Talk:Imaginary unit

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The calculation rule

root(ab) = root(a)root(b)

is only valid for real, non-negative numbers a and b.

Well, actually this equation is valid as long as you stay in the principal branch of the log function, i.e. as long as a and b don't lie on the negative real axis. But that might confuse people unnecessarily. 128.111.88.229

Contents

[edit] What on earth is this?

(From the "i and Euler's formula" section):

You could also compute it using Bernoulli's logarithme imaginaire (imaginary logarithms):

pi/2 = 2log[(1-i^{(1/2)i} \times (1+i)^{(-1/2)i]}

That expression isn't even written correctly. And it doesn't lead to anything. Were more equations supposed to follow?

I'm removing it from the main page, but I invite someone more knowledgable about imaginary logarithms than I (or the original contributor) to rework it and put it back in. --Ardonik 03:45, 2004 Aug 8 (UTC)

Damn you Wikipedia! I'll never finish this story if you keep sucking all of my time into your hideous vortex of knowledge. Damn you. DAMN YOU!

[edit] What is \sqrt{-1}?

Well, what is \sqrt{-1}???

Are you asking a philosophical question, or are you wondering what simpler terms it can be reduced to? In the latter case, it doesn't get any simpler.

[edit] What is \sqrt{i}?

What I meant to ask is: what is \sqrt{i}?

I suppose this question is asked by someone being curious, rather than by someone being subtle (in a way beyond my comprehension).
In elementary math, the squareroot is a single-valued function, and e.g. 25 has only the squareroot +5. In complex analysis, multi-valued functions are allowed, and 25 has two squareroots +5 and -5. This comes about in the following way:
The squareroot of x is defined as any number y satisfying y2 = x. Thus, if y is a squareroot, so is -y (since (-y)\times (-y) = y\times y). Thus, any number except 0 has two squareroots.
Thus, i has two squareroots. One is \frac{1+i}{\sqrt{2}}; the other one is minus that number. You can verify this in the following way:
\frac{1+i}{\sqrt{2}}\times\frac{1+i}{\sqrt{2}} = \frac{(1+i)(1+i)}{\sqrt{2}\sqrt{2}} = \frac{1\times 1+1\times i+i\times 1+i\times i}{2} = \frac{1+2i-1}{2} = i.
When dealing with products, quotients, powers and roots, it's often convenient to represent complex numbers in polar form, i.e. by their distance in the complex plane from the origin, and their direction angle. i is at a distance of 1 from the origin, at an angle of 90 degrees. \sqrt{i} is also at a distance of 1 (i.e. at the unit circle), but at an angle of 45 degrees or 45+180 degrees.
But I don't think any of this belongs to the article on i. Do you?
--Niels Ø 09:05, Mar 22, 2005 (UTC)

I suppose not. I was just wondering.

[edit] This does not produce a false result.

-1 = i \cdot i = \sqrt{-1} \cdot \sqrt{-1} = \sqrt{-1 \cdot -1} = \sqrt{1} = 1

Is \sqrt{1} not -1 also. A square root has two results.

In standard elementary math as taught by competent teachers at all pre-university levels, the squareroot is a function, and as such cannot give two results. That's why the quadratic formula must have "plus/minus" in front of the squareroot sign to give both solutions; the squareroot itself only gives one. In this context, \sqrt{-1} does not exist.
But in complex analysis, it is customary to work with multivalued functions, and the squareroot of any complex number (including 1 and -1, excluding only zero) is two-valued. So, as any paradox in math or logic, understood properly there is no contradiction. The purpose of the paradox is to raise awareness about the difficulties involved in "understanding properly".--Niels Ø 20:35, Jun 15, 2005 (UTC)


It seems to me that the error in this paradox is improperly explained in the article (under "Warning"). The error is not this step: \sqrt{-1} \cdot \sqrt{-1} = \sqrt{-1 \cdot -1}, but this step: i \cdot i = \sqrt{-1} \cdot \sqrt{-1}
In fact, I think there is no error in the first equation at all because taken as a complex square root the following is actually true: \sqrt{-1} \cdot \sqrt{-1} = \pm i \cdot \pm i = \pm 1 = \sqrt{1} = \sqrt{-1 \cdot -1}
The second equation; i \cdot i = \sqrt{-1} \cdot \sqrt{-1} is wrong because it either
a) Uses the squareroot as a function, in which case \sqrt{-1} does not equal anything, it is meaningless, or
b) Uses the squareroot in it's complex form, in which case there are two solutions to the squareroot, and you cannot say i \cdot i = \sqrt{-1} \cdot \sqrt{-1}, but must instead say \pm i \cdot \pm i = \sqrt{-1} \cdot \sqrt{-1}. --cesoid 24 October 2005
The unsophisticated reader should be warned that writing i as a single-valued result of the expression \sqrt{-1}, and use the the rules of calculation for the ordinary square roots also on 'square roots' of negative numbers, leads to contradictions. Some might argue 'okay, but it is just notation, don't worry'; however, a notation using a well-known symbol invites the use of the rules that normally apply for that symbol; so, in that case the reply is by reductio ad absurdum with the discussed 'proof'; here the step \sqrt{-1} \cdot \sqrt{-1} = \sqrt{-1 \cdot -1} assumes that normal calculation rules of square roots also apply to (single valued) square roots of negative numbers, and that step becomes now the source of the contradiction. However, calculation rules of square roots do not apply to negative arguments. Therefor the use of the notation \sqrt{-1} should be discouraged. The square root is a function, that is only defined for non-negative numbers. Bob.v.R 01:37, 27 October 2005 (UTC)
  -1 = i \cdot i = \sqrt{-1} \cdot \sqrt{-1} = \sqrt{-1 \cdot -1} = \sqrt{1} = +/-1

So its correct!--Light current 03:02, 30 October 2005 (UTC)

There is no doubt that 1=-1 is wrong. Where the error is depends on exactly how \sqrt{x} is interpreted. If \sqrt{x} is taken to mean the real square root function then the peice makes no sense. If \sqrt{x} is taken to mean the principle branch of the complex square root function then the rule \sqrt{a} \cdot \sqrt{b} = \sqrt{a \cdot b} is invalid. If \sqrt{x} is taken to be a multi valued function then the statements i \cdot i = \sqrt{-1} \cdot \sqrt{-1} and \sqrt{1} = 1 are half truths due to the fact that there are now two different values for \sqrt{-1} and \sqrt{1}. Plugwash 03:22, 30 October 2005 (UTC)

Yes but \sqrt{1} = +/-1. So the right answer is there if you wish to see it!--Light current 03:30, 30 October 2005 (UTC)

Light current, the reason that isn't true has already been explained several times in this discussion. \sqrt{1} \neq \pm 1. because the the radical symbol refers to the Principal, or positive, square root. \pm\sqrt{1} = \pm 1, \mbox{ but } \sqrt{1} does not. He Who Is 21:39, 10 June 2006 (UTC)

[edit] "Better" Notation

Just to add something really quickly, the square root sign can best be seen as a fractional exponent. This is another way of thinking of it. So the sqaure root of -1 can also be written as (-1)^(1/2). (Sorry, I need to polish up on math tags.) The cube root of 8 can be seen as (8)^(1/3). I'm sorry if I'm boring someone; after all, you can read about it here. Now, if you want to multiply two numbers with the same base, you add the exponents. So (-1)^(1/2) · (-1)^(1/2) is (-1)^(1), or just plain negative one. Perhaps this entire talk section can be somehow integrated into the article. Gracenotes T § 01:58, 7 March 2006 (UTC)

x^(1/y). being the Y root of X only works for postive numbers (for x). -1^(1/2) is equal to one which is not equal to i. (because any even root (positive) has at least two parts a negitive and a positive. X^(1/2) only has one part which is the positive.)

[edit] Non-italic notation of the imaginary unit

in the form i or j is frequently used in the literature to prevent confusion with e.g. indexes like i or j and also to prevent potential conflicts with the alternating current i - mostly used in physics and electrotechnology. The non-italic notation of i in math related sciences appears to be the best solution to prevent these potential conflicts. Further details see discussion of complex number. [Wurzel, 2005-june-23]

For now, it has been agreed that the imaginary unit continue to be represented by an italic i. Please see Talk:Complex number. Paul August ☎ June 30, 2005 15:39 (UTC)

[edit] Elementary question

After having read and re-read the article, and I'm sure I've been thinking about this too much but I can't get my mind around:

i x -i

Please excuse a question from a novice. 24.150.198.36 03:24, 21 January 2006 (UTC)

i x -i = - i2 = - (-1) = 1
Was this what you meant to ask? Bob.v.R 19:02, 22 January 2006 (UTC)

Yes, Thanks Bob JamesWakil 04:00, 24 January 2006 (UTC)

[edit] Copyvio "History" section removed

I removed the following text from the article, added by IP 213.64.127.19 on Feb 5th, which was copied from the cited text:

(Start of removed text)
(The following is taken from Visual Complex Analysis by T. Needham (Oxford, 1997), p.1–2.)
Girolamo Cardano's Ars Magna, which appeared in 1545, is conventionally taken to be the birth certificate of the complex numbers. Yet in that work Cardano introduced such numbers only to immediately dismiss them as "subtle as they are useless". As we shall discuss, the first substantial calculations with complex numbers were carried out by Rafael Bombelli, appearing in his L'Algebra of 1572. Yet here too we find the innovator seemingly disowning his discoveries (at least initially), saying that "the whole matter seems to rest on sophistry rather than truth". As late as 1702, Leibniz described i, the square root of -1, as "that amphibian between existence and nonexistence". Such sentiments were echoed in the terminology of the period. To the extent that they were discussed at all, complex numbers were called "impossible" or "imaginary", the latter term having (unfortunately) lingered to the present day. Even in 1770, the situation was still sufficiently confused that it was possible for so great a mathematician as Euler to mistakenly argue that \sqrt{-2}\sqrt{-3} = \sqrt{6}.
The root cause of all this trouble seems to have been a psychological or philosophical block. How could one investigate these matters with enthusiasm or confidence when nobody felt they knew the answer to the question, "What is a complex number?"
A satisfactory answer to this question was only found at the end of the eighteenth century. Independently, and in rapid succession, Wessel, Argand, and Gauss all recognized that complex numbers could be given a simple, concrete, geometric interpretation as points (or vectors) in the plane: The mystical quantity a+ib should be viewed simply as the vector connecting the origin to that point.
(End of removed text)

It would be good have this content, in our own words, added back to the article (I might do it if I get around to it). Paul August ☎ 16:16, 12 February 2006 (UTC)

Here is a link to the book on Amazon (which can be searched online). And here is the copied text. Paul August ☎ 16:23, 12 February 2006 (UTC)

[edit] Cleanup

IMHO, the article can be improved in the following places:

  • Intro: The introductory paragraph is vague and imprecise. A better intro could be The imaginary unit i is a designated root of -1 in the field of complex numbers. which can also serve as a definition.
  • Definition: The definition is vague and imprecise.
  • i and -i: There are a couple of important references, but I think the section should be rewritten. If the definition is fixed above, there is no ambiguity. The algebraic, i.e. Galois theoretic approach could be emphasized here.
I strongly feel that there is no place for appears [...] ambigous, most precise explanation, appears to occur etc. in a math article. The last paragraph in that section is confusing; if one wants to mention something like this, one should note that in any ring containing a root x of -1, -x is also a solution to X2 + 1 = 0.
  • Warning: The equation does not hold. So there should be a \neq in the appropriate place. In general, I think that giving cookbook recipes such as always use plusminus in front of a root is not a good idea. Math isn't a collection of recipes, we should try to help readers think and argue mathematically.
  • Powers of i: Could be replaced by a section on or reference to roots of unity in general, is IMHO as such not of interest.
  • i and Euler's formula: Delete. There is no such thing as raising an equation to the power i. There is no generally accepted definition of ii (even though it is possible to give one, but then one would have to talk about branches of log, which I think is not such a good idea). Deducing Euler's identity from this seems to me similar to proving that zero is a real number because it is equal to \lim_{n \to \infty} \frac{\pi \cdot n + 3.7}{n^{2}}.
  • A section Generalizations giving a reference to Galois theory would be nice.

Any volunteers? --Tobi (141.20.53.108 14:03, 6 March 2006 (UTC))

I think "... which can also serve as a definition" is somehow problematic, I would leave it out. As to the rest, I agree to all you say; please be bold and do it... (I'm already involved in too many other controversies... ;-). — MFH:Talk 21:36, 27 March 2006 (UTC)

[edit] How to arrive at it?

Currently the article reads:

" From the above identity

eiπ / 2 = i

one arrives elegantly at Euler's identity

eiπ + 1 = 0

" My question is, how is that done? What are the steps in between? Thanks Kirbytime 03:13, 30 March 2006 (UTC)

Square both sides and add one --Glengarry 14:11, 24 April 2006 (UTC)

[edit] Not sure about beginning of article

I quote - "Since there is no real number that squares to any negative real number, we imagine such a number and assign to it the symbol i."

I'm not sure if this is a very good or productive explanation - especially without "Real Number" linking to a definition of a real number, to a non-mathematical reader the definition encourages the myth that Complex Numbers are somehow "less real" than real numbers. Maybe I'm being too pedantic, but both are equally valid based on their uses - negative numbers or irrational numbers (all of which are real) need to be "imagined" just as much as an imaginary number (using "imaginery" to mean "not real") does. Matt 13:10, 9 May 2006 (UTC)

[edit] Simple question not covered in article.

This article fails to address the concept of \sqrt{-i}

Is there an easy way to explain this? I'm a bit unclear on the concept myself, having been unable to find information about it on the web. My elementary experimentation with it - and I am no mathematician - seemed to show that there is no way to handle it except to add another axis to the two dimensional visualization of complex numbers, and hence to have a new set of imaginary numbers which would deal with it. Is this acceptable? Does anybody know of a place on the web where I could find a decent explanation of this?

Thank you for your time.

the square roots of a complex number are complex and we can find them by putting the number into polar form. In the case of your question the principle branch of the square root can be found as:
\sqrt{-i} = \sqrt{e^{-\frac{1}{2}\pi i}}=e^{-\frac{1}{4}\pi i}=\cos(-\frac{1}{4}\pi) + i\sin(-\frac{1}{4}\pi)= \frac{\sqrt 2}{2}-\frac{\sqrt 2}{2}i
And the other square root can be found either by negating the result or adding 2πi to the power in the first complex exponential. Plugwash 10:23, 10 May 2006 (UTC)

To add to that, that is certainly acceptable. In fact, it's what's already used. It's called the complex plane or an Argand diagram. An as an extension of the reals, denoted \mathbb{R}, \mathbb{C} exists to denote the set of complex numbers. Complexes are numbers of the form ai+b, where a and b are constants. Another important factor to consider is that complexes aren't ordered. They can't be compared. No complex number is larger of smaller than another unless they have the same imaginary part. Then the real part is used to compare them. Because the complexes are represented on a plane, a complex function requires four dimensions to properly represent. Because of this, we generally use two seperate 3-dimensional graphs. The output dimension of one is the real part of the output, and the other represents the imaginary part of the output. He Who Is 21:55, 10 June 2006 (UTC) P.S. Sorry if that was a bit condescending. I don't know the extent of your prior knowledge, so I just desided to go over all the basics.

The root of any complex number can be expressed as another complex number. Another method which requires a bit more smacking but a bit less knowledge on the subject:
Let root(-i) = a + bi
-i = a^2 - b^2 + 2abi
Equating coefficients, a^2 - b^2 = 0, 2ab = -1
a^2 = b^2, ab = -1/2
It's fairly obvious that a = 1/root(2), b = -1/root(2) or vice versa.
root(-i) = +/- 1/root(2) -/+ i/root(2)
Square it and see. --Generalebriety 08:07, 16 July 2006 (UTC)

[edit] Finally!

I simply want to commend whoever contributed to this article for their near-complete lack of use of i = \sqrt{-1}, instead of i2 = − 1. People don't seem to realize that sqrt(-1) has no solutions, real or complex. Imaginaries aren't ordered, and radicals denote poitive square roots. i isn't poitive or negative, so it isn't a solution to the first. The only place in the article I see the "incorrect" notation is in the false proof that 1 = -1, which is admittedly false anyway. Thank you all. He Who Is 22:15, 10 June 2006 (UTC)

It's OK to speak about i = \sqrt{-1} if you're careful. But of course, you have to be careful. The section on square roots of i is pretty fucked. -lethe talk + 23:17, 10 June 2006 (UTC)
Your thinking of i being a number that you can see however it is not like that. i is used in vectors (in physics) and it is proven. "i exist and yet it does not exits, at the same time" is a mediphor used to explain i. (i exist in one sence but it does not in another)

[edit] MERGE?

Imaginary number and Imaginary unit are two different articles, with a lot of overlap...I can easily see them being combined into a concise article. --HantaVirus 14:09, 28 July 2006 (UTC)

The imaginary unit is a very specific complex number that plays an important role. A separate article on it seems justified to me. Bob.v.R 20:49, 28 July 2006 (UTC)

[edit] Permalink for WIkipedia research

Hello, editors of Imaginary unit! I am currently working on an essay on Wikipedia, part of which will feature a comparison of articles of Wikipedia and Encyclopaedia Brittanica. To ensure that I send reviewers articles that have not been recently vandalized or have not been involved in an edit war, I would like, by December 31st, a revision of this article to be listed at User:Chrisisme/Research-permalinks that is not vandalized and/or is generally at peak quality. Thank you! Chrisisme 20:05, 20 October 2006 (UTC)

Use the edit history Quantum Burrito

[edit] i^i?

I'm unsure if i^i could massaged into anything useful. Just out of curiosity, is it? If so, what does it equal? thadius856talk 07:24, 18 November 2006 (UTC)

See article [i and Euler's Formula]. DVdm 10:51, 18 November 2006 (UTC)

[edit] cube root of i

You menthend the square root of i but what about other roots? —The preceding unsigned comment was added by Bearmancorn (talk • contribs) 01:26, 11 January 2007 (UTC).

[edit] i and -i

The entry states that there is no quantitative difference between i and -i, and that if you were to replace all instances of i with -i in a given textbook or set of theorems, all theorems would still hold true. Is this true even in the case of the Fourier Transform? The Fourier Transform, I believe, is quite dependent on the sign of the exponent, as the only difference between the FT and the inverse FT is the sign of the exponent and a scale factor. —The preceding unsigned comment was added by 70.171.5.244 (talk)

If i was substituted with -i the theorems of the FT would still hold true but the application would be change in that the forward transform would take you from frequency to time domain. --Philbarker 16:35, 22 February 2007 (UTC)

[edit] powers of i

Perhapse it should be noted that :i^{x} = i^{x mod 4}\, (where mod is the modulo operation). This article section does not seem to cover this very much. - 71.31.226.115 02:47, 17 March 2007 (UTC)

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