Problem. The real part of an analytic function $f(z)$ is given by $3x^2y-y^3$. Find the imaginary part. Find the analytic function of $z$.

Attempt. I begin by applying the Cauchy-Riemann Equations. As $f(z)$ is given to be analytic, we may conclude that:

$$\frac{\partial u}{\partial x} = \frac{\partial v}{\partial y}\;\;\;\;\;\;\frac{\partial u}{\partial y} = -\frac{\partial v}{\partial x}$$

Where $u(x,y)$ and $v(x,y)$ are the real and imaginary parts of $f(z)$, respectively ($x,y\in\mathbb{R}$). Hence,

$$\frac{\partial v}{\partial y} = 6xy \;\;\;\;\;\;\; -\frac{\partial v}{\partial x} = 3x^2-3y^2$$

Integrating these, we obtain:

$$f(z) = \int 6xy\;dy = 3xy^2 + C(x)\;\;\;\;\;\; f(z) = \int -(3x^2-3y^2)\;dx = 3xy^2-x^3 + C(y)$$

Then, the imaginary part of $f(z)$ is given by the sum of the two integrals, $6xy^2 - x^3$, hence:

$$f(z) = 3x^2y-y^3 + i(6xy^2-x^3)$$.

Unfortunately, I seem to have erred somewhere. Where have I gone wrong? Is this a valid approach?

  • $\begingroup$ Why are you adding the two $f(z)$s? $\endgroup$
    – sku
    Commented Mar 14, 2022 at 3:13
  • $\begingroup$ @sku Oops! Thanks for pointing that out, not sure what I was thinking there. $\endgroup$
    – 10GeV
    Commented Mar 14, 2022 at 4:23

2 Answers 2


$u(x,y) = 3x^2y-y^3 $

$u_x = v_y =6xy $

$ u_y=-v_x = 3x^2-3y^2$

$ dv = \frac{\partial v}{\partial x}dx + \frac{\partial v}{\partial y}dy$

$ dv = -\frac{\partial u}{\partial y}dx +\frac{\partial u}{\partial x}dy$ $v = (-3x^2+3y^2)dx +6xydy$

this is an exact differential equation( check it) so $v = \int_{y=constant}3y^2-3x^2dx + \int_{x=0}6xydy$ \

$v=3y^2x-x^3 +c_1 $

$f(x,y)= u(x,y)+v(x,y)=3x^2y-y^3 +i(3xy^2-x^3-c_1) $

  • $\begingroup$ there are also other methods to do this. $\endgroup$ Commented Mar 14, 2022 at 10:21

It is $v(x,y)=3xy^2+C(x)$. We know that $v_x(x,y)=3y^2-3x^2$ so $C'(x)==-3x^2$ Hence $C(x)=-x^3+A$ where A is a complex constant. Now$f(z)=u(x,y)+iv(x,y)$


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