To resolve your confusion here, you have to distinguish between two-dimensional and three-dimensional flow.
The complex potential is a very useful tool for analyzing two-dimensional flows that can be characterized as potential flows -- that is, inviscid, irrotational and incompressible. If the velocity field in terms of Cartesian components is $(u,v)$ then we find in two-dimensional potential flow, the complex velocity $u - iv$ is related to a complex potential $w(z)$ by
$$u - iv = - \frac{dw}{dz}$$
where we introduce the complex variable $z = x+iy \in \mathbb{C}.$
For a two-dimensional line source at the origin, the contribution to the complex potential due to the source is indeed $-m \log z.$
However, the problem appears to pertain to three-dimensional flow affected by a point source, where complex analysis is not generally applicable.
The notion of a velocity potential in three-dimensions arises when the
flow is irrotational. In this case, the curl of the velocity field vanishes (as it would in two-dimensional flow as well), and the velocity field can be expressed in terms of the gradient of a real-valued scalar potential function
$$\mathbb{u} = - \nabla \phi.$$
Ignoring the uniform stream which is handled in a straighforward way, if a point source emits incompressible fluid at a volumetric rate $4\pi m$, then conservation of mass requires that the flux through any spherical surface centered at the source is constant and
$$\tag{*} 4 \pi m = \int_0^{2 \pi} \int_0^\pi\mathbb{u} \cdot \mathbb{e}_r r^2 \sin \theta \, d\theta \,d \phi. $$
By symmetry only the radial component $u_r$ is non-zero and depends only on the radial coordinate $r$. Whence, (*) is satisfied for every $r > 0$ if
$$- \nabla \phi = \frac{d \phi}{dr } \mathbb{e}_r = \frac{m}{r^2} \mathbb{e}_r,$$
and the source contribution to the velocity potential is
$$\phi = \frac{m}{r}.$$
Best Answer
Yes you could do that, but it's probably easier to use the expression of the gradient in polar coordinates:
$$ u_r =\frac{\partial\phi}{\partial r} =\frac{1}{r}\frac{\partial\psi_z}{\partial \theta} , u_\theta=\frac{1}{r}\frac{\partial\phi}{\partial \theta}=-\frac{\partial \psi_z}{\partial r} $$
Both $u_r$ and $u_\theta$ should be zero.
As you probably recognized, this is the potential flow of a source/sink at the origin and a free stream parallel to $x$.