Eigenfunctions of squared angular momentum from differential equation

In spherical coordinates, the eigenfunctions of the operator $hat{L}^2$ can be obtained by solving the problem
$$
-dfrac{hbar^2}{sintheta}dfrac{partial}{partialtheta}left(sinthetadfrac{partial Y(theta,varphi)}{partialtheta}right) – dfrac{hbar^2}{sin^2theta}dfrac{partial^2 Y(theta,varphi)}{partialvarphi^2} = lambda Y(theta,varphi) .
$$
I want to solve this directly without using extra information, such as the eigenvalues and eigenfunctions of the operator $hat{L}_z$.

Using separation of variables, $Y(theta,varphi) = Theta(theta)Phi(varphi)$, I can separate the partial differential equation in two problems by using a separation constant $mu$:
$$
-dfrac{hbar^2sintheta}{Theta}dfrac{partial}{partialtheta}left(sinthetadfrac{partial Theta}{partialtheta}right) – lambdasin^2theta = dfrac{hbar^2}{Phi}dfrac{partial^2 Phi}{partialvarphi^2} = -mu
$$

I know that the problem regarding $Theta(theta)$ can be reduced to an equation whose solutions are the associated Legendre polynomials if $lambda = l(l+1)hbar^2$, so there is no problem with that. My problem is when solving the equation for $Phi(varphi)$, which can be written as
$$
dfrac{partial^2Phi}{partialvarphi^2} = -dfrac{mu}{hbar^2}Phi = -m^2Phi ,quadmu = m^2hbar^2 .
$$
The general solution to this equation is
$$
Phi(varphi) = Ae^{imvarphi} + Be^{-imvarphi} ,
$$
but I know from the eigenfunctions of $hat{L}_z$ that it should be reduced to just $Phi(varphi) = Ae^{imvarphi}$, and that the official solutions for the eigenfunctions of $hat{L}^2$ are $$Y_{l,m}(theta,varphi) = Theta_{l,m}(theta)Phi_m(varphi)propto P_{l,m}(costheta)e^{imvarphi} .$$
Applying the periodicity condition to the angle $varphi$, I have that
$$
Phi(0) = Phi(2pi) Rightarrow A + B = Ae^{i2pi m} + Be^{-i2pi m} ,
$$
which, as far as I know, it only tells me that $minmathbb{Z}$. So, ¿how can I get $B = 0$ to fulfill with the official solution?