z^5 = 1; \Rightarrow L = \left\{ E(k \cdot \frac{2\pi}{5}) | k \in \mathds{N} \cap \left[1, 5\right] \right\};z5 = 1;⇒ L = E(k ⋅2π 5 )∣k ∈N ∩1,5;
