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14 June, 02:48

Is 2i a real or complex root?

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  1. 14 June, 03:13
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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.
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