● Note that `color(blue)(i^2 = –1)` and `color(blue)(( – i)^2) = i^2 = – 1`
Therefore, the square roots of `– 1` are `i, – i.`
● However, by the symbol ` sqrt(−1)` , we would mean `i` only.
Now, we can see that `color(blue)(i)` and `color(blue)(–i)` both are the solutions of the equation `color(blue)(x^2 + 1 = 0)` or `x^2 = –1.`
● Similarly
`color(blue)((sqrt3 i)^2) = (sqrt3)^2 i^2 = 3(-1) = -3`
`color(blue)((-sqrt3 i)^2) = (-sqrt3)^2 i^2 = -3`
The square roots of `color(red)(–3)` are `color(red)(sqrt (3) * i)` and `color(red)(− sqrt(3) * i)`
● Generally, if `a` is a positive real number, `sqrt(−a) = sqrt(a)* sqrt(−1 )= sqrt(a)* i ,.`
We already know that `color(blue)(sqrt(a) ×sqrt( b) = sqrt(ab))` for all positive real number `a` and `b.`
This result also holds true when either `a > 0, b < 0` or `a < 0, b > 0.`
● What if `color(red)(a < 0, b < 0 \ \?) ` Let us examine.
Note that
`i^2 = sqrt(−1) * sqrt(−1) = sqrt((−1) (−1)) ` (by assuming `a × b = ab` for all real numbers) `= sqrt(1) = 1,` which is a contradiction to the fact that `i ^2 = − 1.`
● Therefore, `color(red)(sqrt(a) × sqrt(b) != sqrt(ab))` if both `a` and `b` are negative real numbers.
Further, if any of `a` and `b` is zero, then, clearly, `sqrt(a) × sqrt(b) = sqrt(ab) = 0.`
● Note that `color(blue)(i^2 = –1)` and `color(blue)(( – i)^2) = i^2 = – 1`
Therefore, the square roots of `– 1` are `i, – i.`
● However, by the symbol ` sqrt(−1)` , we would mean `i` only.
Now, we can see that `color(blue)(i)` and `color(blue)(–i)` both are the solutions of the equation `color(blue)(x^2 + 1 = 0)` or `x^2 = –1.`
● Similarly
`color(blue)((sqrt3 i)^2) = (sqrt3)^2 i^2 = 3(-1) = -3`
`color(blue)((-sqrt3 i)^2) = (-sqrt3)^2 i^2 = -3`
The square roots of `color(red)(–3)` are `color(red)(sqrt (3) * i)` and `color(red)(− sqrt(3) * i)`
● Generally, if `a` is a positive real number, `sqrt(−a) = sqrt(a)* sqrt(−1 )= sqrt(a)* i ,.`
We already know that `color(blue)(sqrt(a) ×sqrt( b) = sqrt(ab))` for all positive real number `a` and `b.`
This result also holds true when either `a > 0, b < 0` or `a < 0, b > 0.`
● What if `color(red)(a < 0, b < 0 \ \?) ` Let us examine.
Note that
`i^2 = sqrt(−1) * sqrt(−1) = sqrt((−1) (−1)) ` (by assuming `a × b = ab` for all real numbers) `= sqrt(1) = 1,` which is a contradiction to the fact that `i ^2 = − 1.`
● Therefore, `color(red)(sqrt(a) × sqrt(b) != sqrt(ab))` if both `a` and `b` are negative real numbers.
Further, if any of `a` and `b` is zero, then, clearly, `sqrt(a) × sqrt(b) = sqrt(ab) = 0.`