# Why was $i$ introduced to satisfy this $\sqrt{-1}$?

Can someone explain to me why $$\sqrt{-1} = i$$ I love math and I'm looking at doing it to higher levels. I know that we can NEVER have a square root of a negative number as per my reading hence if I was given $$\sqrt{-4}$$ the answer would be $$2i$$ and if I was given $$\sqrt{4}$$ the answer would be $\pm2$, can I get sources explaining why $i$ was introduced?

Thanks.

• Are you looking for an explanation here (as in your first sentence) or a source (as in your next-to-last sentence)? – Rory Daulton Jun 30 '15 at 21:08
• Yes that is what i want – user249811 Jun 30 '15 at 21:19
• @user249811 which is it? – The Great Duck Jun 22 '16 at 18:04

Technically, $i \neq \sqrt{-1}$. $i$ is simply a symbol that we use to denote a solution to the equation $z^2 + 1 = 0 \iff z^2 = -1$. This symbol is called the imaginary unit and there isn't really a reason why it represents a solution to $z^2 = -1$, it's more that it was defined to be that way. This imaginary unit is quite important in the field of mathematics. It extends the real numbers and generates the complex number system that has quite a significant impact on mathematics. Examples include a whole branch of analysis termed complex analysis or provides algebraic closure for polynomials in $\mathbb{C}$.

Secondly, $\sqrt{4} \neq \pm 2$. The square root function denoted by the radical sign refers to only the positive value, which is why technically $i \neq \sqrt{-1}$ and why $\sqrt{4} = 2$ and not $-2$.

As for why the imaginary number was introduced, there are a multitude of reasons. For example, the introduction of such a number means that the field of real numbers is algebraically closed. Specifically, the algebraic closure of the real numbers is the field of complex numbers that are generated by including $i$. A lot about the usage and technicalities of the imaginary unit $i$ can be gleaned from its Wikipedia page here, that I would encourage you to read.

• This is wonderful info. Thanks – user249811 Jun 30 '15 at 21:24

on the history, my impression is that people were fairly happy having quadratic equations without (realnumber) roots. Then for cubic equations, Cardano, about 1545, introduced a rudimentary version of complex numbers, because his method lead directly to using cube roots of one of these objects. https://en.wikipedia.org/wiki/Complex_number

good book: http://www.maa.org/publications/periodicals/convergence/an-imaginary-tale-the-story-of-1 from the book description:

The solution of cubic equations in the sixteenth century forced mathematicians to deal with the square root of -1, and in the first major chapter of the book, the author handles this episode judiciously and leisurely.

• I am Fascinated by mathematics and i really want complex numbers not to give me hard time. Hence i decided to look into this. Thanks for the link – user249811 Jun 30 '15 at 21:26
• From my recollection of the history, part of the issue was that algebra was not exactly well-developed in Italian mathematics at that point. Instead they typically framed things geometrically; for example, the quadratic equation $x^2-3x+2=0$ might have instead been interpreted as the equality in area of different combinations of squares. That makes it easy enough to dismiss equations like $x^2+1=0$, since one would simply say "no such arrangement exists." (cont). – Semiclassical Jun 30 '15 at 21:38
• What forced the issue re: imaginary numbers was that, when one tackles the cubic case, there are cases which 1) Cardan's formula introduces square roots of negative numbers, and yet 2) if one treats $\sqrt{-1}$ seriously, then once one carries out all the algebraic manipulations then one ends up with real solutions. And that's not so easily dismissed, since these would amount to genuine geometric solutions. So imaginary numbers, even as they seemed nonsensical, were nevertheless quite useful. This link nicely discusses the history. – Semiclassical Jun 30 '15 at 21:45
• @Semiclassical, I thought you were leading up to en.wikipedia.org/wiki/Casus_irreducibilis which the chapter does not mention, or I did not recognize it there. – Will Jagy Jun 30 '15 at 23:17
• The casus irreducibilis shows up in section 4.3 of the linked pdf, though only in passing. But it plays the central role in the third part of that same set of notes. – Semiclassical Jul 1 '15 at 2:52