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Quater-imaginary base
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==Z-order curve== The representation :<math>z = \sum_{k\ge n} z_k \cdot (2i)^{-k}</math> of an arbitrary complex number <math>z\in \Complex</math> with <math>z_k\in \{0,1,2,3\}</math> gives rise to an [[Injective function|injective]] mapping :<math>\textstyle \begin{array}{llcl} \varphi \colon & \Complex & \to & \R \\ & \sum_{k \ge n} z_k \cdot (2i)^{-k} & \mapsto & \sum_{k \ge n} z_k \cdot r^{-k}\\ \end{array} </math> with some suitable <math>r \in \Z</math>. Here <math>r = 4</math> cannot be taken as base because of :<math>\textstyle \sum_{k>0} 3\cdot (2i)^{-k} = \tfrac{-3-6i}5 \; \; \; \; \ne \; \; \; \; 1 = \sum_{k>0} 3 \cdot 4^{-k}.</math> The [[Image of a function|image]] <math>\varphi(\Complex) \subset \R</math> is a [[Cantor set]] which allows to linearly order <math>\Complex</math> similar to a [[Z-order curve]]. Since the image is disconnected, <math>\varphi</math> is not [[Continuous function|continuous]].
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