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Is 2i a real or complex root?

User Troels
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By definition, i = sqrt(-1) because it is used to solve the equation x^2+1 = 0 or x^2 = -1, which has no real number solutions.

Since i is imaginary or complex, then so is 2i

You can express it as 2i = 0+2i to be in the form a+bi with a = 0 as the real part and b = 2 as the imaginary component.

User Pedro Amaral Couto
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