On finding a holomorphic function given its derivative's real part

Question

Suppose we have that $$\Re \left \{ \frac{df}{dz} \right \}=3x^{2}-4y-3y^{2}$$ I used the following handy theorem :

If $f$ is a holomorphic function in a simply connected region, and $z_{0}$ is a fixed point in that region, then $z \mapsto F\left ( z \right )=\oint_{z_{0}}^{z}f\left ( z \right )dz$ is a holomorphic function too. This way, $F\left(z\right )$ doesn't depend on the path of the contour integral, and is unambiguously determined. And then, knowing that a holomorphic function is infinitely differentiable I found $f$ using Cauchy-Riemann's equations for $f^{'}$ and then integrating $f^{'}$:$$f\left ( z \right )=z^{3}+i2z^{2}+iC_{1}z+C_{2}$$ However, I've only been given the information that $f\left(1+i\right)=0$, so I can determine the function only up to a constant. Would you suggest a different approach?

Answer 1

I would determine $f'(z)$ first and then integrate it symbolically. From the problem statement we have $$ \Re(f'(z)) = 3x^2-3y^2-4y $$ We can match the $3x^2$ by $3(x+iy)^2 = 3x^2-3y^2-6xyi$, and subtract the real part of that both sides, giving $$ \Re(f'(z) - 3z^2) = -4y $$ and then match $-4y$ by $4i(x+iy) = -4y + 4xi $, and subtract on both sides again to get $$ \Re(f'(z) - 3z^2 - 4iz) = 0 $$ An analytic function whose real part is zero everywhere must be a pure imaginary constant, so $$ f'(z) = 3z^2 + 4iz + c_1i $$ Now we can integrate this term by term, and get $$ f(z) = z^3 + 2iz^2 + c_1iz + c_2 + c_3i $$

However, as in your solution, knowing just one function value is not enough to determine all three constants. This is not just a failing of the solution method, however: Each choice of $c_1$ leads to particular values of $c_2$ and $c_3$ that will satisfy your original equation and function value, so the problem genuinely does not have a unique solution.