I can't quite figure out the last part of this problem…
Find a harmonic conjugate of the harmonic function $u(x,y)= x^3-3xy^2$. Write the resulting analytic function in terms of the complex variable z.
One harmonic conjugate of u is $v(x,y) = 6x^2y-y^3$ (probably wrong, but here's the calculation):
$v(x,y) = \int_{(x_0,y_0)}^{(x,y)} -u_t(s,t)ds + u_s(s,t)dt = \int_{(x_0,y_0)}^{(x,y)} 6st ds + (3s^2-3t^2)dt$
Choosing $(x_0,y_0) = (0,0)$ (since we're finding a harmonic conjugate):
$v(x,y) = 3x^2y+3x^2y-y^3 = 6x^2y-y^3$
So, the resulting analytic function is $f(z) = x^3 – 3x^2y + i(6x^2y-y^3)$ ($z = x+iy$)
How do you write this in terms of $z$? If it was $f(z) = x^3 – 3x^2y + i(3x^2y-y^3)$, it would be just $f(z) = z^3$, but with the $6$ in there, I don't know where to go.
Best Answer
Indeed, that calculation is wrong. With the choice of $(x_0, y_0) = (0, 0)$ as initial point we have $$ v(x, y) = \int_{(0, 0)}^{(x,y)} 6st \, ds + (3s^2-3t^2) \, dt \\ = \int_{(0, 0)}^{(x,0)} 6st \, ds + \int_{(x, 0)}^{(x,y)} 6st \, ds + \int_{(0, 0)}^{(x,0)} (3s^2-3t^2) \, dt + \int_{(x, 0)}^{(x,y)} (3s^2-3t^2) \, dt \, . $$ The first integral is zero because $t=0$ on the path. The second and third integrals are zero because $ds=0$ respectively $dt=0$. It remains $$ v(x, y) = \int_{(x, 0)}^{(x,y)} (3s^2-3t^2) \, dt = \int_0^y (3x^2 -3t^2) \, dt = 3x^2 y- 3 \frac{y^3}{3} = 3x^2 y - y^3 \, . $$ That is consistent with your observation that $$ z^3 = (x+iy)^3 = x^3 - 3x^2y + i(3x^2y-y^3) \, . $$ is an analytic function with the given real part $u$. The general solution is obtained by adding an arbitrary imaginary constant.