Implied Functions

For complex numbers, the formula



is not valid, because of the discontinuous nature of the square root in the complex plane.

I haven't really worked with complex numbers, but when I plotted it, I don't see how the discontinuities affect the function other than separate the real from the imaginary parts:

 Spoiler

No. No solutions exist to that equation. And no, imaginary numbers were not made to handle sqrt(-1), they were made to handle equations on the form x^2 = -y. The definition of i is not sqrt(i) = -1, but rather, i^2 = -1. You cannot really use the former, though it sometimes work.

i = sqrt(-1)

How is it not used to handle sqrt(-1)?

sqrt(i) = a + bi
i = (a+bi)^2 = a^2 + 2abi - b^2

a^2 - b^2 = (a+b)*(a-b) = 0
a = +/- b

2ab i = i
2*b^2*i = i
2*b^2 = 1
b = sqrt(2)/2 = a

sqrt(i) = sqrt(2)/2 + sqrt(2)/2*i

How can sqrt(i) = -1 sometimes work, when sqrt(i) is already defined?

the square root function only returns complex numbers with non-negative coefficients for the imaginary term, therefore 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

I'm actually not sure what you mean by the bolded. The range of n*cis(θ in the interval [0,π is just [-n,n) in the real plane and [0,n] in the imaginary plane. Wouldn't the real part imply that sqrt(a) can equal -a, because it's in the range?