Question

does multiplication of two imaginary numbers imaginary or real? if i multiply 2i and 3i then it is = -6 which is real .if √-3 and √-2 are multiplied then it gives - √6 which is real . but if i do the same thing like this √-3 * √-2 =√{(-3)*(-2)} =√6 which is also real . so why the ans is -√6 . we know multiplication of two complex numbers is a complex number . if i write 2i as (0+2i) and 3i as (0+3i) and then multiply these two should come a complex number as definition . but why it's real?

Answer 1

Answer on Question #52950 – Math – Complex Analysis

does multiplication of two imaginary numbers imaginary or real? if i multiply 2i and 3i then it is = -6 which is real. if √-3 and √-2 are multiplied then it gives - √6 which is real. but if i do the same thing like this √-3 * √-2 = √{(-3)*(-2)} = √6 which is also real. so why the ans is -√6. we know multiplication of two complex numbers is a complex number. if i write 2i as (0+2i) and 3i as (0+3i) and then multiply these two should come a complex number as definition. but why it's real?

Solution

$$(i)^2 = (-i)^2 = -1 \text{ so } \sqrt{-1} = \pm i.$$

Thus $\sqrt{-3} = \pm i\sqrt{3}, \sqrt{-2} = \pm i\sqrt{2}$

and $\sqrt{(-3)(-2)} = (\pm i\sqrt{3})(\pm i\sqrt{2}) = \pm (i)^2\sqrt{6} = \pm \sqrt{6}$.

Every real number is a complex number with imaginary part equal to 0.

$$(0 + 2i)(0 + 3i) = 0 * 0 - 2 * 3 + (0 * 2i + 0 * 3i) = -6 + 0i$$

So we have the complex number with imaginary part equals 0, i.e. the real number.

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