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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} ) \ $.

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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... –  draks ... May 5 '12 at 12:04
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@draks: the dilogarithm, to be precise. –  J. M. May 5 '12 at 12:40
    
@draks: The text gives the hint: $z^n -w^n = (z-w)\ ( z^{n-1}+z^{n-2}w + \ldots + w^{n-1} ) \ $. –  WLOG May 5 '12 at 13:54
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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. –  TMM May 5 '12 at 14:31
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up vote 3 down vote accepted

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 $|z|, |w| \leq 2/3$.

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