Find the complex potential function for V

Question

Find the complex potential Φ(z) for v.

Let v(x, y) = ⌈ 3x²− 3y²⌉

⌊ −6xy ⌋

The answer is Φ(z) = z³

^{I need help to solve it i am not sure of how they got the answer}

Answer 1

Let Φ = f + ig. Then f_x = 3x² - 3y² (so that f(x,y) = x³ - 3xy² + h(y) for some function h) and f_y = -6xy, i.e., -6xy + h'(y) = -6xy. We deduce h'(y) = 0, forcing h(y) to be constant. If we assume Φ0,0) = 0, then f(0,0) = 0, implying h(y) = 0. Hence f(x,y) = x³ - 3xy².

Since Φ is analytic, the Cauchy - Riemann equations are satisfied, so

(1) g_x = -f_y = 6xy

(2) g_y = f_x = 3x² - 3y²

Integrating equation (1) yields g(x,y) = 3x²y + k(y) for some function k. Then (2) gives 3x² + k'(y) = 3x² - 3y², or, k'(y) = - 3y². Therefore k(y) = -y³ + C for some constant C. The condition Φ0,0) = 0 implies g(0,0) = 0, whence C = 0. Consequently, g(x,y) = 3x²y - y³, and

Φ(z) = (x³ - 3xy²) + (3x²y - y³)i = x³ + 3x(iy)² + 3x²(iy) + (iy)³ = (x + iy)³ = z³

as desired.

Note: All complex potentials associated to v are of the form z³ + C where C is an arbitrary constant. Prescribing a value of Φ at a point then uniquely determines Φ. This is what I've done here, assuming Φ(0,0) = 0.