2015年9月20日 · The principal square root of minus one is i. It has another square root -i. I really dislike the expression "the square root of minus one". Like all non-zero numbers, -1 has two square roots, which we call i and -i. If x is a Real number then x^2 >= 0, so we need to look beyond the Real numbers to find a square root of -1. Complex numbers can be thought of as …

-1 is 1 rotated over $\pi$ radians. The square root of a number on the unit circle is the number rotated ...

In a field such as $\,\mathbb Q,\ \mathbb R,\ \mathbb C,\,$ we have $ \ x^2 = 1 \iff (x-1) (x+1) = 0\iff x = \pm 1.\, $ In rings that are not fields there can be more than two square-roots, e.g. modulo $15$ there are two additional roots $ \ (\pm\,4)^2\equiv 1\pmod{\!15}.\,$ In some contexts authors define single-valued square-root functions that uniformly select one of the roots, e.g. the non ...

Also a wanted property is that it is continuous except for the non-positive reals. And I guess it's also a wanted property that for all numbers except of the negative reals, the square root of the conjugate is the conjugate of the square root (for the negative reals, it's not possible to achieve that). $\endgroup$ –

2015年1月9日 · To get a continuous "branch of square root" it's necessary to remove enough of the plane that "the domain of the square root doesn't encircle the origin". The customary choice is to remove the non-positive reals. (Ironically, this explicitly excludes $-1$ from the domain of the square root.) A common alternative choice is to remove the non ...

2018年3月9日 · 1/2 We have: sqrt(1/4) Using a simple observation, we got: =sqrt((1^2)/(2^2)) Using the properties, (a^b)/(c^b)=(a/c)^b. So, we got =sqrt((1/2)^2) =+-1/2 Since we are talking about real numbers, we will only take the principal (positive) square root, …

2017年11月25日 · The answer is that there are two square roots of $-1$. This is no different than with real numbers; for example, there are two square roots of $4$: $2$ and $-2$. The main difference is that the complex numbers don't have a good way to single out one of the two square roots as the "special" one.

2018年3月12日 · Algebra Properties of Real Numbers Square Roots and Irrational Numbers. 1 Answer . F. Javier B.

$\begingroup$ The incorrect step is the assumption that $\sqrt{ab} = \sqrt{a} \sqrt{b}$ While this holds for nonnegative numbers, it does not hold for negative numbers.

2015年3月28日 · There is a bit more complicated, but more thorough explanation, however, involving complex analysis. The problem lies in trying to take fractional exponents of negative numbers, e.g. $(-1)^{1/2}$.