1^i

[@PapaFlammy69]

*_Oi, mah boi (and grill), hope you enjoyed today's video :v If you did so, why not check out some of today's relevant links to support the channel?_* Tired of learning for yourself? :) Why not motivate yourself and join me and over 130k more people on Study Together then? =D www.studytogether.com/join Handcrafted products, puzzles and more :0 stemerch.eu/collections/handmade-by-stemerch-eu i double dot Merch: stemerch.com/collections/cursed-math-memes-i-double-dot

[@anshulrai677]

Nice

[@nachiketakumar9645]

Nice

[@oni8337]

∮

[@Giogro]

Hey papa flammy, maybe it's not really math related but i'd 100% buy a chess set made by you, think about it :)

[@PapaFlammy69]

Yas, already figuring everything out! =)

[@kobosakos1801]

π = pa φ = fa

[@Lolminport]

i = a

[@wherestheshroomsyo]

I'm so tired of him pronouncing things weird on purpose

[@vedants.vispute77]

I was going to write this lol

[@jf8442]

I was looking for this very comment

[@peterluger1400]

paaahhhh fahhhh

[@theevilmathematician]

Andrew Dotson's quote in REAL LIFE living in the basement: "It was cold... dark... smelled like sauerkraut... and Jens would frequently come down and scream at me; that's how he sleepwalks..." - Andrew Dotson July 4th, 2021

[@gilmonat3155]

Take the limit as k->infinity and you get that 1^i=0

[@caesarinchina]

Which means that if you rotate infinitely many times, you'd end up in the center (0,0). Makes sense. You would get dizzy and, from the perspective of dizziness and eternity, everything would look like zero. Makes sense to me.

[@Assault_Butter_Knife]

Woah what? So does this mean that a real power function is divergent at infinity, but a complex power function is convergent at infinity? How does this make sense? Can someone explain this to me please?

[@dijkstra4678]

@@Assault_Butter_Knife he's talking about the limit as k->inf of e^(-2kpi) which does indeed go to 0 because as k increases 1/e^(2kpi) decreases and will get closer to 0.

[@alejrandom6592]

@@caesarinchina Not exactly. You rotate just to measure the angle, but once you measure it you take a step further closer to the origin. {1, e^-2π, e^-4π, e^-6π...} I know ur joking but I just want to clear up the visualization of this problem.

[@alejrandom6592]

@@Assault_Butter_Knife e^cx is divergent for c>0 and convergent to 0 for c

[@AndrewDotsonvideos]

love is a battlefield

[@kathanshah8305]

And your love is in basement

[@nachiketakumar9645]

Right sir

[@nachiketakumar9645]

@@kathanshah8305 you don't have any respect towards intelligent people.

[@kathanshah8305]

@@nachiketakumar9645 that’s why I respect you

[@Eigenbros]

Damn.. 😭😭

[@angelmendez-rivera351]

4:02 - 4:13 I know this is nitpicking, but if a and b are just arbitrary nonzero complex numbers, then log(a^b) = b·log(a) is not true in general. As far as complex "exponentiation" is concerned, you would simply say that a^b is a short-notation for exp[b·log(a)], where log and exp are well-defined, and then call it a day.

[@AdityaKumar-ij5ok]

noway, last time i was this early, flammy solved riemman hypothesis

[@akselai]

yes

[@PapaFlammy69]

yeye

[@devd_rx]

Profile pic checks out

[@nachiketakumar9645]

Dr. Kumar Eswaran already had solved Riemman hypothesis

[@duncanhw]

@@nachiketakumar9645 No, hoax lol

[@timurpryadilin8830]

7:58 by the same logic, log(e) can take multiple values depending on the branch, i.e. 1+2πni. however, it is better to move straight from log(e^2πki) to 2πki *by definition*. you do not need to write extra log(e) for that.

[@PapaFlammy69]

ye, but I didn't want to complicate it further :'D

[@albertoreyabuelo2504]

@@PapaFlammy69 shouldnt it be ln (x) instead of log (x)?

[@riade-yet1489]

@@albertoreyabuelo2504 it’s just question of notation. ln is neperian logarithme that means the inverse function of e(x). Log without indication of the base is ln.

[@duncanhw]

@@albertoreyabuelo2504 Sometimes log(x) means base e (often in maths), sometimes 10 (physics, engineering, calculators), sometimes 2 (computing)

[@licht4808]

@@riade-yet1489 why dont people specify the base everytime? cant be that hard.

[@user-cr3iv7uf6z]

don't know if this is right but... we know that e^(πi) = -1 and (e^πi)^0 = (e^0)^πi so (-1)^0 = 1^(πi) 1^(πi) = 1 so 1^i = π-th roots of 1

[@justinnettnin9209]

I just noticed from this video that for the natural log, you use log, instead of ln. Is that the way you prefer it?

[@PapaFlammy69]

yup

[@MemeLord42]

@@PapaFlammy69 the problem with that is that log(x) is log base ten and ln(x) is log base e

[@infoeducardo6915]

@@MemeLord42 depends on who's using it A mathematician uses base e by default A computer scientist used base 2 by default

[@j_blue6784]

@@MemeLord42 i was pretty sure until now that lg(x) is base 10. Edit: turns out that its not strictly defined. The most internationally used version is log = base 10, ln = base e and lg = base 2 ... kinda confusing that germany uses lg as base 10 and sometimes log as base e...

[@m-linko]

Who's the chad in your profile pic?

[@DumblyDorr]

"Aah times eee to the ah pah"... you turned into a sweet old southern lady there ...

[@malexmartinez4007]

Papa flammy becoming yummier and yummier with each video.

[@PapaFlammy69]

:D

[@SultanLaxeby]

Held a talk at an international mathematics conference recently and of course I started with "Good morning fellow mathematicians" :D

[@ivan-v-morozov]

Hello, I received your tensegrity sculpture today, and I want to thank you for making this wonderful product. It's a beautiful piece, thank you for your work!

[@PapaFlammy69]

Glad you like it!!! Really glad actually :D

[@WhattheHectogon]

0:27 Gesund.

[@PapaFlammy69]

Deutschland!

[@thisguy3572]

9

[@kathanshah8305]

@@PapaFlammy69 83,86,89 which one ?

[@quantum1861]

Saw the title then did this. IDK about complex log or principle branch or whatever lol but here's what I came up with. It do be 1. i^(2/π)=e^((2/π)(ln(i))=e^((2/π)(iπ/2))=e^i therefore i^(2/π)=e^i then do algebra i^(2/π)/e^(i)=1 (i^(2/π))^i/(e^(i))^i=1^i rewrite (i^i)^(2/π)/e^-1=1^i substitute e^(-π/2) for i^i e^-1/e^-1=1^i 1=1^i

[@timkw]

I got the metal puzzle for my birthday a few days ago and it was lots of fun trying to solve them! ( out of the 3 i I have 1 left to solve )

[@PapaFlammy69]

Glad you liked them!!! =) The wooden puzzles are also lots of fun :3

[@kathanshah8305]

i am very complex

[@Azirapheil]

Fun one Papa, but one unfortunate error in the derivation at 7:39. :( We cannot "drag the exponent in front of the log" once the domain of our logarithm (regardless of branch) allows for complex inputs. For example: Log(e^(2pi i)) = Log(1) = 0 =/= 2pi i = 2pi i * 1 = 2pi i Log(e).

[@alejrandom6592]

ICHTSTBKIMHIHABS

[@user-xh9pu2wj6b]

We can tho. "Log(1) = 0" part is wrong. Log(1) = 2n*i*pi.

[@Azirapheil]

@@user-xh9pu2wj6b Not so. Log(z) is not a multi-function (it only assumes ONE value, otherwise it wouldn't be a function). Maybe I should have been more clean, but here Log is a very particular choice of logarithm, it is a convenient way to denote the principle branch of the logarithm. Context: Any logarithm (let's denote it by lambda(z)) in the complex plane has the form lambda(z)=ln|z|+i*a(z), where ln is the usual natural log for real numbers (|z| is a real number) and a(z) is a continuous angle function. The a(z) function determines what angles are "acceptable", so it could give the angle of z being between 0 and 2pi for example, but if we want it to be *continuous* then we can't include either 0 or 2pi! Meaning, if we choose to use the angle function a(z) that measures angles *open* interval ]0,2pi[ then we can't allow the logarithm lambda(z) that uses a(z) to be defined on the positive real axis. I.e. the domain of our lambda(z) in this case is the complex plane BUT we have cut out the positive real axis. Additionally, the image of lambda(z) is a "horizontal strip" in the complex plane; the real part of the output can be whatever, but the imaginary part is stuck between 0i and 2pi*i, since those are the angles we have chosen. Sidenote: If you want your logarithm lambda(z) to be a *continuous function*, then you have to cut out some half-line from the origin, but you get to pick what half-line. These half-lines are called *branch cuts* and the different logarithms you get by choosing different angle functions are called *branches* Now, what it means for the Log(z) to be the principle branch is that the particular angle function we have chosen is Arg(z) [sometimes arg(z), it stands for *argument*, which is a fancy name for *angle from the positive real axis*], which measures angles (strictly) between -pi and pi; meaning it is defined one the complex plane, but we have cut out the negative real axis. The exact formula for Log(z) is then Log(z) = ln|z| + i*Arg(z) Now let's examine Log(1). |1|=1, so ln|1|=0. Also, the complex number 1 (=1+0i) has only 1 angle available *on the interval ]-pi,pi[*, namely 0, so Arg(0)=0. Thus, Log(1)= ln|1|+i*Arg(1) = 0 +i*0 = 0. :)

[@Azirapheil]

​ @КП-92 Прищепа Дмитро Small addition: You can certainly choose n different lograrithms lambda_n(z) such that lambda_n(1)=2n*pi*i, but you cannot choose a *single* logarithm so that your proposed identity holds. EDIT: In fact, since inverting the exponential function becomes impossible without restricting yourself in some way, you get "weird" behaviour like Log( e^(2n*pi*i) ) = 0, for *all* n. ( Wow! ) You've probably seen this same behaviour, but maybe not realized it, in a different context. Take a close look at the inverse trigonometric functions, how they have to restrict their domain, and what, for example, sin^(-1) ( sin(2n*pi) ) turns out to be. ;)

[@user-xh9pu2wj6b]

@@Azirapheil I messed up Log and log, hah. In the video Papa used log, not the principal branch of it, so I think it's a legal derivation.

[@kathanshah8305]

Bold joke made by a married man I’m expecting mrs fehlau’s comment in this video

[@PapaFlammy69]

:D

[@kathanshah8305]

@@PapaFlammy69 more like :x this be safe papa

[@jonathanbaxter4611]

Flammable maths: 1^i is 1 but not so fast its also 1 but actually its also ...e^4pi, e^2pi, 1, e^-2pi, e^-4pi....,

[@TrimutiusToo]

Usually when you mean log(z) with all branches instead of just primary branch it is written as Log(z)

[@PapaFlammy69]

Nah, Log is the principal one usually I believe

[@TrimutiusToo]

@@PapaFlammy69 indeed, did it completely backwards...

[@Astromath]

Did you know that ln(pi^e) ~ pi And logpi(e^pi) ~ e Therefore pi ~ e/logpi(e) And e ~ pi/ln(pi) Pretty amazing

[@Astromath]

Well, only the first decimal place is right but never mind

[@Astromath]

And on top of that both approximations are off by the same factor!

[@Astromath]

Pi/(e/logpi(e)) = (pi/ln(pi))/e Which is equivalent to: logpi(e^pi)/e = pi/ln(pi^e)

[@Astromath]

Please explain someone

[@Mothuzad]

You could also approach this by representing 1 as i^0, i^4, i^-4, etc. You'd get 1^i = e^(4ki*ln(i)). What's ln(i)? Taking i*pi/2 as the principle value happens to suffice, since k in the other expression gives all the same values anyway.

[@ro5197]

it is the same as e^(2npi) when n E Z (n is an element of the the integers)

[@emanuellandeholm5657]

Yep, the negative sign in the exponent is redundant.

[@mrhiisduh477]

1= e^(2*pi*i*k) ==> euler formula i^i = e^(-2*pi*k) ==> done

[@ramansb8924]

Can we use Euler’s identity as (- e^ iπ)^i = 1^i and find the value using cis form?

[@mr.unknowngamer2109]

Ur content is osome , this is one of my fav. channel

[@PapaFlammy69]

[@johndoh1000]

Wow that’s a lot of math for nothing to change. I love it.

[@PapaFlammy69]

:D

[@the_red_wolf8244]

Video starts at 3:42

[@user-ho1hg4pw5d]

So from 1^i=e^-2Kpi, K is an integer we got that: i=-2Kpi and 1=e

[@lordstevenson9619]

Your intros never disappoint.

[@PapaFlammy69]

[@ImranHossain-zu1nr]

Sir,i want to send a question,how i can....?

[@santoshgarg2625]

Hey Papa flammy Could you make a series on higher standard maths like Calculus and trigonometry etc.That would be great.By a fan from india. ❤️🙏🏻

[@eitanoidos6304]

Thank you papa, very flammable.

[@shaheerziya2631]

I always lose it at "eee to aah faa" and "aah faa paah"

[@yavuz1779]

Til papa flammy is 27 years old

[@theprofessionalfence-sitter]

Is using a^b=e^(b*log(a)) a fair way to define it? I would have defined 1^i using the analytic continuation of x|->1^x, which makes it clear that it should be 1.

[@ivangospodarski80]

1^i=1^sqrt(-1)=1^(-1)^0.5=1^0.5= +/-1 *mic drop*

[@KdEAG1112]

I thought about this like that. when i is root(-1), than 1^i would be 1^(-1)^(1/2) which is the same 1. is that okay or is there a mathematical flaw

[@chainns7852]

Namely, he is using namely quite a lot.. to be namely

[@PapaFlammy69]

yeye

[@kathanshah8305]

Tell andrew to take kelly out of his basement

[@PapaFlammy69]

nah

[@matej_grega]

Basementception????

[@napa31427]

Basment in a basement?!

[@korayacar1444]

My issue with this evaluation of 1^i is that the principal Log is the right inverse of the exponential in the complex numbers, and since id = exp(Log), 1^i=exp(i*Log(1))=exp(i*0)=1. The log branches matter when looking for a left inverse to the exponential (which is where they originate, as exp(2ki*pi)=1 for all integers k), but they are certainly irrelevant to the right inverse of exp, which is Log and certainly no other branch. 1^z=1 for all complex z, and i^i=e^(-pi/2), as id=exp(Log). log(i) has a variety of values, as Log(exp) is not the identity, and other branches of log(exp) allow for other solutions.

[@kenanwood6916]

Something doesn't seem right. 1 is equal to an infinite number of things which are obviously not equal to 1? If that is the case, "=" is not even an equivalence relation on \mathbb{C}. I have always been bothered by this. Can someone precisely explain what is going on? BTW I've not taken Complex Analysis yet.

[@IkikaeruRaimei]

1 can only be equal to itself, the only change is the representation of that value.. So a complex can represent in an infinite way possible a unique real. In the case of the video, just so you know, there's only one solution that would work based on the formula he gives.

[@kenanwood6916]

@@IkikaeruRaimei I wonder if the issue is that 1^i is not well defined. We can represent it as 1 and as e^2pi, both are valid, but they are unique real numbers. See what I mean?

[@HAL-oj4jb]

You know that the video's going to be really interesting when you think you can solve the problem in a few seconds and just know there's some weird secondary branch you missed

[@PapaFlammy69]

:)

[@canr772]

1^i = cos(ln1) + isin(ln1) = = cos(0) + isin(0) = 1 + 0i = 1

[@ErikTheViking92]

1^i = 1^sqrt(-1) = 1^(-1)^(1/2) = 1^((1/2)*(-1)) = 1^(-1/2) = 1 Would've been my first approach. Gutes Video, flammy!

[@kaylo1680]

StemMerch creation-process video papa?

[@PapaFlammy69]

Hopefully soon :)

[@thisguy3572]

Where are community post

[@pokefan-oh1gq]

dont you love it when he says somefing

[@lazymello6778]

But instead of writing it as e^(log(1^i)), can't we just substitute the value of 1 in the original? 1^i = (e^2kπi)^i = e^-2kπ That's much more direct I'd say,not much difference but still

[@igorbondarev5226]

Hey, I don't get it why multiple answers. We are not solving an equation "x = 1^i" to have multiple x'es, we have computational question, and computational questions are bound to have one answer. The same we say square root of two is 1.414... and we never say it's -1.414..., but when we solve an equation x² = 2 then we have two answers.

[@dr.rahulgupta7573]

Excellent presentation. VOW !!

[@TI5040]

Hello zens, nice example of principle branch of logarithm.

[@PapaFlammy69]

:)

[@Pacvalham]

Before watching, I think 1^i = 1.

[@alejrandom6592]

Principal branch

[@mastershooter64]

guud morning fellow mathematicians whaalecum back toanatha video

[@dorian4387]

For small enough values of i, should converge to Jens constant, did it via proof by engineering

[@AboodXD]

You could have raised that other definition you found for 1 at 7:16 to the i'th power instead of substituting in the original equation you made at the beginning of the video.

[@SidneiMV]

Now I understand why they are called "Complex" 😎

[@RoboticusMusic]

My anime figurines are going to look so good on that tensegrity stand on my shelf.

[@PapaFlammy69]

Nice! :D Which one you got? :)

[@dumbmaths9748]

Thanks for informing us about studytogether! Appreciate it.

[@PapaFlammy69]

[@oni8337]

you should do a video talking about real numbers to the ith power

[@mujtabanadeem3901]

Bro can u define your path to math? I want to get my math level to advanced math. Also what is your ug course?

[@of8155]

Iota My crush....

[@PapaFlammy69]

:D

[@xerostat8961]

this was good papa :) ten kew

[@PapaFlammy69]

[@nachiketakumar9645]

@@PapaFlammy69 she's your daughter?

[@dsylexic]

@@nachiketakumar9645 we're all his children :)

[@nachiketakumar9645]

@@dsylexic yes, but how

[@xerostat8961]

@@nachiketakumar9645 yeye :D

[@hameedamathtuber]

Good effort. Keep going.

[@pedrosso0]

1^i=e^(2pi n) n is an integer

[@tszhanglau5747]

d(1^i)/dt =1^(i double dot)

[@FutureWaitHim]

y = 1^i Log y = log 1^ i Log y= o Y= 1

[@dhruvdubey3450]

what's √(a+b)?

[@hameedamathtuber]

That was interesting. Thank you

[@foreverlikeatheorem]

I have a question. Does using the equivalence a = e^(ln(a)) add other solutions? Like saying 1 = 1^(4/4) = (1^4)^(1/4) = 1^(1/4) = {1, i, -1, -i}. For this reason how can we be sure not to have accidentally added solutions? I think that 1^i might only have {1} as solution.

[@nachiketakumar9645]

It has √-1 number of solutions, but they exist in different number space dimension

[@gtziavelis]

How REAL MEN evaluate 1^i

[@pauljackson3491]

Can you do a video of math books to teach yourself various topics? I've done basic Calc but what about things like linear and all the weird elliptic curve.

[@idjles]

"pi", not "pa".

[@PapaFlammy69]

nah, it's pa

[@ilprincipe8094]

@@PapaFlammy69 ππflammy

[@idjles]

@@PapaFlammy69 that’s Überheblichkeit, please show some humility and learn from your community.

[@telotawa]

wow chill to loafy music? i love loafy music

[@ryanjagpal9457]

This terrifies me in every way, i’m gonna fail calculus when I learn it at school, as well as everyone in my year

[@DatBoi_TheGudBIAS]

Wat if u put sum formulas togeder? Like (√3)^√i

[@ivangospodarski80]

If a>b, then which is bigger - a-th root of a or b-th root of b?

[@asonjarossa2023]

I wonder if you could go back to giving problems for the audience to solve? Like from the cos(3 degrees) video you did a week ago. I found it extremely helpful and immersive to the viewing pleasure; I can take on an integral problem like the one you made 3 years ago!

[@CrittingOut]

[Andrew ∈ Basement]

[@zaloreyes830]

love you papa.

[@orenfivel6247]

0:12 god bless to start HW: evaluate [1,0;0,1]^[0,-1;1,0] 9:57 god bless to stop

[@ricardoparada5375]

This is a really neat observation Flammy ma boi. Good work!

[@alialam1893]

do flammable maths reply?

[@PapaFlammy69]

yeye

[@pedrokahn5717]

do you say "Gut Morgen" at the very beginning?

[@nachiketakumar9645]

What is the value of i^n in n-dimensional space???

[@PapaFlammy69]

depends

[@nachiketakumar9645]

@@PapaFlammy69 depends on what

[@nachiketakumar9645]

@@PapaFlammy69 father, when I had solved then i get 1

[@nachiketakumar9645]

@ཀཱ what!......you are a diaper brand

[@miscellaneouscrap1237]

Love for your vids from Australia

[@PapaFlammy69]

[@jackhandma1011]

Blackpenredpen enters the chat.

[@PapaFlammy69]

why?

[@iamgoosht1734]

can we say that : 1 = - e^πi -> 1^i = (-e^πi)^i -> 1^i = -1/e^π ????

[@warrickdawes7900]

Without this special consideration, we would never have e^(i.pi) + 1 = 0.

[@mohammadburhanmanzoor3806]

Papa flammy! A request to do some advanced math topics series with hw exercises for viewers. Is that possible?

[@officialyashsharma]

wth

[@noreaction1]

Why is 1 not equal to 1.000[inf]0001 or 0.999 repeating

[@user-xh9pu2wj6b]

It is, though. Btw, 1.000[inf]0001 isn't something you can really have in real numbers. If you meant it to be something like lim(x->inf)(1+1/x) then it's precisely equals to 1, otherwise - not a valid real number.

[@AhmedMahmoud-tv9vw]

Sorry if I am wrong but, shouldn't it be Ln as in natural logarithm and not just Log(base 10). Nice video btw. I also like your merch. ❤

[@aweebthatlovesmath4220]

You are right! but sometime log means ln it depends on which notation you want to use.

[@AhmedMahmoud-tv9vw]

@@aweebthatlovesmath4220 thanks 👍

[@devd_rx]

Sus power

[@plus-sign]

The principal value of 1^i is still 1 though

[@PapaFlammy69]

yeye, sure

[@SidneiMV]

better intro impossible 😅😂😅😅😅

[@user_2793]

exp(-2nπ) ?

[@alperenerol1852]

There is something unsettling about that shirt.