Finding imaginary part of a function $f(x+iy)=u(x,y)+iv(x,y)$ For this question it is assumed that $f(x+iy)$ is differentiable on $C$. For a function $f(x+iy)=u(x,y)+iv(x,y)$, the real part $u(x,y)=x^3y-xy^3$. Now, I am trying to find the imaginary part of the function, so I used the Cauchy Riemann equations and I got $\frac{\partial u}{\partial x}=3x^2y-y^3=\frac{\partial v}{\partial y}$ and $-\frac{\partial u}{\partial y}=3xy^2-x^3=\frac{\partial v}{\partial x}$.
 A: Your very close to the correct answer. When you compute the integrals the 'constants' are constant with respect to the variable of integration. For example, your first integral should be:
$$v=\int 3x^2y-y^3 dy = \frac{3}{2}x^2y^2-\frac14y^4 + \phi(x)$$
where $\phi(x)$ is a function of $x$ to be determined.
Now differentiating with respect to $x$, then using the Cauchy-Riemann equations gives:
$$\frac{\partial v}{\partial x} = 3xy^2+\phi'(x) \quad \Rightarrow \quad \phi'(x) = -\frac{\partial u}{\partial y} - 3xy^2 = -x^3$$
So that $\phi(x) = -\frac14 x^4+C$, for constant $C$. Thus we have:
$$v=\frac{3}{2}x^2y^2-\frac14y^4-\frac14x^4 + C$$
Notice that this means that we can only determine $f$ up to a constant.
A: A non-systematic answer: One immediately sees that $f$ must be purely imaginary on both the real and the imaginary axis. Also, the real part at least is purely a fourth degree polynomial of in the components of $z$. Together this suggests a function of the form $f(z)=ciz^4$ for some real $c$ ...
(One lesson to take home here, apart from serendipity, is if one of the coordinate functions for a differentiable complex function has a nice algebraic expressions, then there is almost always also an nice expression for the full function that uses purely complex operations. Searching for that form can give a much more satisfying answer than one phrased purely in terms of $x$s and $y$s).
