8
$\begingroup$

This is an exercise from Remmert.

The power series $\sum\limits_{n=1}^{\infty} \frac {z^{n} }{ n^{2}} \ $ has radius of convergence $1 \ $. Show that the function it represents is injective in $\{ z \in \mathbb{C} | \ \ \lVert z \rVert < \frac{2}{3} \} \ $.

The text gives the hint: $z^n -w^n = (z-w)\ ( z^{n-1}+z^{n-2}w + \ldots + w^{n-1} ) \ $.

$\endgroup$
  • 1
    $\begingroup$ Do you know the reason for the $||z||<2/3$? BTW, your series describes the polylogarithm $\text{Li}_2(z)$, but I think you knew that... $\endgroup$ – draks ... May 5 '12 at 12:04
  • 3
    $\begingroup$ @draks: the dilogarithm, to be precise. $\endgroup$ – J. M. isn't a mathematician May 5 '12 at 12:40
  • $\begingroup$ @draks: The text gives the hint: $z^n -w^n = (z-w)\ ( z^{n-1}+z^{n-2}w + \ldots + w^{n-1} ) \ $. $\endgroup$ – WLOG May 5 '12 at 13:54
  • 1
    $\begingroup$ The hint suggests you should try to prove that $$\sum_{n=1}^{\infty} \frac{z^n}{n^2} - \sum_{n=1}^{\infty} \frac{w^n}{n^2} = \sum_{n=1}^{\infty} \frac{z^n - w^n}{n^2}= (z - w) \sum_{n=1}^{\infty} \frac{z^{n-1} + z^{n-2} w + \ldots + w^{n-1}}{n^2} = 0$$ if and only if $z = w$, which would imply injectivity. $\endgroup$ – TMM May 5 '12 at 14:31
  • $\begingroup$ @WLOG: Can we find the sum? $\endgroup$ – Subhash Chand Bhoria Aug 31 '15 at 3:34
4
$\begingroup$

As @TMM has written, start with the hint. If $z\neq w$, then $$ \sum_{n=1}^\infty \frac{z^{n-1}+\cdots + w^{n-1}}{n^2} =0 \enspace. \quad (\star) $$ The first term (with $n=1$) in the sum is actually $1$. The other terms are bounded, whenever $|z|, |w| \leq 2/3$, by $$ \sum_{n\geq 2} \frac{n(2/3)^{n-1}}{n^2} = \frac32 \sum_{n\geq 2} \frac{(2/3)^{n}}n = \frac32 \left(-\log\left(1-\frac23\right) - \frac23\right) =\frac32 \log3 -1 <1 \enspace, $$ therefore $(\star)$ is never true if $|z|, |w| \leq 2/3$.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.