In the given you wrote -2, whereas it should be -2i
z = a + bi and in trigo form | z |.(cos Ф + i.sin Ф).
z = 0 -2i → | z |.(cos Ф + i.sin Ф)
Now let's calculate z:
z² = a²+b² → z² = 0² (-2.i)² → z = -2(i)² →z= -2(-1) → z = |2|
tan Ф = b/a = -2/0 → tan Ф → - ∞ ↔ Ф = -90° or Ф = 270°
then the - 2 I ↔ |2|(cos 270° + i.sin 270°)