2011-03-27, 09:31 AM
N is a non-negative real number equal to the square root of the norm of the complex number you're taking the sqrt of.
Here's a better way to explain: Unless I'm being stupid, the square root function only returns complex numbers with non-negative coefficients for the imaginary term, therefor if you wish for two square roots to add to a real number, both numbers would need square roots in non-negative reals (because the range of the square root function takes the form n*cis [0, pi) )
P. S. idk why I decided to use degrees in my last post
EDIT: @ Noah, sqrt(i) = sqrt(2)/2 + sqrt(2)/2 * i, sqrt(-1) = i, sqrt(a) = -1 has no solution because -1 is outside the range of the square root function.
Here's a better way to explain: Unless I'm being stupid, the square root function only returns complex numbers with non-negative coefficients for the imaginary term, therefor if you wish for two square roots to add to a real number, both numbers would need square roots in non-negative reals (because the range of the square root function takes the form n*cis [0, pi) )
P. S. idk why I decided to use degrees in my last post
EDIT: @ Noah, sqrt(i) = sqrt(2)/2 + sqrt(2)/2 * i, sqrt(-1) = i, sqrt(a) = -1 has no solution because -1 is outside the range of the square root function.