= Solution
These potentials give the uniform <magnetic field> $\mathbf B=B\widehat{\mathbf z}$ and zero <electric field>. With the signed frequency $\omega_c=qB/m$, the <Lorentz force law> becomes
$$
\ddot x=\omega_c\dot y,
\qquad
\ddot y=-\omega_c\dot x,
\qquad
\ddot z=0.
$$
Thus
$$
\dot x+i\dot y=(\dot x_0+i\dot y_0)e^{-i\omega_ct},
$$
and explicitly
$$
\begin{aligned}
x(t)&=x_0+\frac{\dot x_0\sin(\omega_ct)+\dot y_0[1-\cos(\omega_ct)]}{\omega_c},\\
y(t)&=y_0+\frac{\dot y_0\sin(\omega_ct)-\dot x_0[1-\cos(\omega_ct)]}{\omega_c},\\
z(t)&=z_0+\dot z_0t.
\end{aligned}
$$
The transverse trajectory is a circle about the fixed <guiding centre> $(x+\dot y/\omega_c,y-\dot x/\omega_c)$. The positive <cyclotron frequency> is
$$
\boxed{\omega_B=\frac{|qB|}{m}}.
$$
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