# Cauchy Riemann equations, application

For $$u: \Bbb C \to \Bbb R$$ with $$u(x + iy) = 2x^3 - 6xy^2 + x^2 - y^2$$ find a function $$v: \Bbb C \to \Bbb R$$ s.t. $$f = u + iv$$ is holomorphic.

I see that I have to verify the Cauchy Riemann equations, but I did not get the right solution...

• What did you have and how did you get it? We might be able to find why it didn't work. Feb 23, 2020 at 0:44

$$\dfrac{\partial u }{\partial x}=6x^2-6y^2+2x=\dfrac{\partial v}{\partial y}$$

$$\dfrac{\partial u}{\partial y}=-12xy-2y=-\dfrac{\partial v}{\partial x}$$

$$v=6x^2y-2y^3+2xy+f(x)$$

$$v=6x^2y+2xy+f(y)$$

Can you take it from here?

• I got it now, thanks. Feb 23, 2020 at 2:43

Using the Cauchy-Riemann equations we have:

$$f'(z)=\frac{df}{dz} =\frac{\partial u}{\partial x}-i\frac{\partial u}{\partial y}=6x^2-6y^2+2x+12ixy+2iy$$

But $$f'(z)=f'(x+iy)$$, so replacing $$x$$ by $$z$$ and $$y$$ by zero in the last equation we get:

$$f'(z)=6z^2+2z\;\Rightarrow\;f(z)=2z^3+z^2+C$$

where $$C$$ is a constant. But $$u(0,0)=0$$ so the constant has to be purely imaginary $$C=i\alpha$$ for some real $$\alpha$$. Taking the imaginary part of $$f(z)$$ we get the most general $$v(x,y)$$.

$$v(x,y) = \mathrm{Im}(2z^3+z^2+i\alpha)=6x^2y-2y^3+2xy+\alpha.$$