How to prove the following claim

Question

I find an interesting claim in my research:

symbols

Suppose $V$ is a unitary matrix, $U$ is defined to be $$U_{\alpha\beta}=V_{\alpha\beta}e^{i\phi_\beta},$$ where $\boldsymbol{\phi}\equiv(\phi_1,\phi_2,...\phi_n)$ ($n$ is dimension of $V$) is a real valued vector regarded as free variables of $U$. Since $U$ is also a unitary matrix, it can be diagonalized as $U=Q\Lambda Q^{\dagger}$, where $\Lambda_{ij}=e^{i\theta_j}\delta_{ij}$. We denote root of $\text{det}(I-U)$ ($I$ is identity matrix) as $\bar{\boldsymbol{\phi}}$. One of the eigenvalue $e^{i\theta_\bar{j}}$ must be $1$ at $\bar{\boldsymbol{\phi}}$, since determinant is just products of eigenvalues. Generally, this eigenvalue is not degenerate, and this is assumed to be true here. In this case $\bar{j}$ is unique.

claim

$$|Q_{\alpha\bar{j}}|^2=\partial_{\phi_\alpha}\theta_{\bar{j}}$$ at $\bar{\boldsymbol{\phi}}$.

numerical verification

I cannot prove this claim. I believe this is true only because I can verify it numerically. Here's the code. (Sadly it's in julia, which maybe not that popular among mathematics community.)

using PyCall
using Optim
@pyimport scipy.stats as ss

ndim = 6
function getU(ϕs, V) # create U matrix
    U = zeros(Complex128, ndim, ndim)
    for i in 1:ndim
        U[:, i] = V[:, i]*exp(im*ϕs[i])
    end
    return U
end
function getD(ϕs, V) # absolute value of determinant
    return abs(det(eye(ndim)-getU(ϕs, Uin)))
end
restϕs = [π/4, π/5, π/6, π/7, π/3]
V = ss.unitary_group[:rvs](dim=ndim) # random unitary matrix
result = optimize(ϕ1->getD([ϕ1; restϕs], V), [0.0], BFGS()) # find ϕ1bar 
ϕ1bar = Optim.minimizer(result)[1]
U = getU([[ϕ1bar]; restϕs], V)
egvals, egvecs = eig(U)
ione = 0 # find the index of unity eigenvalue
for ione in 1:ndim
    if abs(egvals[ione]-1) < 1.0e-5
        break
    end
end
# numerical differentiation
println(angle(eig(getU([[ϕ1bar+0.0001]; restϕs], V))[1][ione])/0.0001) 
println(abs2(eig(U)[2][1, ione]))
# both prints 0.255 in my desktop