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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...

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

2 Answers 2

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

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

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

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

Can you take it from here?

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  • $\begingroup$ I got it now, thanks. $\endgroup$
    – Lucas-L
    Feb 23, 2020 at 2:43
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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 $v(x,y)$.

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

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