find $f(1+i)$ if $\Im[f'(z)]=6x(2x-y), f(0) = 3-2i, f(1) = 6-5i$ Given that $\Im[f'(z)]=6x(2x-y), f(0) = 3-2i, f(1) = 6-5i$, where $z = x+iy$ how to find $f(1+i)$. The answer in answer sheet is $6+3i$. Any hints will be appreciable.

Hints: $f$ differentiable implies $f'$ differentiable. Then $f'$ satisfies Cauchy-Riemann equations - this should tell you a lot about the real part of $f'$. And once you know $f'$, the initial conditions will tell you $f$. 
