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What's bad about calling $i$ "the square root of -1"? 28 Mar 2015 · 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, …
Approximating square roots using binomial expansion. And one can quickly check that $(x_3)^2=2.000006007\dots$, which is pretty much the square root of $2$. Share.
Is $i$ equal to $\\sqrt{-1}$? - Mathematics Stack Exchange 22 Jul 2017 · Indeed we can extend the definition of the square root to any complex number, setting it as the principal value of $\sqrt z:=e^{1/2\ln z}$ with the same result. However is true …
How to compute $\\sqrt{i + 1}$ - Mathematics Stack Exchange 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 …
complex numbers - why is $\sqrt {-1} = i$ and not $\pm i ... 9 Jan 2015 · 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 …
Is the square root of negative 1 equal to i or is it equal to plus or ... 25 Nov 2017 · 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 …
Why the square root of any decimal number between 0 and 1 … 24 Jan 2018 · Think about a decimal number between 0 and 1 as a fraction with its numerator GREATER than its denominator. Say you are taking the square root of the number $1/25$. So, …
algebra precalculus - Square root inside a square root 24 Jun 2014 · The square root of the square root of x is therefore $$\sqrt{\sqrt x} = (\sqrt x)^{1/2} = (x^{1/2})^{1/2} = x^{1/4} = \sqrt[\large 4] x$$ Since the domain of $\sqrt x$ is $[0, + \infty)$ , this …
complex numbers - What is $\sqrt {i}$? - Mathematics Stack … -1 is 1 rotated over $\pi$ radians. The square root of a number on the unit circle is the number rotated ...
Integral $\\int \\sqrt{1+x^2}dx$ - Mathematics Stack Exchange 21 Feb 2018 · I was trying to do this integral $$\\int \\sqrt{1+x^2}dx$$ I saw this question and its' use of hyperbolic functions. I did it with binomial differential method since the given integral is …