- Is the function
\[ f ( x , y ) = x ^ { 3 } + 3 x y ^ { 2 } - 3 x ^ { 2 } y - 1 0 + i ( y ^ { 3 } + 3 x ^ { 2 } y - 3 y ^ { 2 } x + 5 ) \]
complex analytic? Prove that your answer is correct.
Solution
The function is not complex analytic on any nonempty open set.
1. The Cauchy–Riemann equations hold only on two real lines.
Write \(f=u+iv\) with \(u=x^3+3xy^2-3x^2y-10\) and \(v=y^3+3x^2y-3xy^2+5\). Then \[ u_x=3x^2+3y^2-6xy=v_y,\qquad u_y=6xy-3x^2,\qquad v_x=6xy-3y^2. \] The remaining Cauchy–Riemann equation \(u_y=-v_x\) is equivalent to \[ x^2-4xy+y^2=0, \quad\text{or}\quad \bigl(y-(2+\sqrt3)x\bigr)\bigl(y-(2-\sqrt3)x\bigr)=0. \] Thus the equations hold precisely on the union of these two lines. As \(u,v\) are polynomials, their real partial derivatives are continuous; the Cauchy–Riemann criterion therefore also shows complex differentiability at the points of these lines [@SS03].
2. This pointwise differentiability cannot give analyticity on an open set.
A holomorphic function must satisfy both Cauchy–Riemann equations at every point of its open domain [@SS03]. The union of two lines contains no nonempty open disk: in any disk one can fix a horizontal coordinate and choose a vertical coordinate different from the at most two values specified by the lines. Hence no nonempty open set is contained in the differentiability locus. In particular, at \((1,0)\) one has \(u_y=-3\) but \(-v_x=0\), which alone rules out analyticity on the whole plane.