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Simply, what do I get when I take the square root of negative imaginary number? But, it cant be an imaginary number since the answer to the $2^{\mathrm{nd}}$ power must equal the original negative imaginary number. So:


$$ \sqrt{{-(|x|{_i}})} = y $$

where

$$ y{^2} = {-(|x|{_i}}) $$


So, if $ y=\sqrt{-\sqrt{-1}} $, then what is y ?



So, what would it be? (a double imaginary ($x{_{i}}_i$)?) And, how would your answer to the 2nd power equal the original negative imaginary number?

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    $\begingroup$ $x=\frac{1-i}{\sqrt{2}}$ and $y=\frac{i-1}{\sqrt{2}}$ satisfy $x^2=y^2=-i$ and can be considered to be the square-roots of $-i$. $\endgroup$ – Peter Oct 16 '15 at 13:11
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    $\begingroup$ It's not imaginary (i.e. of the form $bi$), but it is complex (i.e. of the form $a+bi$). $\endgroup$ – Akiva Weinberger Oct 16 '15 at 13:17
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The concept that you are missing is the Complex plane. Not surprisingly, given that I often seen it introduced a full year after imaginary numbers.

On the complex plane, taking a square root becomes almost entirely geometric. The non geometric part is, you have to take the square root of the distance from your number to 0. For instance $15+20i$ is at a distance of $25$ from 0 (per Pythagorean Theorem) so the distance of the square root will be 5.

To find the angle of the square root—just note the angle of the original number (measuring counter-clockwise from the real axis—a.k.a. the x axis) and cut it in half.

You may notice that there are two ways to cut the angle in half—clockwise, and counter-clockwise. That is correct; every number except 0 has two square roots.

You asked about negative numbers. Let's take $-i$ as an example. Note that the position of $-i$ on the complex plane is one unit "south" of 0, the same place that $(0,-1)$ is on a rectangular coordinate plane. Since the distance from 0 is 1, and the positive square root of 1 is 1, the distance from 0 for our answer will also be 1, making that part easy.

Note that the angle of that point, measured counter-clockwise from the x axis, can be stated as either $270$ or $-90$. Use both. Divide each by two. That gives the two answers.

Now solving with the pythagorean theorem (or using trig, but in this case the angles are just 45 degrees in the 2nd and 4th quadrants, so we don't need trig), we get the two square roots of $-i$, which are $\frac{1-i}{\sqrt2}$ and $\frac{-1+i}{\sqrt2}$.

Try squaring each of those with pencil and paper to convince yourself they are indeed correct.

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    $\begingroup$ Complex numbers are sometimes introduced a full year later than the imaginary numbers? Who does that, and what possible justification can they have for doing so? $\endgroup$ – David K Oct 16 '15 at 13:34
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    $\begingroup$ No—the Complex plane almost a full year after imaginary and complex numbers. So students struggle with simple things like multiplying, not knowing the geometric method, and fail to catch what should be obvious errors such as $(3+i) \times (1+4i) = -1-13i$, because they don't think geometrically and notice the wrong quadrant. $\endgroup$ – Wildcard Oct 16 '15 at 13:55
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    $\begingroup$ Though evidently the OP has just hit imaginary numbers and had no idea about complex numbers. So there is still a gap there, it's just not what I was referring to. $\endgroup$ – Wildcard Oct 16 '15 at 13:57
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    $\begingroup$ OK, I see the distinction between the arithmetic interpretation of the complex numbers and the geometric interpretation of the complex plane; but it still seems absurd not to introduce the two concepts simultaneously, or at least very close together. I can't remember any such gap in the instruction I received in high school. $\endgroup$ – David K Oct 16 '15 at 16:39
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    $\begingroup$ @DavidK: I agree. Imaginary numbers on their own, without regard to how they permit us to complete the complex numbers, seem bereft of most of their significance. Since the most common meaningful context where high-school students are likely to see imaginary numbers is in the solution to quadratic equations, it seems particularly silly to leave much of a gap between imaginary and complex numbers, and then between complex numbers and the complex plane. $\endgroup$ – Brian Tung Oct 17 '15 at 0:01
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If you are happy with $$ z = x+ iy = r\mathrm{e}^{i\theta} $$ we know that $-1 = \mathrm{e}^{ik\pi}$ for odd itegers of $k$ we can utilise this to find $$ \sqrt{-z} = \sqrt{\mathrm{e}^{ik\pi}r\mathrm{e}^{i\theta}}= \sqrt{r}\mathrm{e}^{i\frac{k\pi +\theta}{2}} $$

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Let's say we want to find the square root of $-2i$.

Well, since: \begin{align} (1-i)^2&=1-2i+i^2\\ (1-i)^2&=1-2i-1\\ (1-i)^2&=\phantom1-2i \end{align} we can conclude that $1-i$ is a square root of $-2i$.

Remember, every number (other than $0$) has two square roots. Indeed, $-1+i$ is also a square root of $-2i$. So $1-i$ and $-1+i$ are both square roots of $-2i$.

So there's no need to invent more than the complex numbers for this.

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I think the following question (with answers) will be helpful to you:

How do I get the square root of a complex number?

In short, some will say that if you view the square root as a function, this isn't well-defined on the complex numbers. Many will tell you that the square root of a number like $-4$ is $\pm 2i$, but this is false. A function (by definition) can only give you one output for each input. Now, in general we do tend to extend the square root to negative real numbers by saying that $\sqrt{-a} = \sqrt{a}i$ for $a\geq 0$. Again, see the above linked question and answers for more on this.

Also, Wikipedia had an article on this question that might be helpful: https://en.wikipedia.org/wiki/Square_root#Square_roots_of_negative_and_complex_numbers

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  • $\begingroup$ thank you for the links, they're really cool! $\endgroup$ – Jack Giffin Nov 18 '15 at 3:14
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You are bit wrong .because value of $i$ itself is $\sqrt{-1}$ which is a negative number. By example if we want square root of $-4$ it's + or $-2i$. I hope I have given the answer if I interpreted your question correctly. Remember $i^3=-i$ which is a negative complex number and not double imaginary.

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  • $\begingroup$ Well, you can mathematically represent it on paper, so there must be some way to classify it. $\endgroup$ – Jack Giffin Jul 29 '16 at 1:51
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On the Argand or Gauss diagram representing complex numbers the pure imaginary number $ (-\sqrt{2}\, i ) $ just goes to two complex points $ 1+i, 1-i.$ by square rooting.

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You need to pick a branch for the square root function. Some branches are more "standard" than others. A very popular one is along the negative real axis, meaning all numbers $a+bi$ with a negative $b$ will have a square root which also has negative $b$ and the rest will be mapped to positive imaginary parts. If we pick this branch the square root of $-i$ becomes $\frac{1}{\sqrt{2}}(1-i)$.

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The square roots of 1 are 1 and -1. These are real numbers.

The square roots of -1 are i and -i. These are imaginary numbers.

Clearly the square roots of i can not be purely imaginary nor purely real.

But it could be a combination of both. We can combine real and imaginary components to make Complex numbers. For example: 3 + 2i is a number that combines 3 "real units" with 2 imaginary units.

If real numbers exist on a 1-dimensional "line" then complex numbers exist on a 2-dimensional "plane". And they each have a "real part" and an "imaginary part".

So how do we add these numbers? (a+bi) + (c + di) = ? Well, (a+bi) + (c + di) = [a + c] + [b+d]i.

How do we multiple these numbers? $(a + bi)(c + di) = ac + bci + adi + bdi^2= ac + [bc + ad]i - bd = [ac - bd] + [bc +ad]i. $ It takes a little bit of getting used to but the "real part" is the difference of the products of the real parts and imaginary parts, while the imaginary part is the sum of products between the real parts mutiplied with the imaginary pars.

So what are the square roots of $i$? If it is $a + bi$ then $(a + bi)^2 = [a^2 - b^2] + 2abi = 0 + 1*i$. We need to find a, b such that $a^2 - b^2 = 0$ and $2ab = 1$. $a^2 - b^2 = 0$ means $a = \pm b$ so $2ab = 1$ means $a^2 = 1/2$ so $a = b = \pm \frac 1{\sqrt 2}$. So $\sqrt i = \pm \frac 1{\sqrt 2} \pm \frac 1{\sqrt 2}i$.

Lets check $(\frac 1{\sqrt 2} + \frac 1{\sqrt 2}i)^2 = \frac 1{\sqrt 2}^2 + 2*\frac 1{\sqrt 2}\frac 1{\sqrt 2}i + \frac 1{\sqrt 2}^2i^2 = 1/2 + 2*1/2i - 1/2= 1/2 - 1/2 + 1i = i$ . Ta-da!

Now, in the future, NEAT things will come from the fact that the complex plane is a plane. For example we will be able to take angles between numbers and get some nifty geometric results. (Such as when we multiply two complex numbers the result's angle will be the sum of the two numbers angle.) But that's for later.

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