Wikipedia:Reference desk/Archives/Mathematics/2017 November 25

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November 25

Square root of complex numbers does not work

I found that the square root of complex numbers do not work. The proof is given as proof by contradiction.

1. step 1. Calculate the sqrt( 1 + 0 i)
2. step 2. Notice that the point 1 + 0 i is on the unit circle (on the complex plane)
3. step 3. Calculate the square root via two paths simultaneously, the first path going anti-clockwise on the unit circle while the second path going clockwise on the unit circle.
4. step 4. The two paths will meet at -1 + 0i but the value of the square root will differ depending on which path you take
5. step 5. CONTRADICTION therefore the aquare root of complex number must be wrong.

Going anticlockwise around the unit circle I get this.

{{0 deg, 1.00}, {30 deg, 0.966 + 0.259 I}, {60 deg, 0.866 + 0.500 I}, {90 deg, 0.707 + 0.707 I}, {120 deg, 0.500 + 0.866 I}, {150 deg, 0.259 + 0.966 I}, {179 deg, 0.009 + 1.00 I}}

Going clockwise around the unit circle I get this.

{{0 deg, 1.00}, {-30 deg, 0.966 - 0.259 I}, {-60 deg, 0.866 - 0.500 I}, {-90 deg, 0.707 - 0.707 I}, {-120 deg, 0.500 - 0.866 I}, {-150 deg, 0.259 - 0.966 I}, {-179 deg, 0.009 - 1.00 I}} — Preceding unsigned comment added by 110.22.20.252 (talk) 05:05, 25 November 2017 (UTC) 110.22.20.252 (talk) 03:30, 25 November 2017 (UTC)

That's not really a question. But you seem a little confused on how to calculate square roots of complex numbers. You could try to take a look at Wikipedia's article on square roots and see if that clarifies your confusion. –Deacon Vorbis (carbon • videos) 03:37, 25 November 2017 (UTC)

You seem to have basically discovered branches in multivalued functions.--108.52.27.203 (talk) 04:59, 25 November 2017 (UTC)

Your question contains a heap of problems. For some reason you appear to be stepping around the twelve roots of unity (at 0°, 30°, 60°, ...), but when converting from polar form you then step around the twenty-four roots of unity. r cis θ = r(cosθ + isinθ), not r(cosθ/2 + isinθ/2). And how did you wind up with +/- 179°? See Floating-point arithmetic#Accuracy problems and Rounding#Rounding to integer (round towards zero vs. round to nearest). If your numerical calculation package told you that 1 - (3 · 3^-1) was nonzero (or that 3 · 3^-1 truncated to zero), would you believe that you found a proof that multiplicative inverse does not work? If you have an actual question here, if would help if you let us know what you are doing your calculations on. -- ToE 18:09, 25 November 2017 (UTC)

What are you talking about, I am not calculating the root of unity but sqrt(a_point_on_unit_circle). You can check my calculation yourself for 30 deg. Which is { deg , sqrt(cosθ + i sinθ) } and which calculates to { 30 deg , sqrt( cos 30 deg + i sin 30 deg ) } and calc to { 30 deg , sqrt( 0.866 + i 0.5 ) } and calc to {30 deg, 0.966 + 0.259 i }. I am trying to show that when you approach the point -1 + 0 i , the value of sqrt(-1 + 0 i) depends on which path you approach it from. 110.22.20.252 (talk) 06:28, 26 November 2017 (UTC)

@110.22.20.252: Your comment, to be blunt, is utterly lacking in rigor. I am unable to follow what you are trying to do; I have a hard time gathering what 30 degrees has to do with a square root of unity, or what you mean by "path you approach it from". You should read about the complex square root before making further comments here.--Jasper Deng (talk) 06:42, 26 November 2017 (UTC)

At second glance, in the comment immediately above, you appear to be trying to do a fixed point iteration of the (principal) square root function. Firstly, the complex numbers $e^{\pi i}$ and $e^{-\pi i}$ are the same, namely -1. Secondly, fixed point iteration of the principal square root is vehemently not the same as calculating a square root. The only fixed points of the principal square root function are 0 and 1 (the only solutions to ${\sqrt {x}}=x$) so repeatedly taking a square root is going to always converge towards one of those numbers or (possibly) not converge at all. Thirdly, as you (apparently) try to investigate the limit of the principal square root function at -1, the fact that you found that this limit doesn't exist is not a problem. That the principal complex square root is discontinuous at -1 is a consequence choosing a branch cut along the nonpositive real axis. There is absolutely nothing wrong here. Finally, when you reply, please be very explicit about what exactly you are doing, and don't use nonstandard notation like { , }.--Jasper Deng (talk) 07:07, 26 November 2017 (UTC)

Thank you for explaining your notation. You have not presented a proof that the "square root of complex numbers does not work", however you have discovered that the principal square root function is not everywhere continuous. I commend you for your discovery!

From Square root#Principal square root of a complex number:

The principal square root function is thus defined using the nonpositive real axis as a branch cut. The principal square root function is holomorphic everywhere except on the set of non-positive real numbers (on strictly negative reals it isn't even continuous).

So yes, we agree with you that the principal square root function is not continuous at -1. -- ToE 13:29, 26 November 2017 (UTC) Kudos to 108.52.27.203 for understanding the question from the start.

Proof of (ab)mod m = (a ( b mod m)) mod m

The proof I have found in the Internet said:
Let $b=m{\text{I}}_{b}+{\text{F}}_{b}$, where ${\text{I}}_{b}$ - integer part; ${\text{F}}_{b}$ - fractional part, ${\text{F}}_{b}<m$;
multiply by $a$:
$ab=ma{\text{I}}_{b}+a{\text{F}}_{b}$;
$\therefore a{\text{F}}_{b}=(ab)\mod m$;
$\therefore a(b\mod m)=(ab)\mod m$.
QED.

But why did they decide that $a{\text{F}}_{b}<m$??
Username160611000000 (talk) 12:43, 25 November 2017 (UTC)

It might not be, but why is that a problem? It is perfectly legitimate to say 18 ≡ 11 (mod 7), for example, even though 11 > 7. All we need to be able to say a ≡ b (mod m) is for a − b = km for integer k. Double sharp (talk) 13:34, 25 November 2017 (UTC)

• I don't understand. In Microsoft Calculator when I type (11 mod 7) I got single result 4. Username160611000000 (talk) 13:54, 25 November 2017 (UTC)

There seems to be some confusion about the notation here. To me, a = b mod m is a relation between the three numbers a, b and m defined as true when m divides b-a. On the other hand the notation (b mod m) as used here seem to be the remainder of b when divided by m. This is sometimes denoted b % m, which I prefer to avoid confusion with the other definition of 'mod'. The difference is illustrated by 10 = 4 mod 3 is true while 10 ≠ 4 % 3 = 1.

To answer the question, it seems like what they're claiming isn't true so the proof is incorrect. Specifically, it's false, in general, that a(b % m) = ab % m. For example 2(2 % 3) = 4 ≠ (2⋅2) % 3 = 1. --RDBury (talk) 13:46, 25 November 2017 (UTC)

That's not what the OP is saying, I think. I think he means [a(b % m)] % m = ab % m, which is indeed true in general. Double sharp (talk) 13:55, 25 November 2017 (UTC)

Though come to think of it, the proof given seems to assume the mathematical definition of mod, which obviously leaves the OP unsatisfied because he wants assurance that the result of mod is a remainder and is therefore less than the modulus. @OP: in number theory, modulo is defined as an equivalence relation. It is not that "11 mod 7" is an operation that tells you to divide 11 by 7 and take the remainder, as you and Microsoft Calculator have it. Instead mod 7 is an equivalence relation that holds between any two integers which are a multiple of seven apart. So {...25, 18, 11, 4, −3, −10, −17, ...} are all representatives of one equivalence class modulo 7, and it is just as legitimate to say 11 ≡ 4 (mod 7) as 11 ≡ 18 (mod 7), 11 ≡ −3 (mod 7), or 11 ≡ 777777781 (mod 7). Your supposed gap in the proof comes from your expectation that the result of modulo must be the remainder, so that you think it has to be less than the modulus. But that's not the standard definition of modulus in mathematics. Double sharp (talk) 14:11, 25 November 2017 (UTC)

I'm studing now RSA algorithm and found this relation here https://en.wikipedia.org/wiki/Modular_exponentiation#Memory-efficient_method . It seems true empirically, but I want to prove it.Username160611000000 (talk) 14:01, 25 November 2017 (UTC)

Your trouble with it seems to stem solely from problems with the definition of mod. When you use the equivalence relation definition, these statements practically prove themselves: mod m acts as an equivalence relation, so if b and b % m are congruent mod m (which they are by definition), then of course they act the same mod m under all operations. Double sharp (talk) 14:14, 25 November 2017 (UTC)

I have heard about modulo as a relation. But for RSA algorithm mod means 'divide and take reminder' in all programming languages. Ok, how then should I prove this relation $(ab)\mod m=(a(b\mod m))\mod m$ (where mod is defined as operation which gives a reminder)? Username160611000000 (talk) 14:24, 25 November 2017 (UTC)

You can do it the same way, since any numbers belonging to the same equivalence class mod m have the same remainder when divided by m. So instead of saying a * (F_b) ≡ ab (mod m), you let a * (F_b) % m = r where r < m (take the remainder). Then a * (F_b) = mq + r. Then ab = ma * (I_b) + a * (F_b) = m * (a * I_b + q) + r, so ab % m = r, same as a * F_b. Then the result follows. Double sharp (talk) 14:59, 25 November 2017 (UTC)

Trigonometric identities

Are there any trigonometric identities for angles $\alpha \beta ,{\frac {\alpha }{\beta }}$ ? יהודה שמחה ולדמן (talk) 21:50, 25 November 2017 (UTC)

Probably the closest you can come to this are multiple angle and fractional angle identities. See List of trigonometric identities#Multiple-angle formulae. --RDBury (talk) 06:28, 26 November 2017 (UTC)

You shouldn't expect there to be a simple formula for such things, because there is no geometric interpretation of multiplying one angle by another. In particular, the "units" are wrong: $2\alpha$ or $\alpha \pm \beta$ are still angular measures, but $\alpha ^{2}$ and $\alpha \beta$ would approximate the square degrees subtended by a spherical rectangle. This objection is weakened, of course, by the natural and dimensionless status of the radian, but note that as $\sin x^{2}$ is aperiodic (and even!) it cannot be expressed in terms of trigonometric functions of x (or any constant multiple thereof) alone. --Tardis (talk) 14:55, 26 November 2017 (UTC)

You can consider the following identity $e^{i\alpha \beta }=(e^{i\alpha })^{\beta }$ and use expressions for trigonometric functions through complex exponents. The resulting expressions will be infinite series. Ruslik_Zero 20:17, 26 November 2017 (UTC)

If one of the angles is an integer, there are List_of_trigonometric_identities#Multiple-angle_formulae such as De Moivre's formula and the sine, cosine and tangent of multiple angles. Inverse trigonometric functions have logarithmic forms, but they don't appear to lend themselves well to expansion with log(ab) and log(a/b) formulas.--Wikimedes (talk) 20:29, 26 November 2017 (UTC)

Any identities for ${\frac {\alpha }{n}},n\in \mathbb {Z}$ ? For example, What is $\cos \left({\frac {\alpha }{n}}\right)$ in terms of $\cos(\alpha )$ ? יהודה שמחה ולדמן (talk) 18:35, 28 November 2017 (UTC)

Still here. יהודה שמחה ולדמן (talk) 10:15, 29 November 2017 (UTC)

There is no simple (easy-to-use) general formula, even of $\cos {\frac {\alpha }{3}}$, nor of $\sin {\frac {\alpha }{3}}$. However, if you don't care about usefulness, then you can write down the following formula: $\cos {\frac {\alpha }{n}}={\frac {{\sqrt[{n}]{\cos \alpha +{\sqrt {-1}}\sin \alpha }}+{\sqrt[{n}]{\cos \alpha -{\sqrt {-1}}\sin \alpha }}}{2}}$. Be careful with it, because the "=" sign here should not be interpreted as usual, just as it should not be interpreted as usual in the formula $3={\sqrt[{2}]{9}}$ , because for every n greater than 1 - there is more than one root of n-th order. Hope this helps. HOTmag (talk) 13:20, 29 November 2017 (UTC)

The truth is, I wanted to solve

${\begin{aligned}x&={\sqrt[{3}]{p+qi}}+{\sqrt[{3}]{p-qi}}\\&=2{\sqrt[{3}]{p^{2}+q^{2}}}\cos \left({\frac {\arctan \left({\frac {q}{p}}\right)+2\pi k}{3}}\right)\ ,\ k\in \{0,1,2\}\end{aligned}}$

just by using trig identities and inverse functions to get the argument. From this I understand it cannot be done. יהודה שמחה ולדמן (talk) 15:39, 29 November 2017 (UTC)

Please notice, that - in many (non-trivial) cases - the formula you've mentioned can still help you. For example:

For $k=1$: ${\sqrt[{3}]{1+i}}+{\sqrt[{3}]{1-i}}=2{\sqrt[{3}]{2}}\cos {\frac {\arctan 1}{3}}=2{\sqrt[{3}]{2}}\cos {\frac {{\frac {\pi }{4}}+2\pi }{3}}=2{\sqrt[{3}]{2}}\left(-{\frac {\sqrt {2}}{2}}\right)=-{\sqrt[{6}]{32}}$.

For $k=0$: ${\sqrt[{3}]{1+i}}+{\sqrt[{3}]{1-i}}=2{\sqrt[{3}]{2}}\cos {\frac {\arctan 1}{3}}=2{\sqrt[{3}]{2}}\cos {\frac {\left({\frac {\pi }{4}}\right)}{3}}=2{\sqrt[{3}]{2}}\cos {\frac {\left({\frac {\pi }{6}}\right)}{2}}=2{\sqrt[{3}]{2}}\left({\frac {{\sqrt {2}}+{\sqrt {6}}}{4}}\right)={\frac {1+{\sqrt {3}}}{\sqrt[{6}]{2}}}$.

For $k=2$: ${\sqrt[{3}]{1+i}}+{\sqrt[{3}]{1-i}}=2{\sqrt[{3}]{2}}\cos {\frac {\arctan 1}{3}}=2{\sqrt[{3}]{2}}\cos {\frac {{\frac {\pi }{4}}+4\pi }{3}}=2{\sqrt[{3}]{2}}\cos {\frac {\left({\frac {17\pi }{6}}\right)}{2}}=2{\sqrt[{3}]{2}}\left({\frac {{\sqrt {2}}-{\sqrt {6}}}{4}}\right)={\frac {1-{\sqrt {3}}}{\sqrt[{6}]{2}}}$.

Additionally, please notice that our article Trigonometric constants expressed in real radicals, gives a list of cosines (or sines) of all angles of the form ${\frac {2\pi }{n}}$ for every integer $n$ up to $n=24$ (of course the list could be continued for many other integers), so the formula you've mentioned is pretty useful in many other non-trivial cases, besides the (non-trivial) case I have already shown above as an example.

HOTmag (talk) 17:12, 29 November 2017 (UTC)