By Helmholtz' theorem, any continuously differentiable vector field that vanishes at infinity can be decomposed into irrotational and soleniodal parts of the form
$$\mathbf{u} = \nabla \phi + \nabla \times \mathbf{a}.$$
In potential flow we assume that the velocity field is irrotational, whence
$$\mathbf{u} = \nabla \phi.
$$
Applying the continuity condition we have
$$\nabla \cdot \mathbf{u} = \nabla^2 \phi = 0.$$
With suitable boundary conditions there always exists a solution of Laplace's equation for the velocity potential $\phi$ and hence, the velocity field. The Euler equation ensures conservation of momentum and closes the system of equations so we can solve for the pressure field (always).
The Euler equation applies to the general class of inviscid flows (incompressible or compressible) where incompressible potential flow is a special case. With respect to (b), there is no steady, incompressible, inviscid flow (irrotational or rotational) that does not satisfy the Euler equation. Only if the velocity field solves the viscous Navier-Stokes equations will it fail to satisfy the Euler equation.
Also, there is such a thing as unsteady potential flow.
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
$$f(z)=U \left( \frac{az-b\sqrt{z^2-a^2+b^2}}{a-b} \right)$$
$$f(z)=U \left( \frac{z+\sqrt{z^2-a^2+b^2}}{2} \right)+ \overline{U} \left( \frac{a+b}{a-b} \right) \left( \frac{z-\sqrt{z^2-a^2+b^2}}{2} \right)$$