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X² = i as X = √i and i = √-1

Let us assume X = A + B i

So, (A+i B)² = i

A² + i² B² + 2 AB i = i

(A²-B²) = 0 2 AB = 1

A = +- B So 2 A² = 1/2 A = B = +1/√2 or -1/√2

Let us assume X = A + B i

So, (A+i B)² = i

A² + i² B² + 2 AB i = i

(A²-B²) = 0 2 AB = 1

A = +- B So 2 A² = 1/2 A = B = +1/√2 or -1/√2

*X = (1+i) /√2 or -(1+i) / √2*(a+/-ib)²=i

a²+i²b²+2abi=i

a²-b²=0 and 2ab=1

a=-+b

therefore sq roots are i+1/√2 and -(1+i)/√2.