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Пікірлер
@jimbo1-c 40 минут бұрын
There's only one method here. The quadratic formula comes from completing the square (the second method).
@alexandermorozov2248 Сағат бұрын
Is there a solution to the equation 1^z = 2i ? When there is a one on the left modulo, and a two on the right?
@luciboi8182 2 сағат бұрын
4:04 wait what? How can you assume that a and b are whole numbers? They could be any real numbers right? 4 and 1 arent the only possibilities Edit: Riight i get it it could be anything. Any 2 numbers that multiply to 4 can satisfy the equation, and so any of the pairs would work.
@Qermaq 4 сағат бұрын
e is making me pie-eyed. ;)
@aplusbi 2 сағат бұрын
😄
@elfeiin 5 сағат бұрын
Haha that's what I got too. I thought that couldn't be it, but you made a good point about roots having multiple answers. I forgot about that.
@trojanleo123 5 сағат бұрын
z = (1+4k)/4n where n,k ∈ ℤ
@Tletna 6 сағат бұрын
I am always wondering what justification people use when they "complexify" numbers or values. Your video isn't necessarily wrong, but it just feels like it is missing something. Also, some 3d graphs showing 1^x = i. That might've looked cool.
@alexandermorozov2248 10 сағат бұрын
Bravo! ❤️👍
@aplusbi 3 сағат бұрын
Thanks
@cav1928 10 сағат бұрын
An interesting question is if n and k should be different, because if you are working with an equality, both numbers (k and n) must be in the same "lap". So only solutions with k=n would be valid. It is only a guess, not a firm statement.
@arekkrolak6320 11 сағат бұрын
The amount of nonsense in this video is overwhelming :) ln1 is not always zero beats all
@user-kr7wk6ex4c 12 сағат бұрын
I think the issue with such problems is that generalising the base introduces extraneous solutions. I think I solved the problem using limt theory instead. I rewrote the problem as lim w^z = i, as w tend to 1, and found the limit of inf*i and -inf*i, in otherwords plus/minus complex infinity. What do think?
@thomaslangbein297 14 сағат бұрын
Since you obviously changed to the dark side of mathematics, what I call the Flatearthers of mathematics: 0^0=1. 1^z=2 has a solution. f(f(x))=e^x exists for f:R->R, I cancelled the abo of your channel. Too sad!
@MrGeorge1896 14 сағат бұрын
I thought of the second method, too, but was not sure if it would work because of | i | not being less than one i.e. the sum does not converge.
@XJWill1 15 сағат бұрын
I don't know why you would think the WA answer is surprising, unless you meant it was a surprise that you mentioned you would cover it and then you failed to show the WA answer. As I have mentioned in many comments to your videos, WA always assumes the principal value of any multi-valued components that appear in equations. This is the only sensible way to handle it, since otherwise you end up with only 25% of a solution as in this problem.
@dorkmania 16 сағат бұрын
Since in i^(4n) = 1, the equation becomes (i^4n)^z = i i^((4n)z) = (i^1)(i^4k) i^((4n)z) = i^(4k + 1) Comparing powers (4n)z = 4k +1 Or z = (4k +1)/4n With k = 0, n = 1 z = 1/4
@thomaslangbein297 17 сағат бұрын
If you define exponentiation as a function 1^(1/4) is definitely 1. A function can never be multivalued. You are confusing it with the equation z^4=1, which of course has 4 solutions. Have a look at De Moivre. I think we had the discussion a long time ago. (e^(2*pi*i))^z = cos(z * e^(2*pi*i)) + i * sin(z * e^(2*pi*i)) only if z is an integer. Else it is (cos(e^(2*pi*i)) + i*sin(e^(2*pi*i)))^z = 1^z. You are caught in a vicious circle. 1 to any complex number is 1. That‘s what in your words believe some. These some are those who know proper mathematics. This topic has become the equivalent of the Flatearthers in the mathematical world. Learn your stuff before you delude poor adepts. If you define exponentiation as an operator this will be another story. Much too ”complex“ for this channel, I am afraid.
@surelyred 15 сағат бұрын
“If you define exponentiation as a function 1^(1/4) is definitely 1. A function can never be multivalued.” Firstly, where in this video did aplusbi define exponentiation as a function? This will be relevant later, keep this in mind. Indeed, the narrow definition of a function specifies that one input has at most one output, but modern mathematical terminology allows for “multi-valued functions”. Perhaps you prefer “multi-valued relations”, but even in math the terminology changes over time. Function is just a more familiar way to say relation, especially when used with "multi-valued", as then it becomes unambiguous. When using strictly the principal-valued function of exponentiation, it’s correct that 1^(1/4) equals 1. "You are confusing it [i.e., 1^(1/4)-surelyred] with the equation z^4=1, which of course has 4 solutions.” As it's sometimes used on this channel, the notation in 1^(1/4) is basically shorthand for the solutions for z in the latter, it seems. "[a few sentences of math-surelyred] You are caught in a vicious circle." What's the "vicious circle" you speak of? “1 to any complex number is 1. That’s what in your words believe some. These some are those who know proper mathematics.” Is this quote also true when exponentiation is defined as an operation (see my note on the final quote)? And what is "proper mathematics"? And in the past was it ever the case that complex numbers were not part of "proper mathematics"? Additionally, your "proper mathematics" sounds dogmatic if it excludes the possibility of notations' meanings changing over time, and if at the same time "proper mathematics" is to be necessarily upheld. “This topic has become the equivalent of the Flatearthers in the mathematical world.” I do not see the similarity between this matter and the flat-earther debate: One deals chiefly with somewhat-arbitrary notation; The other regards the natural world, about which we can gain information using observations such as measurements, at least insofar as we uphold the principles of natural science and philosophical materialism each. “Learn your stuff before you delude poor adepts. [paragraph break-surelyred] If we define exponentiation as an operator this will be another story. Much too "complex" for this channel, I am afraid." When did aplusbi say that exponentiation wasn't used as an operator here? Why might aplusbi have deserved your polemic against him, if up until this point your comment was in the case of exponentiation being regarded as a "function"? Thanks, surelyred
@thomaslangbein297 15 сағат бұрын
You just repeated what I said. You are a smart alec.
@surelyred 15 сағат бұрын
@@thomaslangbein297 I didn’t “just” repeat what you said-kindly read the message again, I asked several questions
@aplusbi 14 сағат бұрын
Thank you! 🥰
@surelyred 17 сағат бұрын
1:58 I like the word “complexify”. It sounds better than “complicate” haha. Also, the way I see it, 1^(1/4) has 4 values, but I think 1 is the principal-value to that exponential. en.m.wikipedia.org/wiki/Principal_value This Wikipedia page says we can use the notation: “ pv 1^(1/4) “ if we really want to clarify that we just wanted the principal-value, 1 answer, namely 1. I remember watching a video by the youtuber “BlackPenRedPen” that was about these types of confusing exponential, but sadly I can’t find it anywhere.
@scottleung9587 17 сағат бұрын
I got almost the same answer for z, except it was (4n+1)/4. How do you know if you need one or two integers?
@Mediterranean81 18 сағат бұрын
e^2Pi*iz = e^iPi/2 2Pi*z = Pi/2 2z = 1/2 z = 1/4
@SweetSorrow777 19 сағат бұрын
Complexifying 1 and writing "i" in polar form seems to be the way to go.
@SweetSorrow777 19 сағат бұрын
1 as e^(2*PI*n*i)
@ACheateryearsago 18 сағат бұрын
Shouldn't that be equal to -1 instead of 1? I could be mistaken though@@SweetSorrow777
@NadiehFan 22 сағат бұрын
It's a shame that you often don't finish your solutions, saying that the video would otherwise be too long. That is not so, there are many 15 to 20 minute math videos on youtube and some are even (much) longer. Completing your first method would have been instructive because the system a⁴ − 6a²b² + b⁴ = −7 a³b − ab³ = 6 has 16 solution pairs (a, b) but only four of these are real, and of course a and b are assumed to be real. Although this is a homogeneous system, setting b = at and solving for t does not seem to be the most efficient approach. You can rewrite this system as (a² − b²)² − 4a²b² = −7 ab(a² − b²) = 6 Now let a² − b² = d ab = p then we have d² − 4p² = −7 pd = 6 and substituting d = 6/p from the second of these equations into the first we get (6/p)² − 4p² = −7 4p⁴ − 7p² − 36 = 0 4p⁴ + 9p² − 16p² − 36 = 0 p²(4p² + 9) − 4(4p² + 9) = 0 (p² − 4)(4p² + 9) = 0 Since a and b must be real, p = ab must be real, so we only need to consider the real solutions p = 2 and p = −2 of this quartic equation in p. With p = 2 we have d = 3 and we get the system a² − b² = 3, ab = 2 which has the real solution pairs (a, b) = (2, 1) and (a, b) = (−2, −1). With p = −2 we have d = −3 and we get the system a² − b² = −3, ab = −2 which has the real solution pairs (a, b) = (−1, 2) and (a, b) = (1, −2). So, there are four complex numbers which have a fourth power −7 + 24i and these complex numbers are 2 + i, −2 − i, −1 + 2i, 1 − 2i. Of course, since i⁴ = 1, all these four complex numbers can be obtained by multiplying any one of these four numbers by consecutive powers of i. But there is only one of these four numbers which can be considered as _the_ fourth root of −7 + 24i, that is, its principal root, and that is 2 + i. Indeed, if you enter ∜(−7 + 24i) in WolframAlpha, it will only give you a single answer, which is 2 + i.
@aplusbi 3 сағат бұрын
I agree and this is a very nice approach.
@yc4742 Күн бұрын
I don't understand why you introduce ln, it's so convoluted. Starting from e^(z pi)=2z, if you divide by e^(z pi) you get 1 = 2z e^(-z pi). Then you multiply by -pi/2 and get -pi/2 = -z pi e^(-z pi). From here you can apply the Lambert W function.
@mcwulf25 Күн бұрын
I went straight for #2. Why ln both sides if you are going to raise e to the powers?
@XJWill1 Күн бұрын
If you made a comment on this post, you may want to logout and see if it still shows up. There seems to be a shadowbanned comment, since before posting this YT was reporting "5 Comments" but I only counted 4 being displayed.
Күн бұрын
Ok
@PseudoPig Күн бұрын
I would like to try to continue the second method. So, z=a+bi If we subtract the equations, we get -z* (I mean conjugate) +z=8i; -a+bi+a+bi=8i; 2bi=8i; b=4; If we add the equations, 3z* +3z +2|z|=28; 3a-3bi+3a+3bi+2√(a^2+b^2)=28 3a+√(a^2+16)=14 By Pythagorean triple (3;4;5), 3^2+4^2=5^2, hence it is kinda logical to input 3 for a, and it works. Hence, z=3+4i
@XJWill1 Күн бұрын
There are an infinite number of solutions. The simplest are i/2 and -i/2 . To visualize the solutions in the complex plane, plot the following two equations, and look at the intersection points of the family of curves. x^2 + y^2 = 1/4 * exp(x * 2*pi) atan2(y , x) = atan2( sin(y*pi) , cos(y*pi) ) If you are using desmos, note that atan2(y , x) is entered as arctan( y , x )
@XJWill1 Күн бұрын
One interesting thing that is obvious from the graph, but is more difficult to recognize from the equation(s) is that there are no solutions that have negative real parts.
@XJWill1 Күн бұрын
If your graphing program cannot handle atan2(y , x), here is another set of equations you can use. x = - 1/pi * ln( 1/2 * sin(y*pi) / y ) cos(y*pi) = 2*x*exp(- x*pi) These families of curves are slightly more complicated than the ones with atan2(y , x), so I suggest using the previous equations if you can.
@Etienne-pq3dx Күн бұрын
W(🍵. e^(🍵)) =🍵 E Z 2 B 🍵 I love the second method, very smart 😉
@barberickarc3460 Күн бұрын
did this with the formula for lambert W and the property that W(-pi/2)= i* pi/2. z=-i/2
@mcwulf25 Күн бұрын
Plus an infinity of other solutions of course
@miguelcerna7406 2 күн бұрын
1. sinz = i*(1-cosz) 2. Multiply both sides by i ==> i*sinz = cosz -1 3. Subtract i*sinz and add 1 ==> cosz-isinz = 1 4. e^(-iz) = 1 5. -iz = ln(1) NO solution
@MMSaths 2 күн бұрын
I love your videos!
@aplusbi Күн бұрын
Glad to hear that!
@0over0 2 күн бұрын
Isn't this a special case of one we did a while ago, based on 2n/n, where n=7?
@aplusbi Күн бұрын
I think so
@SidneiMV 2 күн бұрын
i⁰iⁿ + i¹iⁿ + i²iⁿ + i³iⁿ = iⁿ(1 + i - 1 - i) = 0 (i + i - 1 - i) = C = 0 i⁰C + i⁴C + i⁸C + ... + i²⁴C + i²⁸ = i²⁸ i⁰C + i⁴C + i⁸C + i¹²C - i¹⁵ = -i¹⁵ i²⁸/(-i¹⁵) = -i¹³ = -i¹²i = *-i*
@phill3986 2 күн бұрын
😎🔥👍✌️😎🔥👍✌️
@scottleung9587 2 күн бұрын
I used the first method.
@phill3986 2 күн бұрын
😁✌️👍🔥😁✌️👍🔥
@vzaimo 2 күн бұрын
Hi, Sybermath! Thank you for video! Here's an equation from Russian bear: n = 1+2i-3-4i+...+n*i^(n-1). Could you solve, please? 🐻
@0over0 2 күн бұрын
Your nth term is +n*i^(n-1). ie, involves a power of i. Shouldn't it be something like [ +/- (2n-1) -/+ (2ni) ] ? (no powers of i)
@SidneiMV 2 күн бұрын
S = 1 + 2i - 3 - 4i +5 + 6i - 7 - 8i + 9 + 10i - 11 - 12i + 13 + 14i - 15 - 16i ... Sᵣ = 1 - 3 + 5 - 7 + 9 - 11 + 13 - 15 .. Sᵢ = i(2 - 4 + 6 - 8 + 10 - 12 + 14 - 16 ..) S = Sᵣ + Sᵢ Sᵢ = 2i(1 - 2 + 3 - 4 + 5 - 6 ..) Sᵢ = 2i/4 => Sᵢ = i/2 Sᵣ = 1 - 3 + 5 - 7 + 9 - 11 .. = = (1 - 3 + 5 - 7 + 9 - 11 ..) + (1 - 1 + 1 - 1 + 1 - 1 ..) - 1/2 = = (2 - 4 + 6 - 8 + 10 - 12 ..) - 1/2 = = 1/2 - 1/2 = 0 Sᵣ = 0 S = Sᵣ + Sᵢ *S = 0 + i/2*
@koennako2195 3 күн бұрын
you could also set z=a+bi, substitute, and go from there, but it's pretty messy.
@JamesDuemmel 3 күн бұрын
The answer can restated as |z-1/3|=2/3.
@JefiKnight 3 күн бұрын
1 will work! (I couldn't help guessing and checking.)
@scottleung9587 3 күн бұрын
Nice - I'm not very good at locus problems, but they're sure fun to do! (BTW z=1 would work if you plug it into the original equation.)
@aplusbi Күн бұрын
Thanks for the tip!
@damiennortier8942 3 күн бұрын
±z ±1 = ± 2z ± z ± 2z = ± 1 ± 3z = ± 1 or ± z = ± 1 z = {-⅓, -1, ⅓, 1}
@oenrn 3 күн бұрын
Not how absolute values work with complex numbers.
@scottleung9587 3 күн бұрын
I used the first method - it wasn't that bad for me, since I knew I had expressions for both a and b in terms of c and d. From there, I got z=7-i and w=2-i.
@bobbyheffley4955 3 күн бұрын
This is a circle centered at (1/3,0) with a radius of 2/3.
@phill3986 3 күн бұрын
🔥👍✌️😎😎✌️👍🔥
@FisicTrapella 3 күн бұрын
(2a)🤗
@emmanuelbrittof.4376 3 күн бұрын
I would calculate the modulus and the angle of the expression and then raise it to 1/4. What do u think as a 3rd method?
@bobkurland186 3 күн бұрын
I used 2b (before I saw your equation) and got mystychief's answer.
@mystychief 3 күн бұрын
W = 2 - i and Z = 7 - i .
@Rai_Te 4 күн бұрын
Before going into the hell shown in this video, I use the relatively easy solution via polar coordinates.
@KipIngram 4 күн бұрын
Well, first of all note that this is not a fourth root - it's an eighth root problem. The square root has to be dealt with too. We can write it like this: (-7 + 24*i)^(1/8) and if you convert that to polar form then it becomes trivial: (25 angle(106.25))^(1/8) 5^(1/4) angle(13.2825 + k*45), k = 0..7