The map is conformal whereever f′ is nonzero. If the two curves have direction vectors u and v at p, the new direction vectors are f′(p)×u and f′(p)×v. Multiplication by the common factor f′(p) moves both u and v through a common angle, thus the images of the two curves meet at the same angle in the domain and range.
Next let f′ = 0. For convenience let p = 0, and write f = zmg+q, where g is an analytic function with g(0) nonzero, q is a constant, and m is at least 2. Shifting the image by q isn't going to change anything, so ignore that. The angle of the direction vector u is now multiplied by m, courtesy of zm, and this is multiplied by g, which is close to g(0) near the origin, which is a nonzero constant. Multiplying by this nonzero constant moves the direction vector through a fixed angle. In the end, both u and v are multiplied by m, then rotated by a fixed amount. The angle between u and v is multiplied by m.
A map is conformal across its domain iff f′ is everywhere nonzero.