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Describe all meromorphic functions f(z) in the complex plane with a simple pole at z=1, a simple zero at z=-1, and for which

$$|f(z)|\le M|z|,$$

for $|z|\ge 2$

for some $M>0$.

I know that, since f has a simple pole at z=1, then f(z) must be of the form $$\frac{g(z)}{(z-1)}$$, where g(z) is analytic and non-zero at z=1.

Similarly, since f has a simple zero at z=-1, then f(z) must be of the form $$(z+1)h(z)$$, where h(z) is analytic and non-zero at z=-1.

Combining the two, I have that f(z) must be of the form $$\frac{(z+1)}{(z-1)}w(z)$$, where w(z) is analytic and non-zero at both z=1 and at z=-1.

Using the inequality given, I have that $$|\frac{(z+1)}{(z-1)}w(z)| \le M|z|$$, for $|z|\ge 2$.

How can I proceed from here? ...Or have I started off incorrectly already? I know I haven't said much yet, but, so far, I've used everything that's given in the problem, I think. Also, I wasn't able to derive any new information from moving around some parts in the inequality.

Thanks in advance,

Edit: Perhaps the upper bound is telling me that my function f(z) grows like a polynomial, hence it is a polynomial, but that can't be true, since I am given that f(z) has a simple pole at z=1.

User001
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  • Good start! Hint: $|(z-1)/(z+1)|$ is bounded above on ${|z|\ge2}$. So $w$ satisfies the same shape of inequality that $f$ does. – Greg Martin Dec 05 '14 at 02:00
  • hmm, @GregMartin...so now I have |w(z)| </= C|z|, for some C>0. I don't think that I can use Liouville's Theorem here, since I can't say whether w(z) is entire or even bounded above - the inequality just gives me an idea of the growth of w(z). My next guess would be that w(z) is a polynomial (because of the polynomial growth), and then since it satisfies the same inequality that f does, I conclude that f(z) must also be a polynomial. But I just seem to be repeating my initial guess... – User001 Dec 05 '14 at 02:35
  • hmm...I can give the necessary estimates to show that polynomial growth implies the function is a polynomial - that just comes from the Cauchy Integral Formula. Then w(z) is a polynomial * (z+1), which is still a polynomial, of course. Call this polynomial P. Then I have f(z) = P/(z-1), which is a quotient of polynomials, i.e., a rational function. What do you think, @GregMartin? Should I go even further to derive more estimates and gather more information to describe these meromorphic functions? – User001 Dec 05 '14 at 03:31
  • I can go a bit further and specify that P(z) does not achieve 0 at z=1, which is where f(z) has a simple pole, and also that P(z) must have a root of multiplicity = 1 at z=-1. I think that's about all we can say. Feel free to comment further, if you spot something else in this problem that is worth noting. Thanks so much for the hint :) – User001 Dec 05 '14 at 04:28
  • You're totally right that a function of polynomial growth must be a polynomial (and the degree is at most the exponent of polynomial growth). So you've basically proved (but should convince yourself that you've proved) that $f(z)$ must be of the form $(az+b)(z+1)/(z-1)$ for some complex numbers $a,b$, and that every function of this form (other than the zero function) satisfies the conditions. – Greg Martin Dec 05 '14 at 08:08
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    Ok, got it. Nice to see your factorization to emphasize that w(z) is at most a degree 1 polynomial. Thanks so much for your help, @GregMartin. Have a great night :) – User001 Dec 06 '14 at 00:12

1 Answers1

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This was sorted out in comments, but here's a summary: by assumptions about the pole and the zero, $f$ has the form $$f(z) = \frac{z+1}{z-1}w(z)\tag{1}$$ where $w$ is an entire function. For $|z|\ge 2$ we have $$ |w(z)|=\frac{|z-1|}{|z+1|}|f(z)| \le \frac{|z|+1}{|z|-1}M|z| = \frac{1+1/|z| }{1-1/|z| }M|z|\le 3M|z| $$ According to Entire function bounded by a polynomial is a polynomial, $w$ is a polynomial of degree at most $1$. And conversely, for any such polynomial $(1)$ gives a function with the required properties.

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