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Use the Cauchy-Riemann equations to determine all differentiable functions that satisfy $Re(f(z))=xy$

I think I know how to do this problem. If we let $z=x+iy$, then $f(z)=u(x,y)+iv(x,y)$. We are given $u(x,y)=xy$. The Cauchy-Riemann equations give us, after some calculations, that $f(x,y)=xy+i(\frac{y^2-x^2}{2}+C)$, $C \in \mathbb{R}$.

I'm somewhat unsatisfied that my answer is expressed in terms of $x$ and $y$; I'd like to have it in terms of one complex variable $z$ rather than two real variables $x$ and $y$ [even if they are equivalent].

I've tried doing this with the identities $x=\frac{z+\bar{z}}{2}$ and $y=\frac{z-\bar{z}}{2i}$, but I arrive at something involving $z$ and $\bar{z}$. It's my understanding that differentiable (and hence analytic) functions shouldn't have a $\bar{z}$ in their formulas, so what am I doing wrong?

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2 Answers 2

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What you have is $-\frac{i}{2} z^2 + iC$, and this seems fine. The issue ought to be in the step you did not detail.

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  • $\begingroup$ Right you are. Guess I had a bit of a brain fart. Thanks! $\endgroup$
    – Ducky
    Jan 25, 2015 at 20:23
  • $\begingroup$ No problem. It was a well-written question. $\endgroup$
    – quid
    Jan 25, 2015 at 20:28
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$$\frac{z+\bar{z}}{2}\frac{z-\bar{z}}{2i} + i((\frac{z-\bar{z}}{2i})^2/2-(\frac{z+\bar{z}}{2})^2/2 +C)$$

$$\frac{z^2-\bar{z}^2}{4i} + i(\frac{z^2-2z\bar{z}+\bar{z}^2}{-4}/2-\frac{z^2+2z\bar{z}+\bar{z}^2}{4}/2+C)$$

$$\frac{z^2-\bar{z}^2}{4i} + i\frac{z^2-2z\bar{z}+\bar{z}^2}{-4}/2-i\frac{z^2+2z\bar{z}+\bar{z}^2}{4}/2+iC$$

$$\frac{z^2-\bar{z}^2}{4i} + i\frac{z^2-2z\bar{z}+\bar{z}^2}{-4}/2+i\frac{z^2+2z\bar{z}+\bar{z}^2}{-4}/2+iC$$

$$\frac{z^2-\bar{z}^2}{4i} + i\frac{z^2-2z\bar{z}+\bar{z}^2+z^2+2z\bar{z}+\bar{z}^2}{-4}/2+iC$$

$$\frac{z^2-\bar{z}^2}{4i} + i\frac{2z^2+2\bar{z}^2}{-4}/2+iC$$

$$\frac{z^2-\bar{z}^2}{4i} + i\frac{z^2+\bar{z}^2}{-4}+iC$$

$$\frac{z^2-\bar{z}^2}{4i} + \frac{z^2+\bar{z}^2}{4i}+iC$$

$$\frac{z^2}{2i} +iC$$

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