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I am studying Dirichlet characters and modular functions as part of my research, specifically working through concepts found in the paper "An explicit hybrid estimate for $L(1/2 + it)$" by Ghaith A. Hiary [ 1 ] (MR 3580112, Zbl 1401.11121, arXiv). In the course of my studies, in the proof of Lemma 3.3 of that paper I encountered the following identity that I am attempting to prove: $$ \chi(1 + C_1 x) = \prod_{j=1}^{\omega} \chi_j(1 + C_1(p_j^{a_j}) x)= \exp\left( \frac{2 \pi i \tilde{L} x}{D_1} \right), $$ The following conditions are given: $$ \begin{cases} \chi_j(1 + C_1(p_j^{a_j}) x) = \exp\left( \dfrac{2 \pi i \tilde{L}_j x}{D_1(p_j^{a_j})} \right),\\ \tilde{L} = q \displaystyle \sum_{j=1}^{\omega} \frac{\tilde{L}_j}{p_j^{a_j}}. \end{cases} $$ where

  • $L_j$ is integer for all $j$,
  • $ C_1(q) \cdot D_1(q) = q $,
  • $ C_1(p_j^{a_j}) = p_j^{\lceil a_j / 2 \rceil} $,
  • $ D_1(p_j^{a_j}) = p_j^{\lceil a_j / 2 \rceil - a_j} $,
  • and finally both $ C_1 $ and $ D_1 $ are multiplicative functions.

We need to prove that $$ \exp\left( 2\pi i x C_1(q) \sum_{j=1}^{\omega} \frac{\tilde{L}_j}{p_j^{a_j}} \right) = \exp\left( 2\pi i x \sum_{j=1}^{\omega} \frac{\tilde{L}_j}{D_1(p_j^{a_j})} \right). $$

How can I manipulate this expression and prove the equality? Particularly, how do the multiplicative properties of $ C_1 $ and $ D_1 $ help in establishing this identity?

Any help or guidance would be appreciated!

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  • $\begingroup$ what is unclear in my question $\endgroup$ Commented Dec 1 at 0:11
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    $\begingroup$ Hi Fatima; I believe the downvote was related to your original title, so I’ve edited. If you’d like to revert it please feel free. $\endgroup$
    – Alec Rhea
    Commented Dec 1 at 0:47
  • $\begingroup$ @AlecRhea thanks $\endgroup$ Commented Dec 1 at 0:49
  • $\begingroup$ i don't know anything about those specific values of the function at prime powers, but the fact that you asked particularly about the multiplicative property means you might not know that it sometimes helps to write $a=\sum_{j=1}^w a_j (q/q_j)^{-1} (q/q_j)$ mod($q)$ where $a=a_j (q_j)$ $\endgroup$
    – tomos
    Commented Dec 1 at 21:59
  • $\begingroup$ @tomos . Could you clarify what exactly you mean by $ a_j(q_j) $? Do mean any multiplicitive function $\endgroup$ Commented Dec 1 at 23:54

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