How do you show that if $|\langle \psi|\phi\rangle| = 1$, then $\phi$ and $\psi$, both of dimension $d$, represent the same quantum state?
(Same quantum state iff there exists a $\theta$ s.t. $|\psi\rangle = e^{i\theta}|\phi\rangle$)

I've tried doing given
$|\langle \psi|\phi\rangle| = 1$

$$\Leftrightarrow \left|\sum_{k=0}^{d-1}{\psi_{k}\phi_{k}}\right| = 1$$

$$\Leftrightarrow \left|\exp(i\theta)\sum_{k=0}^{d-1}{\psi_{k}\phi_{k}}\right| = 1 $$
for any $\theta \in \mathbb{R}$, but couldn't go much further.

Answer 1

Hint: use Cauchy–Schwarz inequality.

As quantum states are normalized, we have $$|\langle \psi | \phi \rangle|^2 = 1 = \langle \psi | \psi \rangle \langle \phi | \phi \rangle,$$ so $\phi$ and $\psi$ are linearly dependent. Neither of them is zero, hence one must be a scalar multiple of the other. This scalar is in $U(1)$, as both states have magnitude $1$.