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Recently, in the study of unicity problems in complex analysis, I met a problem that can be stated in the following way,

Let $\{m_i\}_{i=0}^{3}$ and $\{n_i\}_{i=0}^{3}$ be eight integers in $\mathbb{Z}\setminus\{0\}$ such that $m_i\neq n_i$ for $0\leq i\leq 3$.
Then $$ R(z):=\sum_{i=0}^{3}(-1)^{i}\binom{3}{i}\frac{1-(\lambda_i z)^{m_i}}{1-(\mu_i z)^{n_i}}-\frac{1-(\lambda_{0}z)^{-m_0}}{1-(\mu_{0}z)^{-n_0}} $$ is not identically to some constant for any non-zero $\lambda_i$ and $\mu_i$.

Of course, this may be checked using the computer in a very direct way. I wondered is there a nice way to discuss it by hands. Thanks in advance.

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  • $\begingroup$ $$ \begin{align} & \mathbb Z \backslash\{0\} \\ {} \\ & \mathbb Z \setminus\{0\} \\ {} \\ & \mathbb Z \smallsetminus\{0\} \end{align} $$ The first line above uses \backslash, the second \setminus and the third smallsetminus. Those last two provide horizontal spacing proper to a binary operation symbol. I.e. we normally see $3+5$ and not $3{+}5$, whereas +5 with nothing before it appears as $+5$ without that horizontal space (and similarly $3+$). The thing that makes you think one font looks good and another does not consists of many little things like this. $\qquad$ $\endgroup$
    – Michael Hardy
    Commented Jul 23, 2023 at 16:07
  • $\begingroup$ And that is why I edited the question as I did. The \backslash, causing you to see $\mathbb Z\backslash\{0\}$ instead of $\mathbb Z\setminus\{0\}$ will strike people who are aware of things like this about the same way a spelling error does, and others will be affected in the way noted above, about differences between appearances of different fonts. $\endgroup$
    – Michael Hardy
    Commented Jul 23, 2023 at 16:09
  • $\begingroup$ Remember that this software evolved ultimately from the work of the renowned software genius Donald Knuth, who explicitly intended to make attention to this kind of thing possible. $\endgroup$
    – Michael Hardy
    Commented Jul 23, 2023 at 16:11
  • $\begingroup$ Given that $\frac{1-z^2}{1-z}-3\frac{1-z^3}{1-z}+3\frac{1-z^4}{1-z^2}-\frac{1-(-z)^2}{1-(-z)}-\frac{1-(-z)^2}{1-(-z)}\equiv -1$, you may want to revise your question and impose some more restrictions before any good answer to it can be given ;-) $\endgroup$
    – fedja
    Commented Jul 23, 2023 at 23:09
  • $\begingroup$ @fedja Thanks for your comment, fedja. For the last term, whether you mean $-\frac{1-z^{-2}}{1-z^{-1}}$, thanks. $\endgroup$
    – yaoxiao
    Commented Jul 23, 2023 at 23:22

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