Why are negative square roots are on the real number line if square root of a negative number not a real number?

Answer 1

Negative square roots are just the opposite of positive square roots. Since square roots (of positive numbers) are real, the negative square roots are also real.

Square roots of negative numbers are not real.

Note that -1 = exp(Pi*i), so (-1)^(1/2) = exp((1/2)*Pi*i) = i.

Note that exp(i*x) = cos(x) + i*sin(x), for instance by taking derivatives:

(d/dx)(exp(i*x)) = i*exp(i*x), and

(d/dx)^2(exp(i*x)) =(-1)*exp(i*x).

This means that the second derivative of exp(i*x) equals -exp(i*x).

The same property holds for cos(x) + i*sin(x):

(d/dx)(cos(x) + i*sin(x)) = -sin(x) + i*cos(x)

(d/dx)^2(cos(x) + i*sin(x)) = -cos(x) - i*sin(x) = -(cos(x) + i*sin(x)))

Hence cos(x) + i*sin(x)) = C + Dx + exp(i*x), for some C and D.

Comparing the values on both sides for x = 0, we find:

1 = C+1, so C = 0 and for the first derivative:

i = D + i, so D = 0.

So cos(x) + i*sin(x)) = exp(i*x) for all x.

by comparing x=0 for both functions and their first derivative. Since they coincide,