Find v from u = (x-1)³ - 3xy² + 3y²

u = (x − 1)³ − 3xy² + 3y².

Check that u is harmonic

u_x  = 3(x − 1)² − 3y²     u_xx = 6(x − 1)
u_y  = −6xy + 6y           u_yy = −6x + 6

u_xx + u_yy = 6x − 6 − 6x + 6 = 0 ✓.

Find v by C–R equations (v_y = u_x, v_x = −u_y)

Integrate v_y = 3(x − 1)² − 3y² with respect to y:

v = 3(x − 1)² y − y³ + g(x)

Differentiate with respect to x: v_x = 6(x − 1)y + g′(x).

This must equal −u_y = 6xy − 6y = 6(x − 1)y. So g′(x) = 0 and g = constant.

Answer: v = 3(x − 1)² y − y³ + c

The analytic function

f′(z) = u_x(z, 0) − i u_y(z, 0) = 3(z − 1)², so f(z) = (z − 1)³ + c.

Check: (z − 1)³ = (x − 1 + iy)³ has real part (x − 1)³ − 3(x − 1)y² = (x − 1)³ − 3xy² + 3y² = u ✓, and imaginary part 3(x − 1)²y − y³ = v ✓.