# Talk:de Moivre's formula

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## Assert true for all real powers

HI De Moivre's formula is actually true for all complex numbers x and all real numbers n, but this requires careful extension of several functions to the complex plane. <-- Loisel I'm not so sure this makes any sense$.$

From (cosy+isiny)^2=cos2y+isin2y, one obtains cosy+isiny=±SQRT(cos2y+isin2y), or SQRT(cos2y+isin2y)=±(cosy+isiny). Now, let y=x/2. Then (cosx+isinx)^(1/2)=±[cos(x/2)+isin(x/2)]. That is what De Moivre's formula should be when a square root is involved:; the square root must have ± value. One of them is a false root. If so, the example when x=0 and x=2*pi will not be ambiguous as long as the false root is discarded. Therefore, the section that says De Moivre's formula fails for non-integer power has used the wrong form of De Moivre's formula and has drawn an incorrect conclusion from the wrong formula. Can the editor give a legitimate reference to back up his conclusion? 2602:306:CE35:2510:5008:B12F:27A9:9BF1 (talk) 19:53, 18 December 2014 (UTC)Mystified
Plus 1 is a value. Minus 1 is a value. However plus or minus 1 is not a value. it is two possible values. Waving your hands saying one root is false does not fix anything. How do you know which is the 'false' root except in that it doesn't correspond with what you first thought of? Have you tried reading the section De Moivre's formula#Failure for non-integer powers, and generalization? Dmcq (talk) 00:04, 19 December 2014 (UTC)
Yes, I have. That is why I am asking where one can find a widely referenced paper (published in a reputable journal), or a textbook, or a handbook that says De Moivre's formula fails when the power is not an integer. Then and only then do readers accept that Section in Wiki. To me, that section in Wiki is baseless. Sqrt (1) = ±1. One should keep it as such until information is available to make a judgment. That section in Wiki has no basis to pick one value over the other to support the false claim. 108.227.82.81 (talk) 03:18, 19 December 2014 (UTC) Mystified
Are you not making an unduly strong assertion about a false claim? The maths involved is clear enough to most of the experienced maths editors here. One can actually argue this case without reference to de Moivre's formula, only referencing material about exponentiation involving complex numbers. The Exponentiation article gives all the necessary background. —Quondum 03:43, 19 December 2014 (UTC)
The example given by Wiki does not support the claim. When the power of a complex number is 1/2, De Moivre's formula should be either (cosx+isinx)^(1/2)=±[cos(x/2)+isin(x/2)], or (cosx+isinx)^(1/2)=cos[k*pi+(x/2)]+isin[k*pi+(x/2)], k=0 and 1. When x=0 and x=2*pi, either form of the formula gives sqrt(1)=±1, as should be. My point is, Wiki had better use some other example to make the claim. 108.227.82.81 (talk) 13:42, 19 December 2014 (UTC)Mystified
Your point is simply wrong, and your insistence on repeating it over and over again is very dull. The article is correct as written, contains a reasonably clear explanation of why it is correct, and also contains a not-wrong explanation of how to handle non-integer powers. This has been explained to you repeatedly. So please stop wasting the time of other editors by repeating yourself over and over without trying to understand what you have been told. To the extent that your problem is failing to understand the simple point that "±1" is not a number (but rather, a set containing two numbers) there is very little I can do for you, but perhaps you could go bother the people at the math reference desk or math stackexchange instead. --JBL (talk) 14:13, 19 December 2014 (UTC)

Can you answer my request to the point? Give me a reference in a textbook, a handbook, or a journal paper. 108.227.82.81 (talk) 23:13, 19 December 2014 (UTC)Mystified

Thanks for pointing that out; n has to be an integer. x can be complex however, since Euler's formula

e^(ix) = cos(x) + i sin(x)

works for all complex x. --AxelBoldt

Is the function described as a [multivalued function]? in fact a periodic function?

Yes it is. It is 2π periodic, because sin and cos are. --AxelBoldt

Is the function described as a multivalued function in fact a function? If so, what is the domain of this function? z,w being fixed chosen numbers, it is is some sense rather a multivalued constant, isn't it? (I mean, a constant can be seen as a nullary function, but this would be even more nonsense hair-splitting... and again, I can't see on what domain it is defined - don't say the empty set, because it is well known that there is only one universal function defined on this set, and it definetely has not multiple complex values.) MFH: Talk 19:46, 10 May 2005 (UTC)

Hey Loisel. I looked in a few advanced level math books that featured DeMoivre's Theorem, and n is always stated as a RATIONAL, but no proof allows for the irrational component of the reals. The high school textbook I looked at claims that n is a POSITIVE integer. Also, this is the only case where I have seen x used in the statement of DeMoivre's formula -- all the books I have referenced use ${\displaystyle \theta }$. Thelema418 (talk) 22:25, 18 August 2012 (UTC)

I now found a few books that state n is real. And after finer reading of the books that say RATIONAL, it appears that they claim n is real in later analysis. What is truly interesting is that several texts make no statement concerning the domain for n. The PRAXIS exam guides created for the math requirements contains the theorem, but does not give clarification. Also, a few of Larson's texts do not state it either. Thelema418 (talk) 03:58, 19 August 2012 (UTC)

## Page might want tidying

I've just edited the page in large sections but am not too familiar with wiki editing, as such my edit may seem to be laid out in a somewhat messy fashion for the < math > sections. You may also want to check the whole article to check that the tone stays constant throughout as I have not edited the entire page, only the sections that warranted it. Thank you. :) sumandark8600 (talk) 01:01, 17 January 2015 (UTC)

## proof

Is the proof in the article the original proof of the formula? I mean, is it the way DeMoivre obtained or proved it?

Since DM's formula is a lot easier to prove using Euler's formula, I see no other reason for this proof to be in the article (except curiosity)... -- Euyyn (March 26 2005, 2346 GMT)

unreal numbers are things like -20 centimeters and the square route of -5. —Preceding unsigned comment added by 66.189.214.66 (talk) 03:20, 5 May 2009 (UTC)

DeMoivre never explicitly stated this formula; it was named in honor of him, but I'm not certain if he even used it. I added a citation to this article becuase I think that matter is confusing. Thelema418 (talk) 22:05, 18 August 2012 (UTC)

Doing some more research I found that the following 2 page article by De Moivre is cited as his implicit proof. Note that versine was commonly used at the time. http://cerebro.xu.edu/math/Sources/Moivre/de_%20sectione_anguli.pdf Thelema418 (talk) 02:39, 23 August 2012 (UTC)

## first proof

I don't think the first proof is correct, because we are assuming without proof that the exponential law is true also for imaginary numbers.

## cos(x+i)

I think you guys have the parentheses in the wrong places, don't you?

cos(x+i) sin x is not the same thing as cos x + i sin x, and I'm pretty sure the latter is what you mean. When someone who knows how to edit HTML fixes this, feel free to delete my comments here on the talk page. --anon

You are right, there was a bug in the proof. Fixed now. Oleg Alexandrov 19:33, 3 August 2005 (UTC)

## de moivre formula is valid for all n

i dunno why this article says only integers, but i can prove its true for i. all i did is the hyberbolic trig functions and i proved that cis(ix)=e^-pi or something, my point is its valid for all complex n. is there something wrong with what i did? --anon

The big problem is that for nonintegers n, the n-th power is not defined uniquely. If you try to use De Moivre's formula then, you run into trouble. See here:

${\displaystyle -1=\cos \pi +i\sin \pi \,}$

but also

${\displaystyle -1=\cos 3\pi +i\sin 3\pi \,}$

By De Moivre's formula:

${\displaystyle (-1)^{1/2}=\left(\cos \pi +i\sin \pi \right)^{1/2}=\cos {\frac {\pi }{2}}+i\sin {\frac {\pi }{2}}=i\,}$

but also

${\displaystyle (-1)^{1/2}=\left(\cos 3\pi +i\sin 3\pi \right)^{1/2}=\cos {\frac {3\pi }{2}}+i\sin {\frac {3\pi }{2}}=-i\,}$

so you get two different answers which is not good. Does this help? Oleg Alexandrov (talk) 05:51, 16 October 2005 (UTC)

-1 is square of +i or -i. So square root of -1 is naturally +i and -i. -i should be viewed as -(-1)^{1/2}. In this way, the ambiguity is avoided and De Moivre's formula remains applicable. 70.53.228.108 (talk) 02:13, 21 November 2014 (UTC) Cucaracha
The cubic root of -1, obtained by De Moivre's formula, is 0.5+0.866i, -1, 0.5-0.866i. I do not see a problem with the formula when n is a rational number. 70.53.228.108 (talk) 02:38, 21 November 2014 (UTC)Cucaracha
The cube root of −1 is also −1 using your logic and De Moivre's formula so all three are the same by your reasoning. There is a section [[[Failure for non-integer powers]] which explains why it doesn't work for non-integer powers. Dmcq (talk) 09:39, 21 November 2014 (UTC)
De Moivre's formula is valid for all rational n. I said the three cube roots of -1 are 0.5+0.866i, -1, 0.5-0.866i, NOT ALL ARE -1. The section on [Failure for non-integer powers] should be revised. 70.53.228.108 (talk) 16:01, 21 November 2014 (UTC) Cucaracha
I did visit that Section. I found the example of using i=(-1)^(1/2)=(1/-1)^(1/2)=1/{(-1)^(1/2)}=-i is wrong. The correct expression is i=(-1)^(1/2)=(1/-1)^(1/2)=±1/{(-1)^(1/2)}=±i. -i is a fictitious root from taking the square root and should not be reported. The correct result is i=(-1)^(1/2)=(1/-1)^(1/2)=-1/{(-1)^(1/2)}=i. This section should be revised.70.53.228.108 (talk) 16:46, 21 November 2014 (UTC)Cucaracha
Please refer to the section. It doesn't talk about or use i in the example. Dmcq (talk) 17:51, 21 November 2014 (UTC)
PLEASE go to the link "Failure of Power and Logarithm Identities" in that Section to see the example I quoted. I hope someone in charge of editing here would listen to me and would revise this Section. One of the examples this Section based on is flawed. 70.53.228.108 (talk) 21:04, 21 November 2014 (UTC)Cucaracha

──────────────────────────────────────────────────────────────────────────────────────────────────── I can't decipher what Cucaracha is saying, but it is certainly true that the section in question is not written clearly and could use some serious editing. --JBL (talk) 23:26, 21 November 2014 (UTC)

The example in "Failure of Power and Logarithm Identities" that shows i=-i is wrong. It forgets (1)^(1/2) can be ±1. If (1)^(1/2)=-1 instead of 1 is used, i=i will follow. In a math article, how can this kind of mistake not be corrected? 70.53.228.108 (talk) 02:46, 22 November 2014 (UTC) Cucaracha
(1) This is a comment about a totally different article, why are you making it here? (2) You don't understand the example you are objecting; the precise point is that it is not possible to define these functions as honest functions in a consistent way. Saying "but it can take multiple values" is to completely miss the point of the exercise. --JBL (talk) 15:06, 22 November 2014 (UTC)
This article makes reference to the link. The link uses an invalid result to make a point. How can you say the link is irrelevant and not see the need to correct a glaring error? Please do not reply to my comment unless you are an editor. It is the editor, not you, who bears the responsibility of making Wikipedia correct and accurate. 70.53.228.108 (talk) 22:14, 22 November 2014 (UTC) Cucaracha
Everyone can be an editor on Wikipedia. What is said at the linked place is correct. It seems you disagree with the people there and here so what you need to do to get any further is a WP:Reliable source that says what you say. Without that no 'correction' to conform to your ideas will take place. Dmcq (talk) 23:36, 22 November 2014 (UTC)
Come on. n can even be a complex number, let alone a real number. Let us raise e^(ix) to the power i. Then, e^(-x)=cos(ix)+isin(ix)=cosh(x)-sinh(x). So who say n has to be an integer?? Why don't you admit that the example in the link is flawed and n need not be an integer. 70.53.228.108 (talk) 19:56, 23 November 2014 (UTC)Cucaracha
Ok, here's what someone said earlier in TALK: ".... I looked in a few advanced level math books that featured DeMoivre's Theorem, and n is always stated as a RATIONAL...." 70.53.228.108 (talk) 20:10, 23 November 2014 (UTC)Cucaracha
A citation is where you give the name and author of a book and the page number. And preferably for the discussion here say what it says. I'll copy the demonstration of the problem in the section 'Failure for non-integer powers'
For x = 0 the formula gives 1½ = 1
F:or x = 2π the formula gives 1½ = −1.
As that section shows there is no corresponding problem with the complex exponential function equivalent to de Moivre's formula. Basically there are two possible values for 1½ according to de Moivre's formula but a function can only have one value for each input. What you are talking about is in the article as a generalization which can be used to find the various roots of a complex number. Dmcq (talk) 23:02, 23 November 2014 (UTC)
No. The example should be understood as cos(kπ+x/2)+isin(kπ+x/2)=±{sqrt[e^(ix)]}, where k=0 and 1. When x=0, the relation becomes cos(kπ)+isin(kπ)=±1. For +1, k=0. For -1, k=1. The result is one-to-one corresponding, The ± from taking a square root cannot be dropped. When you take the square root of 1, do you not get ±1? Don't you see why I keep telling you the example is flawed? 70.53.228.108 (talk) 00:10, 24 November 2014 (UTC)Cucaracha
Formulae don't understand anything. It is you who are keeping in mind where the 1 comes from. The 1 does not remember anything. Dmcq (talk) 01:37, 24 November 2014 (UTC)
(1) If you stick to i^2=-1, sqrt(1)=±1, and sqrt(-1)=±i, you can never conclude from the example that i=sqrt(-1)=sqrt(1/-1)=-i. (2) Let us assign λ=sqrt(2) in De Moivre's formula [cosh(z)+sinh(z)]^λ=cosh(λz)+sinh(λz). For, say, z=1, we obtain LHS=(1.543081+1.175201)^sqrt(2)=4.113251, and the RHS=cosh[sqrt(2)]+sinh[sqrt(2)]=2.178184+1.935067=4.113251. Next, for, say, z=i, we obtain LHS=[cosh(i)+sinh(i)]^sqrt(2)=[cos(1)+isin(1)]^sqrt(2)=e^[isqrt(2)]=cos[sqrt(2)]+isin[sqrt(2)], and the RHS=cosh[isqrt(2)]+sinh[isqrt(2)]=cos[sqrt(2)]+isin[sqrt(2)]. It is seen, the formula also works for an irrational power. My points are: (A) The article should not restrict the power to integers only. (B) The example used to support the restriction is flawed. 70.53.228.108 (talk) 16:43, 24 November 2014 (UTC)Cucaracha

──────────────────────────────────────────────────────────────────────────────────────────────────── And you are wrong on all points. Specifically, there is no "±" in de Moivre's formula. At n = ½, de Moivre's formula unambiguously assigns the value 1 to 1½ (when we take x = 0) and also unambiguously assigns the value −1 to 1½ (when we take x = 2π). This inconsistency is precisely what is meant by "the formula doesn't work for non-integer values of the exponent." This is explained clearly in the article (well, at least I feel that it is clear after I have rewritten it). --JBL (talk) 17:28, 24 November 2014 (UTC)

So you are the author. Please go to Eq. (1.3-9) in Korn and Korn to see it for yourself. I have decided not to write anymore. Is it not that the title of this Section says de Moivre formula is valid for all n? I happen to agree with it. And you do not. Bye now. 70.53.228.108 (talk) 18:53, 24 November 2014 (UTC)Cucaracha
Wikipedia is a collaborative enterprise, there is no one author.
I just had a look on the web and it seems there are a number of people saying they have proof for all n or all rational n. Those proofs are not correct because they assume that exponentiation works cleanly and is unique for complex numbers. Dmcq (talk) 23:05, 24 November 2014 (UTC)
FWIW, Cucaracha appears to be referencing this book: [1] which I do not have access to from home. --JBL (talk) 23:21, 24 November 2014 (UTC)
Also, a brief run around the internet suggests that at least some sources use the name "de Moivre's formula" for the (correct, but different) formula that appears in the Applications subsection. Possibly this is related to (or is the cause of) the underlying misunderstanding. --JBL (talk)
I don't have that book either, seemingly there's an old Schaum book for schools which also says something like that but doesn't try and give a proof. Amalgamating those sections makes it clearer what is being said I think, thanks. It is worrying when mathematics becomes a question of faith for people. Dmcq (talk) 09:36, 25 November 2014 (UTC)
I apologize for going back on words by writing again. But for the sake of math, I accept to be scolded. I believe the confusion will go away if the example in WP is viewed as follows. cosx+isinx=cos(2kπ+x)+isin(2kπ+x). Then, sqrt(cosx+isinx)=±[cos(kπ+x/2)+isin(kπ+x/2)], k=0, 1. Now, it does not matter if x=0 or x=2π, the above expression will yield sqrt(1)=±1, which is the way it should be. The difference between WP's view and my view is ± after a square root is taken. If my view is correct, then the subject example cannot be used to cast doubt on De Moivre's formula when n is not an integer. An Indian proverb says: "When you lie in bed, it matters not if your head or your feet are at the headboard, your bum is always in the middle of the bed." The moral is the fact cannot change.70.53.231.244 (talk) 17:17, 26 November 2014 (UTC)Cucaracha
Well whatever makes you happy but it would be better if you gave up this idea that one thing can be equal to two different things. It is possible to deal with multivalued functions as giving a set containing the results but that's very messy and people try to avoid anything like that nowadays. They started going out of fashion with Riemann surfaces I believe. Dmcq (talk) 17:42, 26 November 2014 (UTC)

## TeX formatting/Expert plea

I did a lot of formatting with the TeX and the general organization. Since I'm not an expert in the topic, I really hope I didn't mess anything up. So, if anyone who is an expert can give it a check, I'd be very grateful. Foxjwill 19:52, 5 March 2007 (UTC)

## Picture

The picture has the cube roots of unity as (-1/2 +/- i*sqrt(3/2)), where they should be (-1/2 +/- i*sqrt(3)/2).

## Complex Numbers and Trigonometry

"The formula is important because it connects complex numbers (i stands for the imaginary unit) and trigonometry"

I think the reason for this article being so quality impaired is that it takes De Moivre's formula at face value and tries to bring together trigonometric functions and the exponential functions as if they were totally unrelated to begin with. This article reads like a formula sheet and nothing else. The real subject here is the complex exponential function, defined by the power series expansion. Sine and Cosine functions are nothing more than notational conveniences for the imaginary and real parts of said expansion, respectively.

If it weren't for De Moivre stumbling upon this before Euler's contributions, this formula would be nothing more than a novelty of Euler's formula. It would be in the best interest mathematically to merge this article into Euler's formula (because it is just that with integer-multipled arguments) or an article on complex variables but the encyclopedic standards say that it should not. I would try to improve the article but I am not versed in the art of editing. --Fusionshrimp (talk) 05:16, 22 January 2009 (UTC)

It says that the formula doesn't hold in it's simplest form for for fractional exponents. But is there any work on how the sine/cosine representation might "mesh" with Newton's binomial expansion applied to complex numbers? That question may not make any sense - number theory isn't my forte at all. —Preceding unsigned comment added by 216.9.142.188 (talk) 02:55, 3 February 2011 (UTC)

## de moivre's formula

I will always recall my maths teacher explaining the wonder of de moivre who derived the following:

e to the power of pi i is equal to -1 ie: an indeterminant number to the power of another indeterminant number multiplied by an unreal number can exactly equal a negative integer. —Preceding unsigned comment added by 58.160.121.28 (talk) 22:24, 12 February 2009 (UTC)

Ah the joys of math teachers who only confuse themselves and their students. That's the biggest load of garbage I've ever heard. Could you ask your maths teacher to make precise the notion of an 'indeterminate' number for us? Same goes for 'unreal'. --Fusionshrimp 128.194.39.250 (talk) 07:43, 1 March 2009 (UTC)

## Proof for n belonging to the set of Rational numbers!

Hello guys.

I m a mathematician and I want to present before you all a proof on De Moivre's theorem.Well, here is it..

Suppose n= p/q where p,q are integers and q is not equal to 0

we can write cos x + i sin x = cos (q/q)x + i sin (q/q)x

put x/q = a => cos x + i sin x = cos qa + i sin qa

=> cos x + i sin x = (cos a + i sin a)^q [using De Moivre's theorem for integers for q]

Decreasing the power by q on both sides, we have

>>>Sorry, if you are a mathematician (as you claim), how can you "decrease the power by q"? Would you also decrease the power by 2 on both sides of the equality 2^2 = (-2)^2 ? Boris Tsirelson (talk) 19:28, 2 January 2010 (UTC)
>>>He means take the 'q'th root of both sides. if q=2, then the square root, for example. Just like if X^2=9, then X=sqrt(9), +-3. decrease probably isn't the right way of saying it, but his maths certainly looks fine to me. 118.208.134.29 (talk) 21:47, 31 March 2010 (UTC)

(cos x + i sin x)^1/q = cos a + i sin a

Increasing the power by p on both sides,

(cos x + i sin x)^p/q = (cos a + i sin a)^p

Now we know p/q is n, so

(cos x + i sin x)^n = cos pa + i sin pa [Again using De Moivre's theorem for Integers for p]

Now we can write a= x/q,so

(cos x + i sin x)^n = cos (p/q)x + i sin (p/q) x

Since p/q is n,

therefore,

(cos x + i sin x)^n = cos nx + i sin nx

For n belonging to rational numbers... ____________________________________________________________________________________________________

By this proof, I wish to disprove the so called 'MYTH' that n can be an integer only.

I want all the support I can get from you guys.

I am 100% confident about my proof, but if you sill see a flaw, feel free to discuss it at my wikipedia page!

A publication Backup can be here given

Page 294, Chapter 43

Schaum's Outline series Theory and problems of college Maths 2/ed By:- Frank Ayres,Jr.

Philip.A.Schmidt

Schaum's outline 3rd edition says the proof is outside the range of the book. The book has no proof. It states something which is wrong. The proof above is your own Original research and not based on any book. You have not raised it at WP:RD/MA for others to discuss with you like I asked nor found a published proof. Dmcq (talk) 18:55, 2 January 2010 (UTC)

But it is stated there and I have supplied the proofs...so we 2 just shut up for a while and see what others have to say about it! Moreover..that book is for <high school but >Degree college. That's they deleted the proof. Or if we go by you, a first grader would have all the periodicity of sin and cos and the log of e and all that(the thought makes me laugh) —Preceding unsigned comment added by Mathsislife (talkcontribs) 19:02, 2 January 2010 (UTC)

Have a look here for why this "proof" is wrong: http://en.wikipedia.org/wiki/De_Moivre%27s_formula#Failure_for_non-integer_powers Ulner (talk) 19:11, 2 January 2010 (UTC)
Gosh you're quick, thanks, I was just putting in the notice that I'd raised this problem at Wikipedia talk:WikiProject Mathematics#De_Moivre's_formula. Dmcq (talk) 19:16, 2 January 2010 (UTC)
I've put in a couple of references to the generalization section into the leader and the bit about the failure, that deals with using fractional powers to find roots. Dmcq (talk) 09:05, 3 January 2010 (UTC)

In the section 'Failure for non Integer power' you stated e^{iπ} = -1 But for any value, The value of e can never fall below 0. what have you to say for it? —Preceding unsigned comment added by Mathsislife (talkcontribs) 09:18, 3 January 2010 (UTC)

What you are saying about e is only true for real powers. For complex powers e can assume any complex number value except 0. By the way greater than 0 isn't a very meaningful idea for complex numnbers. See the section Exponential function#Complex plane. See also Euler's identity which is possibly the most famous identity in mathematics. Dmcq (talk) 09:32, 3 January 2010 (UTC)

but the thing is..

In the failure for non integer power section..u took Pi as the argument...but I m giving Proof of Rational numbers. But Pi is irrational!! —Preceding unsigned comment added by Mathsislife (talkcontribs) 08:27, 4 January 2010 (UTC)

That was for x. No ones disputing x can be anything. The discussion is about n being a non integer value and n was taken as a half which is a rational non-integer. Dmcq (talk) 15:17, 4 January 2010 (UTC)

That's what I m talking about! In the failure section... When I saw it closely..It revealed some facts

1)When disproving N to b any number, the value of x is not kept constant, which should not be happening because after every period of 2 pi, the values differ...

2)The principal argument of any complex number has to be either -pi to pi or 0 to 2 pi.

Both of them cant happen simultaneously, but in the failure section, u have first taken principle value -pi to pi then in the second example 0 to 2 pi.

The answers are bound to differ.

Please DMCQ read the taylor series of sine and cos functions. They are periodic. —Preceding unsigned comment added by Mathsislife (talkcontribs) 09:59, 7 January 2010 (UTC)

What you are saying sounds a bit similar to what I said in Talk:Exponentiation#Rational Exponents about the principal argument and rational powers if x is restricted to (-π,π]. However a person is quite entitled to specify any value they like at all for x. The business of principal argument only comes into it when converting back to angles from a sine or cosine or whatever, not when handing an argument to sine or cosine. Dmcq (talk) 10:48, 7 January 2010 (UTC)
Well If I would have not known that the values of sine and cosine repeat itself after a period of 2π, I would have blindly agreed with you.

But,but,but I agree a person is free to choose any thing. But if you plot a graph of the x u have, firstly under -π,π then to 0, 2π, you would see the difference! —Preceding unsigned comment added by Mathsislife (talkcontribs) 09:49, 8 January 2010 (UTC)

How about trying it with π/2 and power 5/2. If you take the square root first then you get π/4 and then the fifth power gives 5π/4. However if you take the 5th power first you get 5π/2 which is the same angle as π/2, then taking the square root gives π/4. Dmcq (talk) 20:43, 8 January 2010 (UTC)
Dmcq, I'd like to know where you learnt your "maths" you speak of. First off, √(1/2) = √(2/4) = √2/2; (√2/2)^5 = (√2)^5/2^5 = 4√2/32 = √2/8. Ergo: (√(π/2))^5 = (√(2π))^5/2^5 = (2π)^2√(2π)/32 = (π^2)√2π/8. (Sorry it's so long winded, I wouldn't want you to get lost somewhere, like you did in any lesson with powers). Now, in reverse: √((π/2)^5) = √(π^5/32) = √(π^5)/√32 = (π^2)√π/4√2 = (π^2)√(2π)/8 —Preceding unsigned comment added by 82.8.237.217 (talk) 18:32, 15 January 2010 (UTC)
Furthermore, since I forgot partway the first error I noticed, just because cos and sin have periodicity of 2π, that doesn't mean that 5π=π in EVERY branch of mathematics. —Preceding unsigned comment added by 82.8.237.217 (talk) 18:38, 15 January 2010 (UTC)
I'm sorry but I really can't be bothered discussing this with you any longer, you need to discuss the problems you have with complex numbers with someone who's paid for the job or likes doing it. As I said before you need to cite a book giving your argument for your argument to be taken seriously on Wikipedia.. Square roots of complex numbers have two possible values and cube roots have three etc, they do not have unique values whereas de Moivre's formula gives a single unique value. Thus it can't be true. There simply is no more to it. The generalization gives the bit about that that the value it gives is one possible value. Dmcq (talk) 18:59, 15 January 2010 (UTC)

are you paid for Defending this Article? —Preceding unsigned comment added by Mathsislife (talkcontribs) 15:47, 21 January 2010 (UTC)

No, are you paid to make it wrong? Please go and learn a bit more maths or do some cleanup where maths expertise isn't required. Peoples own work is not suitable for inclusion in Wikipedia even if it is right if it isn't cited somewhere - never mind about if it is wrong. See WP:OWN about complaining about people who defend article if you wish to complain. Dmcq (talk) 17:28, 21 January 2010 (UTC)

I don't wish to complain to anyone, but What I want that the person wrong is you,not me! Please consult an expert, not consider yourself 1! —Preceding unsigned comment added by Mathsislife (talkcontribs) 06:39, 5 February 2010 (UTC)

If some editors disagree about an article and there seems no way for them to resolve the dispute on their own then they should follow the WP:DISPUTE process rather than continuing. I asked at the maths wikiproject for other people to look at it and they agreed with me. You still disagree and I see little prospect of agreement. I was pointing out to you the route to complain about me if you felt I was guarding the article from people correcting it. Please read the dispute policy. You can either ignore this article, raise an action yourself, or change the article and have me raise an action. Dmcq (talk) 09:05, 5 February 2010 (UTC)

## Unwarranted assumptions about whether "n" is odd or even

The section titled "De_Moivre's_formula#Formulas_for_cosine_and_sine_individually" gives a formula for cos(nx) that assumes "n" is even. Likewise, it gives a formula for sin(nx) that assumes "n" is odd. The correct formulae that don't make these unwarranted assumptions are here. I will fix.Anythingyouwant (talk) 05:10, 27 August 2011 (UTC)

As far as I can see the old formulae were correct and the new ones are also correct. I can't say I'm particularly happy with either. Neither are very obvious from the original formula. However the new one seems a bit worse to me as having a funny construction for generating just one or minus one and removing half the entries. I've put a new version at here which should make the correspondence more obvious. Dmcq (talk) 11:37, 27 August 2011 (UTC)
I've commented over there at List of trigononetric identities. I think it would be better to just add a sentence about the effect of the sine function. I'll try that in this article.Anythingyouwant (talk) 16:41, 27 August 2011 (UTC)

## Refernces, etc.

Allegedly, the Abramowitz and Stegun Handbook of Mathematical Functions was used to source this article. Yet, the source only contains 3 lines about De Moivre's formula. This material says that if x is restricted to a 360° range, then n can be any number; if the restriction is not made, then n must be taken as an integer. This does not agree with the Wikipedia article's presentation of the subject. For that reason I am encouraging editors to find more citations for this article. Thelema418 (talk) 17:16, 19 August 2012 (UTC)

## Abraham de Moivre's formula

This formula is named after Abraham de Moivre in which I assert that it should be properly titled de Moivre's formula (lowercase title) as opposed to De Moivre's formula as the formula is not named after Abraham De Moivre. Technical 13 (talk) 18:10, 19 May 2013 (UTC)

Here is a correctly capitalized sentence: "De Moivre's formula does not, in general, hold for non-integer powers." The title "De Moivre's formula" is correctly capitalized for the same reasons the sentence is correctly capitalized. On the other hand, the two capital "D"s bolded in the first sentence should be lower-cased. --JBL (talk) 18:36, 19 May 2013 (UTC)
The exception here is that fact that his name is "de Moivre" and it does not matter where it comes in the sentence since it is a noun. You wouldn't say "EBay's auctions are final." or "IPod's sales are in the multi-millions." It would still be "eBay's auctions are final." or "iPod's sales are in the multi-millions." respectively. So, considering it is a proper noun no different than the other two examples, I maintain that the article should be {{Lowercase title}} and that any usage of it would be properly represented as "de Moivre's formula does not, in general, hold for non-integer powers." Technical 13 (talk) 18:52, 19 May 2013 (UTC)
 Response to third opinion request: While there seems no dispute that "de Moivre" is the correct form when the name is used in the middle of a sentence, I don't think it's obvious that the same applies at the beginning of a sentence. According to WP:MOSCAPS, there are exceptions to the normal capitalisation rules for mathematical notation and for trademarks (such as eBay), but there is no such exception for personal names. While the particular example in the style guideline is of a given name, it would, to my mind, logically also apply to surnames, since they are also "personal names". In this instance, the guideline states that When such a name is the first word in a sentence ... the first letter of the personal name should be capitalized regardless of personal preference. A Google search shows that both forms of capitalisation are used in what appear to be reliable sources. For that matter, even the French WP article uses the capital D where there is a risk of ambiguity (to whit, the double "de"). So, while the MOSCAPS guidelines aren't mandatory, I see no reason to overrule them in this instance. Anaxial (talk) 20:58, 19 May 2013 (UTC)
WP:NAMECAPS supersedes personal names and specifies: "Proper names of specific places, persons, terms, etc. are capitalized in accordance with standard usage: Winston Churchill, John de Balliol, Wales, Tel Aviv, Three Great Gardens of Japan, etc." Based on that, I maintain that the article should use the {{Lowercase title}} template regardless of how it is linked to from other articles. Technical 13 (talk) 21:21, 19 May 2013 (UTC)
Okay, had to save to go find the diff... Just like this talk page has used since July 6, 2010. when NuclearWarfare added {{Lowercase title}} here. Technical 13 (talk) 21:25, 19 May 2013 (UTC)
I would maintain that "standard usage" in this case is to use the capital letter when the name appears at the beginning of a sentence (although not, of course, elsewhere), and see no statement that NAMECAPS "supersedes" the rest of MOSCAPS. Indeed, NAMECAPS doesn't mention the specific circumstance questioned here, while, elsewhere in the same article in which it appears, there is a specific statement to the effect that it doesn't apply in this situation. Although I obviously don't have the final word on this, IMO, the specific guideline would supersede the more general one. Otherwise, why have it? Anaxial (talk) 22:00, 19 May 2013 (UTC)
That is exactly my argument, the policy supersedes the general usage page. The only question here is whether or not the article should have the {{Lowercase title}} template exactly the same as this talk page has as was instituted by an administrator/oversighter/arbitrator. Technical 13 (talk) 22:29, 19 May 2013 (UTC)

Technical13: can you find any evidence anywhere of anyone treating the "de" in a French name the way you suggest at the beginning of a sentence? JBL (talk) 23:54, 19 May 2013 (UTC)

I'm not sure that we on the same page. It doesn't matter where it comes in the sentence. The article should be de Moivre's formula with the {{Lowercase title}} just like an administrator/oversighter/arbitrator set this discussion page almost three years ago. Technical 13 (talk) 23:59, 19 May 2013 (UTC)
That would be true if and only if it is standard to treat capitalization in French names in this way. Do you have any evidence at all to support the idea that this sort of nonstandard capitalization is used anywhere, by anyone? JBL (talk) 00:55, 20 May 2013 (UTC)
So, you want an WP:OTHERSTUFFEXISTS list?

Hey, there we go... Another mathematical thing exactly as in this instance that uses {{Lowercase title}} just like I think this article should... There is your standard to treat capitalization for French names that are used as proper names for mathematical formulas and theorems in this way. I hope this satisfies your request. Technical 13 (talk) 01:18, 20 May 2013 (UTC)

## de Moivre's formula is currently grossly incorrect

To see that De Moivre's formula fails for rational exponents, consider:
${\displaystyle (\cos(2\pi )+i\sin(2\pi ))^{1/4}=1}$
while
${\displaystyle \cos({\pi /2})+i\sin({\pi /2})=i}$
Paul August 22:29, 17 January 2015 (UTC)
The error in your proof was in saying the following was true for all n.
${\displaystyle \left(e^{ix}\right)^{n}=e^{inx}.}$
That is only in general true for integer n when applied to complex numbers. See failure of power and logarithm identities. Dmcq (talk) 23:36, 17 January 2015 (UTC)
@:Paul August, False; ${\displaystyle (\cos(2\pi )+i\sin(2\pi ))^{1/4}=1,-1,i}$ and ${\displaystyle -i}$ as ${\displaystyle (\cos(2\pi )+i\sin(2\pi ))^{1/4}=^{4}{\sqrt {(\cos(2\pi )+i\sin(2\pi ))}}}$ as such ${\displaystyle \cos({\pi /2})+i\sin({\pi /2})=i}$ is correct for 1 value of ${\displaystyle (\cos(2\pi )+i\sin(2\pi ))^{1/4}}$
By amending de moivre from ${\displaystyle (\cos(nx)+i\sin(nx))}$ to ${\displaystyle (\cos(nx+{\frac {2k\pi }{n}})+i\sin(nx+{\frac {2k\pi }{n}}))}$ where x ∈ ℝ, n ∈ ℂ, k ∈ ℤ such that the argument ${\displaystyle (nx+{\frac {2k\pi }{n}})}$ falls between -π and π; i.e, -π < argZπ where Z is a complex number such that ${\displaystyle Z=(\cos(nx+{\frac {2k\pi }{n}})+i\sin(nx+{\frac {2k\pi }{n}}))}$ This is rather simple mathematics so I fail to understand why it is that you find it so hard to comprehend and as such make erroneous and invalid conclusions.
I have provided a ISBN number above from a credible book that cites "de moivre's theorem" to be true for all powers of n where n ∈ ℚ and the subsection in the De Moivre's formula directly below 'Failure for non-integer powers, and generalization aka Roots of complex numbers correctly contradicts that de moivre's theorem fails for powers of n wheren ∉ ℤ as it is clearly true for all n; i.e, n ∈ ℂ(Sumandark8600 (talk) 23:52, 17 January 2015 (UTC))
@Dmcq, Again that is erroneous due to the refusal to use multi-valued functions "sigh" so much bad maths, so many great mathematicians rolling in their graves (Sumandark8600 (talk) 23:52, 17 January 2015 (UTC))
${\displaystyle (\cos(2\pi )+i\sin(2\pi ))^{1/4}=1,-1,i}$ and ${\displaystyle -i}$
But the left hand side is just ${\displaystyle 1^{\frac {1}{4}}}$ and any any positive real raised to a real power is a real number, from e.g. the formula
${\displaystyle a^{b}=e^{b\ln a}}$
so the left hand side is just 1. While as noted
${\displaystyle \cos({\pi /2})+i\sin({\pi /2})=i}$
--JohnBlackburnewordsdeeds 00:03, 18 January 2015 (UTC)
Only for 1 solution as ${\displaystyle 1^{1/4}=^{4}{\sqrt {1}}=1,-1,i,-i}$ ;i.e, ${\displaystyle a^{1/b}=^{b}{\sqrt {a}}}$ which has b solutions. I know my stuff, I am not an idiot (Sumandark8600 (talk) 00:10, 18 January 2015 (UTC))
The section on generalization is what you want if you're talking about multiple values. Even from the point of multiple values the equation
${\displaystyle \left(e^{ix}\right)^{n}=e^{inx}.}$
is wrong when used on complex numbers with n being a half. One side only gives one value whereas the other side gives two possible values. The basic point is that a set of possible values is not a particular value. The formula gives values that are equal. Dmcq (talk) 00:20, 18 January 2015 (UTC)
Following my earlier point that de moivre's theorem can be amended such that the argument ${\displaystyle (nx+{\frac {2k\pi }{n}})}$ falls between -π and π; i.e, -π < argZπ where Z is a complex number such that ${\displaystyle Z=(\cos(nx+{\frac {2k\pi }{n}})+i\sin(nx+{\frac {2k\pi }{n}}))}$ where x ∈ ℝ, n ∈ ℂ, k ∈ ℤ, this can be extended to exponents such that ${\displaystyle Z=e^{i(nx+{\frac {2k\pi }{n}})}}$ I thought that was an intuitive step that needn't be explained to you but obviously I was wrong in assuming that. This is and of itself one of the famous proofs that ii has infinite solutions all of which are real when given no restriction for the argument; i.e, not being limited to the principal argument. The fact that a set of values is not a particular value is beside the point, a multivalued function can be equal to multiple things without them being equal to each other by the nature of it being a function.(Sumandark8600 (talk) 00:36, 18 January 2015 (UTC))
Your amendment is a mess multiplying and dividing by n in the same expression. I think I know what you intended but it doesn't work the way you want. The fact that a set of values is not a particular value is the whole point. An expression with an equals sign means one is asserting the two sides are exactly equal, not that one can choose one side so they are equal. By your reasoning 5=0/0 is true. It isn't. Five is just one possible value of 0/0. One should be able to put the expression on each side into a computer, stick in the parameter, and it produces a value which is exactly the same for both sides. And by the way a multiple valued function is not considered a function in modern mathematics. Dmcq (talk) 00:59, 18 January 2015 (UTC)
5 is not a possible value of 0/0, the only possible values of 0/0 are ±∞. My amendment uses correct formatting and is not a mess, different parts of the expression are being multiplied or divided by n, not the entire thing (which would make it a mess). Just because f(x) = a, b, c, d that does not imply that a = b = c = d but only implies that f(x) can take multiple values. Your reasoning is illogical, have you even checked my reference yet? Anyway I'm in the UK where it is 01:15 so I'm off to bed for now, night. (Sumandark8600 (talk) 01:18, 18 January 2015 (UTC))
See Indeterminate form about 0/0. I don't have access to that Edexcel school textbook but the fact that you had to make up your own proof rather than it giving one might indicate something to you. If it says it is true for all n then I'm afraid it is wrong and you're getting some strange ideas about equality and functions in your effort to try and make sense of it. Dmcq (talk) 02:04, 18 January 2015 (UTC)

## This is not true

The "x" is NOT a complex number. It is the angle of the vector representing the actual complex number, which is "r*(cos(x)+i*sin(x)". An important part of De Moivres formula is "r" which is the length of the vector, because without "r" only complex numbers with numerical value of 1 are represented, and the formula holds for ANY complex number. 87.52.19.19 (talk) 19:07, 17 May 2018 (UTC)

This is a difficult one. The statement in the lead is actually true, sine and cosine have been extended so they are defined for complex numbers. However using complex numbers instead of reals in this context is unusual, and especially in elementary maths. De Moive certainly wouldn't have meant complex numbers. I think the mention of x possibly being a complex number should be removed from the lead and x just called an angle, and x possibley being complex should just be mentioned as an extension in the article. Dmcq (talk) 22:38, 17 May 2018 (UTC)
This would be ok with me. This only addresses one of the two issues IP raised; the other one (about adding r) is less interesting. --JBL (talk) 23:29, 17 May 2018 (UTC)
Good to hear both a clarification and an agreement. Must admit I didn't know of, nor consider the possibility of complex angles. I would still insist on adding "r", because it is both an essential part of De Moivres formula and an absolute necessity for the notifcation of complex numbers in polar form. 87.52.19.19 (talk) 17:28, 18 May 2018 (UTC)