I'm not quite sure what you are saying with your conditions.
Since #\frac{\partial}{\partial x}(u(y))=0=\frac{\partial}{\partial y}(v(x))# for all #x# and #y#, the first of the Cauchy-Riemann equations doesn't put any conditions on the functions in the form given.
Since #\frac{\partial}{\partial y}(u(y))=u'(y)# and #-\frac{\partial}{\partial x}(v(x))=-v'(x)#, the second of the Cauchy-Riemann equations implies that #u'(y)=-v'(x)# for all #x# and #y#. Since the LHS is a function of #y# and the RHS is a function of #x#, it follows that #u'(y)=-v'(x)=A# for some (real) constant #A#.
Integration now implies #u(y)=Ay+B# and #v(x)=-Ax+C# for some (real) constants #B# and #C#.
Therefore, functions of the form #f(z)=(Ay+B)+i(-Ax+C)# are the functions of the form #f(z)=u(y)+iv(x)# that are analytic. This corresponds to functions #w=f(z)# of the form #f(z)=-iAz+(B+iC)# where #A, B#, and #C# are real.
The geometric effect of such a function is to dilate a given complex number/vector #z# by #|A|#, reflect it through the origin if #A<0#, rotate it clockwise by #90^{\circ}# (because of multiplying by #-i#), translate horizontally by #B# units (right if #B>0# and left if #B<0#), and then translate vertically by #C# units (up if #C>0# and down if #C<0#).
