Is $e^{i\beta} R_Z(-2\beta)$ equivalent to $U_1(2\beta)$?

Question

As far as I know the single qubit gate

$$ e^{i\beta\sigma_z} = \begin{bmatrix} e^{i\beta} & 0 \\ 0 & e^{-i\beta} \end{bmatrix} = e^{i\beta} \begin{bmatrix} 1 & 0 \\ 0 & e^{-i2\beta} \end{bmatrix} = e^{i\beta} R_Z(-2\beta). $$

However, I have seen the above gate implemented using $U_1(2\beta)$, where $ U_1(\lambda) = \begin{bmatrix} 1 & 0 \\ 0 & e^{i\lambda} \end{bmatrix} $

Is $e^{i\beta} R_Z(-2\beta)$ equivalent to $U_1(2\beta)$?

Update:

As Davit explains below, $e^{i\theta/2} R_Z(\theta) = U_1(\theta)$, so with $\frac{\theta}{2}=\beta$ we have $e^{i\beta} R_Z(2\beta) = U_1(2\beta)$. Note the difference: in this case the $Z$-rotation is positive, whereas in my original question it is negative.

Answer 1

$R_z$ gate from M. Nielsen and I. Chuang textbook (page 174):

$$R_z(\theta) = e^{-i\theta Z/2} =\begin{pmatrix} e^{-i \theta /2} &0 \\ 0&e^{i \theta /2} \end{pmatrix}$$

If we use this definition for $R_z(\theta)$ then:

$$R_z(\theta) = e^{-i \theta/2} \begin{pmatrix} 1 &0 \\ 0&e^{i \theta} \end{pmatrix} = e^{-i \theta/2} U_1(\theta)$$

Therefore, $R_z(\theta)$ and $U_1(\theta)$ can be regarded as equivalent gates, because the global phase $e^{-i \theta/2}$ can be neglected. Note that this comparison is not true for their controlled versions: $cR_z(\theta)$ and $cU_1(\theta)$ can't be regarded as equivalent gates.