There is two complex square root of -8i. what is the one with a positive real part in a+ib form, after as much simplifying.?

z = a + ib

z² = (a + ib)²

z² = a² + 2aib + i²b² → where: i² = - 1

z² = a² + 2aib - b²

z² = (a² - b²) + i.(2ab) → given that: z² = - 8i

(a² - b²) + i.(2ab) = - 8i → you compare both sides → you obtain 2 equations:

(1) : a² - b² = 0

(2) : 2ab = - 8 → ab = - 4 → a = - 4/b → a² = 16/b²

You restart from (1)

a² - b² = 0 → recall: a² = 16/b²

(16/b²) - b² = 0

(16 - b⁴)/b² = 0 → where: b ≠ 0

16 - b⁴ = 0

b⁴ = 16

b² = ± 4 → as b is a square, b² is a positive value

b² = 4

b = ± 2

Recall (2): a = - 4/b

When: b = 2 → a = - 2 → z₁ = - 2 + 2i

When: b = - 2 → a = 2 → z₂ = 2 - 2i ← this is the complex number with the positive real part