+128475 z^2 = 2i
(a + bi)^2 = 0 + 2i
a^2 + 2abi - b^2 = 0 + 2i
(a^2 - b^2) + 2abi = 0 + 2i equate coefficients
a^2 - b^2 = 0 (1) and ab = 1 (2)
(a - b) ( a + b) =0
(a - b) = 0 or ( a + b) = 0
a = b or a = - b (reject because a and b will have opposite signs and ab will be negative if this is true )
If a = b then (2) becomes a^2 =1 ⇒ a = ±1
And if a = 1 then b = 1
And if a = -1, then b =-1
So....the solutions are a = 1, b = 1 ⇒ z = 1 + i or a =- 1 , b = -1 ⇒ z = -1 - i
