What is above complex numbers?

Question

A complex number is the square root of -1.

What is the name for the square root of a complex number?

If this is extended as far as reality will allow, does it end? I had heard that at one point getting the root of the "impossible" numbers will result in a single positive number

Answer 1

A good question and a good thought, but the progressive types of numbers do not go on and on the way you think.

With positive integers, we can construct equations which need negative integers for their solution.

With positive and negative integers, we can construct equations which need rational or irrational numbers for their solution.

With integers, rationals, and irrationals, we can construct equations which need complex numbers for their solution.

BUT . . .

Every equation constructed with complex numbers has solutions which are complex numbers. For example, the square root of a complex number is just some other complex number.

You can imagine how relieved mathematicians were when they had proved this.

Answer 2

Just to clarify, an imaginary number is something like the square root of -1, which would equal i. A complex number is a combination of a real number AND an imaginary number. Usually they are of the form a + bi, where a and b are real numbers and i is the imaginary unit.
As far as I know, you can take the square root of a complex number, but it does not have a new, special name. If you take the square root of i, you basically have a fourth root of -1. So, I don't think this will be a positive number. I'd be intrigued if it did though. Hope this was helpful. Good luck.

Answer 3

In some rather important sense the complex numbers C is as "far up" as one can get: C is a field which is also algebraically closed, meaning: every complex polynomial has a root within C itself, which is far from being true in the reals R (for ex., the real polynomial x^2 + 1 has NO root within R itself).
There exist another algebraic structures that contain C, in particular the Quaternion algebra , but it is NOT a field, or the field C(x) of rational complex functions, which of course is a field but it is NOT algebraically closed.
About impossible numbers I don 't know what you mean: the square root of -1 is as "impossible", imagin ary or real as the number -1, 4 or 47.3334, and besides this: EVERY complex number has both its square roots within C itself.

Regards
Tonio
Pd. Pay attention to the fact that in C we have NO order: there are no numbers greater than others, or positive or negative. This applies to real numbers, NOT to complex ones.

Answer 4

All numbers have a graphical meaning...
Like if I said 5, I could be looking at graphing a point that's five away from a given point.
In general, that'd be 5 marks to the right of 0.

Complex numbers also have a graphical interpretation, that has to do with an angle and a distance, so rather than just counting up and down or left and right, you'd be at an angle.

When you do your math that way, (like if you said 5 - 3 and went to the right 5 and left 3, you'd get 2, another integer)
then you'd be able to see clearly that the root of a complex number is another complex number.

The property is known as the closure of complex numbers.

Answer 5

The fourth root of -1.
They're called imaginary, or complex numbers.
Complex numbers have an imaginary part denoted by i, such as a + bi is a complex number.
Because √-1 = i, i² = -1; i³ = i² * i = -1 * i = -i; and i^4 = i² * i² = -1 * -1 = 1. This repeats every fourth power.
The square root of i (or a + bi) would be the 4th root of -1, so it wouldn't necessarily be a different kind of number. Then again, who knows? Maybe they'll be called "sub real" numbers or something.
Well, that's my two cents.

Answer 6

considering that i is unquestionably an intensive, i.e. squarert of -a million, then what needs to be accomplished is to rationalize the denominator, this is multiply the denominator via itself to sparkling out the unconventional. yet, with the intention to no longer replace the cost of the whole expression, the numerator would desire to be more suitable via the comparable huge type. So, what you're unquestionably doing is multiplying the completed expression via a million, and changing its style interior the approach. 2/i = 2 / squarert (-a million) = 2 [squarert (-a million)] / [squarert (-a million)] = 2 [squarert (-a million)] / -a million = -2 i A be responsive to clarification is mandatory for the final step. bear in mind that dividing one huge type via yet another is the comparable element as multiplying the huge type interior the numerator via the reciprocal of the denominator. So, as quickly as we divide a huge determination via -a million, we are unquestionably multiplying the numerator via a million/-a million = -a million. it quite is how the unfavourable sign is presented into the numerator.

Answer 7

To take the square root of a complex number, first you express it in polar form:
a+bi
=(a^2+b^2)^(1/2)*exp(i*theta)
where
cos(theta)=a/(a^2+b^2)^(1/2)
and
sin(theta)=a/(a^2+b^2)^(1/2).
Then (a+bi)^(1/2)
=(a^2+b^2)^(1/4)
*exp(i*theta/2)
or
-(a^2+b^2)^(1/4)
*exp(i*theta/2).

In your case, a=0 and b=1,
so theta=pi/2
and thus
sqrt(i)
=1/sqrt(2)+1/sqrt(2)*i
or
-1/sqrt(2)-1/sqrt(2)*i

Answer 8

That's as far as it goes. As for your hear-say, I am not aware of that comment.