how is e^e^x=1 solvable??

Рет қаралды 627,336

8 ай бұрын

The exponential equation e^x=0 has no solutions, not even in the complex world, but e^e^x=1 does have solutions! I was surprised to see how WolframAlpha actually gave the solutions to this seemingly impossible equation and I would like to show you how to solve it! The trick is to write 1 in the complex polar form. Subscribe to ‪@blackpenredpen‬ for more math for fun videos. #math #complexnumbers #blackpenredpen #fun #tutorials
🛍 Euler's Identity e^(iπ)+1=0 t-shirt: amzn.to/427Seae
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Thank you all!

Пікірлер: 728

@blackpenredpen 8 ай бұрын

Can 1^x=2? Solution here: kzfaq.info/get/bejne/b916bLxz3K68d30.htmlsi=dx-XGv_Wf3_0VDH2 🛍 Euler's Identity e^(iπ)+1=0 t-shirt: amzn.to/427Seae

@ISoldBinLadensViagraOnEbay 8 ай бұрын

Can you solve (e^i)+(i^e) pls

@MATHMASTERPRO 8 ай бұрын

@@ISoldBinLadensViagraOnEbay where the equation?

@ISoldBinLadensViagraOnEbay 8 ай бұрын

@@MATHMASTERPRO I am asking for the solution of the answer, this is not an equation, since e and i are numbers

@MATHMASTERPRO 8 ай бұрын

@@ISoldBinLadensViagraOnEbay (e^i)+(i^e)=cos(1)+cos(i*pi/2)+i(sin(1)+sin(i*pi/2))

@xinpingdonohoe3978 8 ай бұрын

@@ISoldBinLadensViagraOnEbay cos(1)+cos(πe/2)+i sin(1)+i sin(πe/2) It won't really contract into any nice form.

@PauloChacal 8 ай бұрын

Number “1” will never be the same after your explanation

@doodle1726 8 ай бұрын

Real true man now i can't get it out of my head arrrrghh😱😱🤯

@onesecondbaobab 8 ай бұрын

I giggled out loud xD

@Hardcore_Remixer 8 ай бұрын

You should watch Animation vs Math. The fun begins when it comes to -1 being replaced by e^(i(pi)) which has the same value as -1 😂

@threeuniquefingers 8 ай бұрын

Ahh I gave this comment the 1000th like…the transition from 999 to 1k seemed surreal

@Questiala124 7 ай бұрын

Instead of saying “I am number one” say “ I am number e^(2i(Pi)n)”.

@MathFromAlphaToOmega 8 ай бұрын

There's a theorem (called Picard's little theorem) that says that any non-constant holomorphic function on the whole complex plane can miss at most one value. Since e^e^x definitely can't equal 0, it must hit 1 somewhere. :D

@pluieuwu 8 ай бұрын

thats so cool!!!

@Isvakk 8 ай бұрын

Are you sure it's holomorphic?

@MathFromAlphaToOmega 8 ай бұрын

Yes, it's infinitely differentiable everywhere.

@jakobullmann7586 8 ай бұрын

@@Isvakk It’s a composition of holomorphic functions, this holomorphic itself.

@blackpenredpen 8 ай бұрын

Wow! This is super cool!!

@DanoshTech 5 ай бұрын

I love how passionate he is he makes math seem cool and interesting

@alexeynezhdanov2362 2 ай бұрын

Because it is - cool and interesting.

@DanoshTech 2 ай бұрын

@@alexeynezhdanov2362 some aspects aren't at the moment I am doing 'math methods' and 'specialist' math' its Australian senior math and essentially it is calculus, geometry, algebra the top senior maths that is done in America and we just covered permutations and combinations and damn they are boring

@haashir7312 8 ай бұрын

I graduated six years ago and I used to watch your videos back then, just stumbled across this now and it takes me back. Thank you for making these videos with so much enthusiasm.

@acykablyatley 8 ай бұрын

i agree it is very straightforward to see now that e^2ipi = 1 and e^ln(2ipi) = 2ipi so that e^e^ln(2ipi) = 1 is true. but only unintuitive because it is easy to forget that e^ix is periodic in the complex plane...

@seja098 6 ай бұрын

lol, you definitely would have never guessed this answer, be humble.

@_cran 4 ай бұрын

It's not really easy to forget there are so many proofs and techniques over moivre formula 😭 I'm pretty sure you learn it in calc lessons and the start of complex analysis starts with it since you use it to solve everything in that whole class. If anyone reading who doesn't know what's moivre formula is e^ix=cosx+isinx

@CertifiedOrc 4 ай бұрын

@@seja098when you're not specified whether the solution is purely real, it is instinctive to check both in the real and complex plane, especially for someone studying higher grade maths, also don't talk shit to people you don't know

@acm-gs6bl 2 ай бұрын

i like your funny words magic man

@acykablyatley 2 ай бұрын

@@seja098 i did not say that i guessed that answer, and my comment clearly says it was easy to see /after/ watching the video.

@Sg190th 8 ай бұрын

It's nice seeing the complex world being used more.

@okWishFull 8 ай бұрын

I love your excitement! Loved every moment of this!!!

@stephanelem822 8 ай бұрын

Each time I watch one of your video, I discover one more time, the set of constraints I used to know to solve an equation is largely incomplete. I've no idea to discover without Wolfram I'd be wrong.

@ayssinaattori9313 8 ай бұрын

Thank you gor this video! I've been thinking about complex exponents recently and this was something really interesting I hadn't thought of before.

@dentonyoung4314 8 ай бұрын

Wow. That was an amazing explanation.

@dougdimmedome5552 8 ай бұрын

The greatest thing about complex analysis is slowly overtime making more insane infinite expressions to approximate 1.

@camelloy 8 ай бұрын

I want to make this very clear this video rescued my desire to learn math. It gave me the first visualization of what e^ipi was instead of rote memorization that drove me up a wall. I actually understand what the complex plain is after being told repeatedly by my professor not to bother looking into it. Can’t wait to dive in further.

@leoniekrenzer7716 8 ай бұрын

This is really cool! Also a great way to show why we can't just hit complex functions with the logarithm (since the exponential function is not injective on the complex plane)

@Josp101 7 ай бұрын

Wow okay so this is the reason why the simple approach misses solutions!

@dethfr491 4 ай бұрын

That's why there is term called "principal logarithm" of complex numbers .

@timothyrosenvall1496 8 ай бұрын

I’ve worked with an equation in the past that seemed to reduce to e^x = 0. I wondered if x was always just undefined but I have the vaguest memory of reducing it from e^e^x = 1. This is a phenomenal result

@algorithminc.8850 8 ай бұрын

There are many great bits on this channel ... but I really loved this one. Still chuckling ... sincere thanks. Cheers

@garywalker6216 8 ай бұрын

Love your videos!

@ddsqadod2994 8 ай бұрын

The way a complex solution appeared out of nowhere is by avoiding the e^nothing=0, and instead to find a certain number that's equivalent to 0 when on the power of e. Which, leads to a fact that e^2cπi = e^0 = 1, works for every integer c.

@robsmith9696 7 ай бұрын

For anyone missing why the 1 was added back in on the fourth line, it’s because the 1 is still there and multiplied in. When you take the ln() of both sides, ln(1) shows up and gives you the initial issue.

@gheffz 8 ай бұрын

Love it! Thank you.

@blackpenredpen 8 ай бұрын

So glad!

@johncirillo9544 8 ай бұрын

This made me smile! 😊

@isilverboy 8 ай бұрын

@3:00 instead of convert 1, I would prefer change i into e^i(pi/2+2pi c2). In this way you do not have the log of a complex number in the solution.

@xinpingdonohoe3978 8 ай бұрын

If you rewrite a complex number, you still have a complex number, only written differently. Don't be a coward; the log of a complex number is real man business.

@isilverboy 8 ай бұрын

@@xinpingdonohoe3978I have no problems with logs of complex numbers, but imho they still need to be simplified: the log of a complex number can be further simplified by using ln(i) = i(pi/2+2 pi c2).

@vadimpetruhanov4150 8 ай бұрын

Log of a complex number is many-valued function, therefore it is preferable not to use it when it is possible

@XJWill1 8 ай бұрын

The "method" of solving complex-valued equations by randomly converting constants to exp(i*something) is not a reliable way to do it. It may work on some simpler equations, but it will fail on other equations. A more reliable way is to use the multi-valued complex natural logarithm function, which is written log() in complex analysis. exp(exp(x)) = 1 log(exp(exp(x)) = log(1) exp(x) + i*j*2*pi = 0 + i*k*2*pi where j, k are any integer, this is because log() is multi-valued exp(x) = i*m*2*pi where m is any integer log(exp(x)) = log(i*m*2*pi) x = log(i*m*2*pi) + i*n*2*pi where m and n are any integer

@rainerzufall42 2 ай бұрын

Agreed! See above: x = i π (4 c_2 + 1) / 2 + ln(2 π c_1) and c_1 !=0 and c_1 element Z and c_2 element Z Clear real and imaginary parts, no complex log, just real ln!

@bendeguz_ 8 ай бұрын

loved the video!

@user-zg9mo8oi2m 4 ай бұрын

That was beautiful thank you

@nicklanders5178 8 ай бұрын

Fascinated by the way you hold and switch between markers

@michaelz2270 8 ай бұрын

You can do this systematically. Let w = e^z. Then you are solving e^w = 1, solved by w = 2pi i n for an integer n. So you wish to solve e^z = 2pi i n. For n > 0 one has 2 pi i n = e^(i pi /2 + ln 2pi n). Then e^z = e^(i pi /2 + ln 2pi n) is solved by z = i pi /2 + ln (2pi n) + 2pi i m for integers m. This works for n < 0 too if you replace ln (2pi n) by pi i + ln (2pi |n|). Stated in terms of the multivalued logarithm, these are log(log(1)).

@TheDoh007 8 ай бұрын

I simply brute-forced my way to e^(e^(0.5*pi*i+1.8378770664093455)) lmao

@vivianriver6450 8 ай бұрын

If I'm not mistaken, the reason that e^x = 0 has no solution, but e^(e^x) = 1 has a set of complex solutions is because e^x is periodic with period 2*pi*i, but ln(x) is computed using *only one* period. It's similar to how x^2 = 1 has *two* solutions, but sqrt(1) is always evaluated to 1.

@vibaj16 8 ай бұрын

I don't think that's the full explanation. sqrt(1) is always evaluated to 1 because that's just how sqrt(x) is defined: the positive number that when squared equals x.

@vivianriver6450 8 ай бұрын

@@vibaj16 x^2 = 1 has two solutions, but *one* of those solutions vanishes when you take the square root of both sides of the equation. Likewise, e^(e^x) = 1 has a family of solutions that disappear when you take the log of both sides of the equation because ln(x) evaluates to only *one* value, even tho e^(ln(x) + 2*pi*i) also evaluates to x.

@kazedcat 8 ай бұрын

@@vivianriver6450It's a mathematical trick called equivalence classes. The solutions do not disappear you are just picking one value that represent and infinite set of solutions. The reason this is needed is because functions by definition must have unique mapping.

@mtaur4113 8 ай бұрын

ln(2m pi i) is also multivalued and could be broken down into more elementary parts. I suppose the definite integral of 1/z from 1 to 2m pi i works, but it's not something you can plug into a standard scientific calculator, or learned about in most algebra or Calc 1 classes.

@donwald3436 8 ай бұрын

It's 2am why am I watching this lol.

@deathmight2uuotba987 7 ай бұрын

Same bro

@michaelbaum6796 7 ай бұрын

Really crazy🙈- great👍

@arsalmathacademy 8 ай бұрын

Great job great tricks he has Thanks Sir

@et427gamer9 8 ай бұрын

I understood very little of this as a high school student but it was very enjoyable

@WhosBean 5 ай бұрын

This comes about because when you are using the analytical continuation of the ln function all outputs have a +2pi*n at the end. For example ln(e^2) = 2+2pi*n. So ln(1) = 0 + 2pi*n.

@SuperDeadparrot 8 ай бұрын

Ln( 2pi*i*c1 ) = Ln( 2pi * c1 ) + i * pi/2 + i * 2k*pi because ln( i ) = ln( exp( i*pi/2 + 2k*pi*i ) ). Also, in complex functions, ln becomes log.

@lieman7136 6 ай бұрын

ln doesn't become log instead ln becomes Ln and log(a)b becomes Log(a)b - capital letters

@General12th 8 ай бұрын

Hi Dr. Pen! Very cool!

@1dayofmusic748 8 ай бұрын

i guess its just what complex jumbers are all about. you can solve more things at the sacrifice of your result being the only one. just the number 1 can be expressed by a lot similar to like the complex roots having more than one solution etc...

@aliariftawfq5354 8 ай бұрын

Take integral both side to get (c1,c2) value Thank you

@bobh6728 8 ай бұрын

The c’s can be any non-negative integer. How does a integrating find a value?

@epikherolol8189 8 ай бұрын

@@bobh6728Bros just Messing around lol

@greenrocket23 3 ай бұрын

Complex analysis is mind-blowing! I wish I had more time to study that area of mathematics.

@_cran 4 ай бұрын

You can just use the unit circle to find 2pi*i or using moivre formula, is quicker. Your way is a way to approach too but it's kinda longer 😅

@danielmcshane2562 8 ай бұрын

Brilliant!

@jimschneider799 8 ай бұрын

Similar to the way that ln(1) = 2*i*pi*C[1], you also have ln(i) = (4*C[3]+1)*i*pi/2, so ln(2*i*pi*C[2]) = ln(2*pi*C[2]) + (4*C[3]+1)*i*pi/2, making the entire solution into x = ln(2*pi*C[2]) + (4*C[1]+4*C[3]+1)*i*pi/2. And, since C[1] and C[3] are just arbitrary integers, that can be further simplified to x = ln(2*pi*C[2]) + (4*D+1)*i*pi/2, for arbitrary integers constants D and C[2], with C[2] != 0.

@atripathi6349 8 ай бұрын

this is satisfying answer to the equation

@user-eh2ec3rn6w 7 ай бұрын

Nice solution

@Healthsolution.694 8 ай бұрын

Great teacher

@armanavagyan1876 8 ай бұрын

Amazing👍

@samuelatienzo4627 8 ай бұрын

I love the excitement at 6:30 😂

@punpcklbw 8 ай бұрын

The logarithm cannot be defined for the whole complex plane, as exp(z) = exp(z+2πki) for any integer k. You're basically left with log(0) that is also undefined and approaches negative infinity in the limit.

@daniwalmsley611 8 ай бұрын

I feel like this should've been taught in schools, Like Sqrt(x^2] removes negative solutions, we needed a warning for logarithms too I am now slightly scared of how their might be a whole other set of numbers like okaginary but for logs instead of sqrt

@davejohnsondeveloper 8 ай бұрын

Reminds me of this video: kzfaq.info/get/bejne/g7Zjo7JoptWcaKM.htmlsi=T40ITkfSMril8ZGG

@Sidnv 6 ай бұрын

Complex analysis already deals with how to define logarithms for negative numbers (and any nonzero complex number in general). One difference is unlike square root having two values, logarithms are infinitely multi-valued. That is really what this calculation is doing. Any complex number can be represented as re^(i theta) where r is the distance from the origin and theta is the angle the line segment joining 0 to the number makes with the real line. But adding 2 pi to the angle doesn't change the value of the complex number, so it actually has infinitely many possible representations, each separated by 2pi in the angle. When you take the logarithm, you get ln(r) (a positive real number) + i (theta + 2 pi n) where n can be any integer, so you have infinitely many values. The only number for which you cannot definite a logarithm is 0, and there is no way to actually make sense of ln(0), because ln has an "essential" singularity at 0. Here's a specific example, supposed you want to take ln(-1). -1 can be represented as e^(i (2n+1)pi) for any integer n. So taking the logarithm, you get the set {i (2n+1) pi: n is any integer}. So any of these values can be considered a logarithm of -1, and all these logs already exist in the complex plane.

@mschuhler 4 ай бұрын

this was taught in schools.

@user-mv7nc1ki4c 8 ай бұрын

オイラーの定理から e^x＝2iπ　　∴x＝ln2iπ まではすぐわかりましたオイラーの定理から e^（iπ/2）＝i ですから x＝ln2iπ＝ln2＋lnπ＋iπ/2 となるんですね

@mqb3gofjzkko7nzx38 8 ай бұрын

Black pen red pen using a blue pen?

@hungry-sandwitch1355 Ай бұрын

I know how impossible this sounds, but black pen red pen is using a blue marker

@hyperbroli6672 Ай бұрын

Calculate the concentration of opium that is in your bloodstream

@SimpdePaint Ай бұрын

Im waiting for blackpenredpenbluepengreenpenorangepenpinkpen

@johnbutler4631 8 ай бұрын

This is really wild. I actually tried this on a TI-84, which js nowhere near as powerful as Wolfram Alpha, and it worked.

@wabc2336 7 ай бұрын

I thought you would also write out what the ln of the imaginary 2iπc_1 equaled. ln(2iπc) = ln(2πc) + ln(i) ln(i) = ln(e^(iπ/2)) = iπ/2 + 2πc_3 but this is redundant with c_2. So our final answer is apparently ln(2πa) + i(π/2 + 2πb) for any integers a,b

@GSDKXV 4 ай бұрын

Best channel on KZfaq idc

@NibbaHibba 8 ай бұрын

The answer i got when i solved it was pi/2*mi + 2pin where n is any integer and m is any integer congruent to 1 modulo 4. Is this still the same thing? ( i solved for e^x = 2(pi)n(i) and said that angle is pi/2*m and amplitude is 2pi(n) )

@wren.10. 8 ай бұрын

I desperately dream to be as excited to unravel mathematics as he is.

@amirmostafa3143 4 ай бұрын

That omg in the end

@abhijithcpreej 8 ай бұрын

Usually I have a tough time figuring out my own solutions during some of the "math for fun" videos. But this time, I could tell from the thumbnail. So weird😊

@JCCyC 8 ай бұрын

Wait a sec. The first term, ln(2iπC₁), is just ln(i) + ln(2πC₁). But we know ln(i) is many values, namely iπ/2+2πiC₃ -- so the complete solution is iπ/2+2πiC₃+ln(2πC₁)+2πiC₂ -- which reduces to... ln(2πC₁) + i(π/2+2πC₂) -- which I think looks better because it's not expressed in terms of something weird like ln(i).

@fetch7312 8 ай бұрын

i dont know if you read or take suggestions from the comments, but here's something that stumped me and my calc BC teacher while I was trying to prove the derivative of sin(x): I wanted to solve for the sum of sines without using the geometric proof, so I decided to implement euler's formula so I could use properties of exponents and real and imaginary parts to solve for the sum of sines. However, I wanted to first prove the formula, so after solving for the summation representation of powers of e through binomial expansion using lim(n>inf)(1+a/n)^n, i plugged in iz to the sum and separated it into the real and imaginary parts, giving me two taylor series, infsum(n=0)((-1)^n(z^2n)/(2n!)) and i*infsum(n=0)((-1)^n(z^(2n+1))/(2n+1)!). I already knew these series functions would create the equation cos(z)+isin(z) but I wanted to figure out if there was a way to work backwards from the taylor series functions to the original functions assuming that we don't know the equations the taylor series functions correspond to. My efforts were fruitless, so I'm curious if you could take a shot at it.

@OptimusPhillip 3 ай бұрын

Took me a second, but I got it. 0 is not the only solution to z=ln(1), any integer multiple of 2πi will also satisfy it. So e^x just needs to be an integer multiple of 2πi. This would make x=ln(2nπi), when n is some non-zero integer, or roughly (1.838+1.571i)+ln(n).

@danencel157 5 ай бұрын

Great video ! What is the goal of putting at 3:13 the e^i2πC2 ?

@GreenMeansGOF 8 ай бұрын

I agree

@joaninhafumacrack 6 ай бұрын

Math is so f***ing satisfying.

@Eichro 8 ай бұрын

this guy never fails to bring out what i like the most out of arithmetics and also what i hate the most

@TauGeneration 7 ай бұрын

i like that i was surprised that the "let c1 = c2 = 1" did result into 1. of course that's the solution, you tried to prove that to begin with

@maxhenderson1890 8 ай бұрын

Why specifically stop at c_2? You could keep multiplying by 1 an infinite amount of times, so wouldnt c_(n>1) be more fitting for conciseness?

@mtrichie111 8 ай бұрын

Still amazing explanations professor

@user-xr7ou1xf9x 2 ай бұрын

Where did we get the second “1” which is e^i2πC2? Where did it come from?! As initially it did appear in the exponent!

@Black_Hole_Institute 4 ай бұрын

Ln(2*pi*i*c1)=ln(2*pi*c1)+i*pi/2. Without this your solution is incomplete. You should also plot the distribution of answers on complex plane for wide range of c1 and c2.

@Jaymac720 7 ай бұрын

In my mental math, it is equal when X equals -infinity. Take the natural log of both sides. You get e^x = 0 Take the natural log again, but you have to apply limits. As X approaches 0 from the left, ln(x) approaches -infinity. You can also graph it. Y=1 is an asymptote of the function. It will only reach 1 at -infinity. Math is fun. The kinda complication with this though is that the function isn’t symmetrical because you can’t take a log of a negative number. Maybe you can with imaginary numbers, but I don’t know anything about that

@thetaomegatheta 25 күн бұрын

The expression 'e^x' is undefined in this context. Also, functions do not have asymptotes. Curves do.

@martinpechler4122 8 ай бұрын

Since your question is incomplete, you get a bunch of solutions in the range of area for complex numbers. But in the range of area for real numbers there is no solution. Your can spin the arrow in the complex world as many times as you like, but in the real world that does not matter

@wallaceferreira4739 3 ай бұрын

Amazing.

@TFclife 4 ай бұрын

I integrated both sides, e^u =e^u e^e^x = x In both sides e^x = ln x Find derivative of both sides e^x = 1/x In both sides x = In (1/x) Derivative of both sides 1= - Ln x 1= - 1/x -1= x

@thetaomegatheta 25 күн бұрын

And you got the wrong answer.

@ManjulaMathew-wb3zn 2 ай бұрын

The answer could be simplified a bit more. e^x =(2PIn)i =(2PIn)e^(2PIk+PI/2)i Taking ln you get x=ln(2PIn). + i (,2PIk+PI/2) which is in standard a+ib format.

@ManjulaMathew-wb3zn 2 ай бұрын

Now the back substitution of the general solution. e^x=(e^(ln2PIn))*e^(2PIk+PI/2)i =(2PIn)*(cos(2PIk+PI/2)+isin(2PIk+PI/2)) e^x=(2PIn)(0+i)=2PIni e^(e^x)=e^2PIni =cos2PIn+isin2PIn = 1+i*0=1

@GMPranav 8 ай бұрын

Nobody: The number 1 - "Now I am become death, the destroyers of worlds".

@thesugareater8607 8 ай бұрын

If you do fourier series/fourier transforms often, then youd agree with this video. I don't know what the reason for the 2ipi term is though. Ln(2ipi) seems like a solution.

@Sidnv 6 ай бұрын

It is because ln(2i pi) is itself multi-valued. 2ipi can be written as 2pi e^[i(2n + 1/2) pi] for any integer n. This calculation is just taking a fixed value for this logarithm and then adding in all the other solutions.

@mathmachine4266 8 ай бұрын

x=ln(2πN1)+(2N2+1)πi/2, where N1 and N2 are integers

@user-oi3on6on3l 7 ай бұрын

I tried to solve it by raising both sides by e^e^ln(), however, I arrived at the conclusion that if e^e^x =1, then e^e^x = 1. I am truly a mathematical wizard

@danielmoylan3033 8 ай бұрын

Yeah I paused and tried to figure it out myself. So if e^(pi*i) = -1, then e^( e^( ln(pi*i) ) ) = -1, thus ( e^( e^( ln(pi*i) ) ) )^2 = 1, thus e^( 2*e^(ln(pi*i) ) ) = 1, thus e^( e^( ln(2*pi*i) ) ) = 1 (what I did is basically cancel the e/ln pair and recombine), and since 1*1 = 1, we can put a c next to the 2 pi i. Of course that doesn't explain the non-log version.

@TheMemesofDestruction 8 ай бұрын

“True Love, Priceless. For everything else there’s Wolfram Alpha.” ^.^

@nicolastorres147 7 ай бұрын

For the final answer i don't like to be inside a log

@Ligatmarping 8 ай бұрын

Making the mistake of reducing e^(e^x) = 1 => e^x = 0 is like the graduate level version of a^2 = b^2 => a = b hahaha. Nice video!

@zhenhuazhao6100 8 ай бұрын

I am not sure if anyone has commented on it already. There are literally boxes and boxes of "blackpenandredpen" under the table. 😂😂😂

@cristrivera 8 ай бұрын

True😂

@tomctutor 8 ай бұрын

∞ {🖋🖍} 😎

@hadhamalnam 8 ай бұрын

Does it make sense to take the natural log of a complex number though? Wouldnt that have multiple solutions, hence making it not function? For example, I can say that ln(i) is i(pi/2), but it can also be any i(pi/2) +2npi.

@Ironpecker 2 ай бұрын

You don't really need to treat the ln as a function in this case though, it's just an operation you apply, same thing with stuff like arcsin or arcos normally they wouldn't be functions (unless you restrict their codomains) but in an equation it's not important. I'm not 100% sure myself though, but this is the logic that makes the most sense to ms

@misterroboto1 Ай бұрын

Think of the log as the "inverse" of the exponential map. If it makes sense to compute the exponential of a complex number, then it can make just as much sense to ask yourself "what complex number(s) x could I input in the exponential map in order to get a given complex number y as the ouput".

@Ytterbium176 Ай бұрын

I think I came across another, yet slightly different solution: x = ln(2pi*|m|) + pi/2 *n*i for all integers m,n (and m0). Based on the equation e^x = 2pi*m*i, I supposed that x is a complex number of the form a+bi. This lead me to e^(a+bi) = e^a * e^bi = 2pi*m*i. Therefore, e^a = 2pi*m and e^bi = i, resulting in a = ln(2pi*|m|) and b = pi/2 * n. I'm not an expert on complex numbers, though... Is this a valid approach/result?

@timstrhnr8144 2 ай бұрын

understood like 10% but love your passion and energy🫶

@simonekentish7491 8 ай бұрын

I wonder if DeMoivre got this giddy when he discovered complex solutions using the complex plane and polar coordinates.

@lucanina8221 8 ай бұрын

x=ln(|2c1pi|) +i*(2pic2 + pi/2*sign(c1) ) where c1 integer different from 0 and c2 integer and ln the natural log (real to real) is the correct solution. Expressing x in terms of complex logarithm is kind of cheating the exercise.

@xlorrix-6320 8 ай бұрын

why in the third step did you take i2pi*C1 anc multiply it by the polar form of 1 instead of just taking the logarithm?

@markyoung01maccom 8 ай бұрын

Loved it!

@rubikaz 2 ай бұрын

The solution without using complex logarithms is ln(2πn)+(2m-1)πi/2 with n, m integer numbers

@ash95959 6 ай бұрын

I have a different answer: e^e^x = 1 e^e^x = e^i2πn, n is an integer They have the same base so we can assume their exponents are equal e^x = i2πn, n is an integer x = ln(i2πn), n is an integer

@sebmandal 8 ай бұрын

Wouldn't e^2*i*pi = 1 + 0i? Since we can't define 0i as 0 in the real dimension, would it not need to be explicitly stated as 1 + 0i?

@alslaboratory570 8 ай бұрын

I'm confused. Why do we have to bring the one back at 3:11 instead of taking ln? And what is stopping you from multiplying infinite ones to get different solutions?

@shalopo 8 ай бұрын

It confused me as well. I think it was not necessary to multiply like that. However when you perform ln on both sides, THEN you'd get the additional additive term for the solution. If you keep multiplying, you'll get an equivalent result. You're just adding arbitrary constants *2i pi (after ln, it's adding), so they are equivalent to adding one constant * 2i pi.

@trojanleo123 8 ай бұрын

Can I use Euler's Identity for this to conclude that e^x = 0 = e^(i*pi) + 1 Therefore, x = ln[e^(i*pi) + 1] Would that be an acceptable answer?

@wilwdr96 8 ай бұрын

well e^(i*pi) + 1 = -1 + 1 = 0 so thats just the same thing as saying x =ln (0) though, which is just as undefinined as e^x = 0, so you have simply written that there is no solution in a more confusing way

@iananderson5891 8 ай бұрын

When we have f^-1(f(x)) we say it equals x but this makes sense if f has an inverse i.e. f is an 1-1 function. e^x is not an 1-1 function in the complex plane. So how do we do this?

@alipourzand6499 8 ай бұрын

At some point we have: e^x = i.2.pi.c1 i = e^i.(pi/2 + 2.pi.c2) e^x = 2.pi.c1.e^i.(pi/2 + 2.pi.c2) x = ln(2.pi.c1) + i.(pi/2 + 2.pi.c2) Right? Wrong?

@lucanina8221 8 ай бұрын

almost there, note that c1 cannot be negative inside your logarithm

@alexandershapiro28 8 ай бұрын

Interesting how it's only e^e^x=1 iff when there's at least one full revolution, while it's impossible when there's inaction. I think this is because of some deep algebraic topology stuff although pi(SO(2,C)) even at no rotation is well defined, maybe someone expert on the matter would give me a light on my fog of thoughts

@burzum_ 8 ай бұрын

I am no expert but it is easy to see when c1, where c1 is the amount of rotations, is 0; x = ln(2ipic1)+A = ln0 + A. ln0 is undefined therefore in this case e^x^x is undefined.

@Sidnv 6 ай бұрын

Yes there is a connection with algebraic topology here. You can look at at the exponential function as a map from the complex plane to itself minus 0. The fundamental group of the plane minus 0 is Z (and in fact this space is homotopic to SO(2, R), which is the circle S1). So the fundamental group acts on the preimage of the map above any point, essentially via rotation.

@mikejurney9102 8 ай бұрын

Does this also have a solution in the quaternions and octonians? Does i equate to some quaternion number?

@xinpingdonohoe3978 8 ай бұрын

i would equate to the quaternion number i. Quaternions are just complex numbers with the j and k axes added to the mix, and octonions follow that pattern again.