Finding $f(z)$ from $\Im(f'(z))$. The problem reads

If $f(z)$ satisfy the following condition, $\Im (f' (z)=6x(2y-1)$, $f(0) = 3-2i$, and $f(1) =6-5i $, find $f(1+i)$.

I have tried solved this problem, and the result I got is $f(z) = 2z^3-3iz^2+C$, and also, $f(x,y) = (2x^2-6xy^2-6xy)+i(6x^2y+3y^2-3x^2+2y^3) + C$ where $C$ is a complex constant. However, when plugging in the values for $f(0)$ and $f(1)$, I got different values for $C$.
I just want to know whether the result I got is right or I have to try it again.
Thanks in advance.
 A: A holomorphic function on a connected domain it determined by its imaginary part only up to an additive real constant. Here $\operatorname{Im}(f'(z)) = 12xy - 6x$ is given, and the general solution is
$$
 f'(z) = 6 z^2-6iz + A
$$
for $z \in \Bbb C$, with an arbitrary constant $A \in \Bbb R$. Integration gives
$$
 f(z) = 2 z^3 - 3i z + Az + C
$$
with constants $A \in \Bbb R$ and $C \in \Bbb C$.
Now the equations $f(0) = 3-2i$ and $f(1) = 6-5i$ can be solved for $A$ and $C$, and then $f(1+i)$ can be computed.
A: Your answer that $f(z) = 2z^3 - 3iz^2 + C$ is correct. However, when you expanded $f(x + iy)$, it looks like you made some errors on a couple of the terms - both sign errors and exponent errors.
If you work solely from the first version of the function, you should find that you get consistent values of $C$ for both of the given points.
A: OP and Comments confused me.
CorrectioN:
$12xy=\Im(6z^2)$ and $-6x=\Im(-6iz)$, so $6x(2y-1)=\Im(6z^2-6iz)$ and hence $f'(z)=6z^2-6iz+d$ where $d$ is real (Here was the mistake of OP and me.)
Integrating, $f(z)=2z^3-3iz^2+dz+c$.
$f(0)=c=3-2i$ gives $c=3-2i$ and $f(1)=2-3i+d+c=6-5i$ gives $c=3-2i$ and $d=1$. $$f(z)=2z^3-3iz^2+z+3-2i.$$
Then $f(1+i)=6+3i$.
