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For example I have a complex function $f(z)$, $z = x + iy$.

I know that $\operatorname{Re} f(z) = x^3 + 6 x^2 y + A x y^2 - 2 y^3$

and that $f(0) = 0$.

What should I do to find $f(z)$?

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    Entry $f(z)$ means? – 2017-02-24
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    @VikrantDesai, sorry, I meant entirely – 2017-02-24
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    @Harry49, does it mean that there isn't any suitable $f(z)$? – 2017-02-24
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    If $f(0)=2$, then $Ref(0)=2$. But from $\operatorname{Re} f(z) = 2 \cdot y^3 - 6 \cdot x^2 \cdot y$, we get $Ref(0)=0$. hence no such $f$ exists. – 2017-02-24
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    @Fred, OK, it was bad example, I edited the question – 2017-02-24
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    Do we assume $f(z)$ is analytic, for example? Continuous only? Arbitrarty? – 2017-02-24
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    @coffeemath, analytic – 2017-02-24
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    What happens if one tries the Cauchy Riemann equations on this $u(x,y)$ as real part? Do we get a possible $v(x,y)$ for imaginary part? [Also I think the real part of an analytic function must be harmonic.] – 2017-02-24

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First method: Cauchy-Riemann equations. The function $f$ is assumed analytic. Therefore, there is an open set around zero where it is holomorphic. The Cauchy-Riemann equations write \begin{equation} \frac{\partial\, \text{Im} f}{\partial y}(x+\text{i}y) = \frac{\partial\, \text{Re} f}{\partial x}(x+\text{i}y) = 3x^2 + 12xy + Ay^2 \end{equation} and \begin{equation} \frac{\partial\, \text{Im} f}{\partial x}(x+\text{i}y) = -\frac{\partial\, \text{Re} f}{\partial y}(x+\text{i}y) = -6x^2 -2Axy + 6y^2 \, . \end{equation} Due to the equality of mixed partials, one has \begin{equation} \frac{\partial^2\, \text{Im} f}{\partial y\,\partial x}(x+\text{i}y) = \frac{\partial^2\, \text{Im} f}{\partial x\,\partial y}(x+\text{i}y) \, , \qquad\text{viz.}\qquad A = -3\, . \end{equation} Integrating $\partial\text{Im}/\partial y$ with respect to $y$, one obtains \begin{equation} \text{Im} f(x+\text{i}y) = 3 x^2 y + 6 x y^2 - y^3 + B(x)\, . \end{equation} Then, the expression of $\partial\text{Im}/\partial x$ imposes \begin{equation} B'(x) = -6x^2\, ,\qquad\text{i.e.}\qquad B(x) = -2x^3 + C\, , \end{equation} where $C$ is in $\mathbb{R}$. Finally, the condition $f(0) = 0$ yields $C=0$, and \begin{equation} \text{Im} f(x+\text{i}y) = -2 x^3 + 3 x^2 y + 6 x y^2 - y^3 \, . \end{equation}

Second method: Power series. The function $f$ is assumed analytic. Thus, let us look for an expression as a convergent power series in the vicinity of $z=0$: \begin{equation} f(z) = \sum_{n=0}^{+\infty} a_n\, z^n \, , \quad\text{for}\quad |z|