A Complex Summation

If ${A}_{k}=\cos { \dfrac { k\pi }{ 20 } } +{ i }\sin { \dfrac { k\pi }{ 20 } }$ , then

$\large \frac { \sum _{ k=0 }^{ 2018 }{ \left| { A }_{ k+1 }-{ A }_{ k } \right| } }{ \sum _{ k=1 }^{ 673 }{ \left| { A }_{ 2k+1 }-{ A }_{ 2k+2 } \right| } } = Z$

Find $Z^Z$ .

Notation: $i = \sqrt { -1 }$ denotes the imaginary unit .