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Find an explicit analytic function on $\mathbb{C}\setminus[-1,1]$ which is bounded and non-constant.

Suggestions on how to approach this problem?

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  • $\begingroup$ @user254665: ... is not bounded in that domain. Consider $f(iy)$ for $y > 0, y \to 0 $. $\endgroup$
    – Martin R
    Sep 1, 2016 at 19:39
  • $\begingroup$ Sorry.You are right........ $\endgroup$ Sep 1, 2016 at 19:54

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Try to find a holomorphic function from your domain into the unit disk. Proceed in steps and start with $$ T(z) = \frac{z-1}{z+1} $$ which maps $\mathbb{C}\setminus[-1,1]$ into the complex plane without the negative real axis. Continue with a mapping into the right half-plane, and you are almost done.

Spoiler:

$$ f(z) = \dfrac{\sqrt{\frac{z-1}{z+1}} - 1}{\sqrt{\frac{z-1}{z+1}} + 1}$$

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  • $\begingroup$ The equation I get is $f(z)=\sqrt{z^{2}-1}-z$. Is there a nice way to show that this is bounded? $\endgroup$
    – esavaleo8
    Sep 1, 2016 at 18:54
  • $\begingroup$ @esavaleo8: You could write it as $f(z) = z (\sqrt{1-1/z^2} - 1)$ and use the binomial formula to get an estimate. But you have to show that there is holomorphic branch of $\sqrt{1-1/z^2}$ in the domain. The above approach avoids this problem. $\endgroup$
    – Martin R
    Sep 1, 2016 at 19:00
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One approach: If $\phi$ is any function integrable on $[-1.1]$ and you define $f$ to be tha Cauchy integral $$f(z)=\int_{-1}^1\frac{\phi(t)\,dt}{z-t}$$then $f$ will be holomorphic in $\Bbb C\setminus[-1,1]$. If $\phi$ is smooth enough you can show $f$ is bounded...

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$y = \sqrt {z^2-1}$ defines a $2$-fold cover over of $\Bbb C$ with two branch points at $\pm 1$. When you consider one of its branches $y(z)$ on $\Bbb C \setminus [-1;1]$, you only get half of the Riemann surface associated with it. Then you simply want to find a function defined on that Riemann surface whose poles are not seen on that branch.

Here, the Riemann surface has genus $0$ so there are functions with a single pole anywhere you want.

For example, look at $z/ (2 y + \sqrt 3 z)$.

Its poles are included in the roots of $(2y+\sqrt 3z)(2y-\sqrt 3z) = 4y^2-3z^2 = z^2-4 = (z-2)(z+2)$, which are the $4$ points $(z=2, y= \pm \sqrt 3)$ and $(z=-2, y = \pm \sqrt 3)$.

So $z/ (2 y + \sqrt 3 z)$ has two poles on the Riemann surface, located at $(z=2, y = - \sqrt 3)$ and $(z=-2, y = \sqrt 3)$.

Then you simply have to choose the branch where you only see the points $(z=2, y = \sqrt 3)$ and $(z= -2, y = - \sqrt 3)$ to view $z/(2y(z)+ \sqrt 3 z)$ as a bounded holomorphic function on $\Bbb C \setminus [-1;1]$

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