find all $v(x,y)$ so that $f(x+iy)=u(x,y)+iv(x,y)$ is entire I'm practicing to write down solutions clearly and thoroughly. Is this a proper answer to this exercise? How can it be improved?
Let $u(x,y)=x^3-3xy^2$, find all $v(x,y)$ so that $f(x+iy)=u(x,y)+iv(x,y)$ is entire.
The Cauchy-Riemann equations tell us that:
$$\frac{\partial u}{\partial x}=\frac{\partial v}{\partial y} \qquad \frac{\partial v}{\partial x}=-\frac{\partial u}{\partial y}$$
Therefore:
$$\int \frac{\partial u}{\partial x} dy= \int \frac{\partial v}{\partial y} dy = v(x,y)$$
$$\int -\frac{\partial u}{\partial y} dx =\int \frac{\partial v}{\partial x} dx= v(x,y)$$
Using that:
$$\frac{\partial u}{\partial x}=3x^2-3y^2 \qquad \frac{\partial u}{\partial y}=-6xy$$
I find:
$$\int 3x^2-3y^2 \, dy = 3x^2y - y^3 + C_1 = v(x,y)$$
$$\int 6xy \, dx = 3x^2y + C_2 = v(x,y)$$
So:
$$3x^2y - y^3 + C_1 =  3x^2y + C_2 \Longrightarrow C_1 = w, C_2=-y^3 + w \quad w \in \mathbb{C}$$
And:
$$v(x,y)= 3x^2y - y^3 + w \quad w\in \mathbb{C}$$
 A: You should write $C_1(x)$ instead of $C_1$ and $C_2(y)$ instead of $C_2$:
$$v(x,y)=3x^2y-y^3+C_1(x)=3x^2y+C_2(y)$$
Then, $C_2(y)=C_1(x)-y^3$. This implies that $C_1(x)$ is constant. Therefore,
$$v(x,y)=3x^2y-y^3+C$$
for some $C\in\Bbb C$.
A: Note
$$z^3 = (x+iy)^3 = x^3 + 3x^2(iy) +3x(iy)^2 +(iy)^3 = x^3  -3xy^2 +i(3x^2y-y^3).$$
Since $z^3$ is holomorphic, one answer to the problem is $v(x,y) = 3x^2-y^3.$ You still have to show any other answer has the form $v(x,y) + C$ for some real $C,$ but at least this shows where the problem comes from.
A: I think our OP falibali's argument is essentially correct and well-written, but could be improved slightly by the inclusion of the suggestions made in axotatje's answer.  One need always remember that, when integrating a multivariate function with respect to only one of its arguments, the so-called "constant of integration" will in actuality be a function of the remaining (non-integrated) variables.  And I think for the sake of clarity and communication it is often best to explicitly show these functional dependencies.  These caveats aside, the OP's proof looks fine.
These things being said, there are other approaches which lend themselves to clean, concise expression:
Assuming we're looking for real functions $v(x, y)$, here's another way to look at it:  with $z = x + iy$, recall that
$z^3 = (x + iy)^3 = x^3 - 3xy^2 + i(3x^2y - y^3), \tag{1}$
by the binomial theorem, or by just grinding it out if one pefers.
Since $z^3$ is entire (as indicated in zhw.'s answer), and the real part of $z^3$ is $x^3 - 3xy^2 = u(x, y)$, we see that $v(x, y) = 3x^2y - y^3$ is one possible solution for $v$.  Now let $w(x, y)$ be another function with $f(z) = u(x, y) + iw(x, y)$ entire. Then $z^3 - f(z)$ is also entire; but
$z^3 - f(z) = u + iv - (u + iw) = i(v - w); \tag{2}$
that is, the real part of $z^3 - f(z)$ is zero.  Then by Cauchy-Riemann applied to $z^3 - f(z)$, we see that
$\nabla (v - w) = 0; \tag{3}$
that is, $v - w$ must be constant.  Thus
$w(x, y) = v(x, y) + C = 3x^2y - y^3 + C, \;\; C \in \Bbb R; \tag{4}$
thus every requisite $v(x, y)$ is of the form
$v(x, y) = 3x^2y - y^3 + C \tag{5}$
for some real $C$.  
