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2 similar questions

  1. Find all possible $u(t)$ and $v(x,y)$ for which the function $f(z) = u(xy) + iv(x, y)$ is holomorphic.

  2. Let $f$ be an entire function (analytical in $\mathbb{C}$) of the form $f(x,y) = u(x) + iv(y)$. Prove that it is a constant.\ (there may be a typo in this question)

I'm pretty sure the process is, take partial derivatives and use the Cauchy-Riemann equations to then integrate. The question is, what do $u(xy), u(x), iv(y)$ mean?

Thus-far in my complex analysis course, everything appears in the form f(z) = u(x,y) + iv(x,y). I can guess that u(x), iv(y) refer to functions that are only have x or y in them*, but I'm at a loss as to what $u(xy)$ means.

*(not sure if this interpretation is correct, and if so, would it also include polynomials, rationals, and exponentials?)

Sunday
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  • See this for correct statement about 2. – Prism Sep 18 '13 at 18:07
  • $u(x)$ means that $u$ is a function of $x$ only. In other words, it does not depend on $y$. (Although in general, $u$ does depend on both $x$ and $y$, as you wrote $f(z)=u(x,y)+i v(x,y)$). Your intuition is correct: $u(x)$ can be thought of as a function that "have" only $x$ in it, but note that in general functions don't need to have explicit formulas. – Prism Sep 18 '13 at 18:09
  • With $u(xy)$, it is meant that there is a function $u \colon \mathbb{R}\to\mathbb{R}$ such that $\operatorname{Re} f(x+iy) = u(x\cdot y)$, the real part of $f$ is the composition of $(x,y) \mapsto x\cdot y$ and $u$. – Daniel Fischer Sep 18 '13 at 18:11

1 Answers1

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All $u(xy)$ means is that $x$ and $y$ only appear in the function in pairs, multiplied by each other. In other words there is some function $ u: \mathbb R \to \mathbb R \ \ \ t \mapsto u(t)$ and the real part of the complex function $f(z)$ can be expressed as $u(xy)$. It would make sense to write $u(x,y)$. But this would carry less information. $u(xy)$ could be $(xy)^4 - xy$ for example, but it cannot be $x^2 y$.

You can answer this question by requiring the Cauchy Riemann equations to hold for $f$ and then writing $u(xy)$ as $u(\alpha(x))$ where $\alpha(x) = xy$ and employing the chain rule $\frac{\partial}{\partial x} u(\alpha(x)) = \frac{\partial u}{\partial t} \frac{\partial \alpha}{\partial x} $.

Daron
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