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I'm working through a problem set in Complex Analysis and have encountered the following question:

Problem

Write down the Cauchy-Riemann equations and explain their connection with holomorphic functions. [Completed]

Let $U=\mathbb{C} \diagdown (-\infty , 0]$. Show that for $z=x+iy\in\mathbb{C}$, we may define $u(x,y), \ v(x,y)$ by;

$$\sqrt{2}u=\left(x+\sqrt{x^2+y^2} \right)^{\frac{1}{2}}$$ $$v=\frac{y}{2u}$$

where we take positive square roots in the definition of $u$.

I can't see how we can identify a computation for $u$ and $v$ when they're seemingly arbitrary in the question. Is this a misprint, or am I missing something?

Regards,

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Have you tried assembling $u+iv$? –  J. M. Dec 9 '11 at 14:09
    
Hang on, I've confused myself there. Can we say that as $$u(x,y)+iv(x,y)=\frac{x+\sqrt{x^2+y^2}}{\sqrt{2}(x+\sqrt{x^2+y^2} )^{1/2}}+\frac{iy}{\sqrt{2}(x+\sqrt{x^2+y^2} )^{1/2}}$$ and the $u$ and $v$ we have satisfy those conditions, then the result holds? Or is that just circular because we're given the $u$ and $v$ in the problem? It seems as though it may be. –  Mathmo Dec 9 '11 at 14:44

1 Answer 1

up vote 1 down vote accepted

I assume that you want to prove that $f(z) = f(x + iy) = u(x,y) + i v(x,y)$ is holomorphic in $\mathbb{C} \setminus (-\infty,0]$. Now you can either check that $u$ and $v$ satisfy the Cauchy-Riemann equations, or you can show that $f(z)$ is actually some well-known holomorphic function. I'll go with the second option. Now as you wrote in a comment,

$$f(z) = \frac{x + \sqrt{x^2 + y^2}}{\sqrt{2}(x + \sqrt{x^2 + y^2})^{1/2}} + \frac{iy}{\sqrt{2}(x + \sqrt{x^2 + y^2})^{1/2}},$$ which simplifies to

$$f(z) = \frac{z + |z|}{\sqrt{2(x+|z|)}}.$$

Now let $g(z) = f(z)^2$. Then

$$g(z) = \frac{z^2 + 2|z|z + |z|^2}{2(x + |z|)} = \frac{z(z + 2|z| + \bar{z})}{2(x + |z|)} = z.$$

This means that $f$ is a branch of the square root, valid as $2(x + |z|) \neq 0$, or in $\mathbb{C} \setminus (-\infty,0]$.

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Thanks! That's exactly what I needed. –  Mathmo Dec 9 '11 at 16:22

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