# Finding the analytic function of $z$, given the real part

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?

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

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

$$u_x = v_y =6xy$$

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

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

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

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

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

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

• there are also other methods to do this. 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$$f(z)=u(x,y)+iv(x,y)$$