i^i

What is i to the i-th power, namely i^i? Is it real? Is it possible to have imaginary^imaginary=real?
This is a classic complex numbers question and in fact i^i is real!
How about i^i^i? Check this out: • tetration of i^i^i = ?
Are you wondering about (a+bi)^(c+di) now? Here's the video • the tetration of (1+i)...
#blackpenredpen #math #complexnumbers
💪 Support this channel, / blackpenredpen

[@TheUnlocked]

I'm watching this instead of doing math homework.

[@mishikookropiridze]

same :D

[@Jj-or5ix]

Angel's Of Revelation , tf

[@amirmazor3670]

We all

[@Wild4lon]

This is better than homework

[@clyde__cruz1]

Way better

[@kfftfuftur]

Euler had a hard time understanding negative numbers, but with complex numbers he is just fine.

[@pettanshrimpnazunasapostle1992]

He had a hard time with negatives since he was a positive person ;)

[@georgegkoumas5026]

It was just that his personality made him look for the root of what he didn't understand.

[@HanzCastroyearsago]

Imao

[@louisyama9145]

You could say he was a complex person

[@SteelHorseRider74]

​@@louisyama9145 so he was an i person ^_^

[@lP41N]

Another simple way to get the same result: We know that: e^iπ = - 1 (e^iπ)^1/2 = (-1)^1/2 e^i(π/2) = i So if we raise to the i power we get: e^(-π/2) = i^i :)

[@xjetfirex3956]

Got the same thing when tried to solve it ;)

[@GarGlingT]

Pi/2 radian is already i vector.

[@divyajyotibose5769]

It's the first thing that struck me

[@albertstern3006]

e^i(π/2+2nπ) is always i So the other solutions are: i^i=e^-(π/2+2nπ)

[@TyroneSteele]

Me too :)

[@iaagoarielschwoelklobo6342]

0:03 My friends when I talk about mathmatics

[@tcocaine]

This is so accurate!

[@Matrixician]

IAAGO ARIEL SCHWOELK LOBO relatable

[@subinmdr]

Same here 😂😂

[@blackpenredpen]

IAAGO ARIEL SCHWOELK LOBO sadly I have to agree too. This made me laugh so hard lollllll

[@scitwi9164]

They don't run away if you have the seed ;> (And yes, I'm speaking metaphorically right now ;> )

[@pneujai]

me: i is complex my English teacher: no "i AM complex"

[@nilsastrup8907]

Lol

[@empty3293]

bruhther

[@easds7881]

@@damuddohonson2282 wtf

[@lyrimetacurl0]

*I (capital i)

[@nightytime]

@@damuddohonson2282 bro…

[@JJ_-fp9sj]

This absolute madlad pulled out another blackboard from the ceiling. Most badass thing I've ever seen on a math class

[@anglaismoyen]

You'd love the MIT open course ware videos.

[@potatopassingby]

blackshirtredshirt :D

[@blackpenredpen]

it is!

[@coc235]

@@blackpenredpen Isn't it?

[@padoraye]

Redchalkwhitechalk

[@uselessiSpace]

@Fred The Llama yes

[@itsuki_jeff]

lmao

[@ammarbarbhaiwala9908]

Hey I just watched this video yesterday and it came in my mathematics exam today Nobody but me solved it

[@blackpenredpen]

Wow. Nice!!!!!!

[@ammarbarbhaiwala9908]

@@blackpenredpen Thanks In love your Videos

[@justarandomdudelol7702]

@אהבה יהוה not sure what you’re trying to express when your workings are clearly wrong

[@TrueSpeak-TS]

Nice hahaha

[@poteresurreale5781]

@@justarandomdudelol7702 why?

[@eduardvanbeeck9539]

Hey man I love your videos, the way you explain the problems and also how much you enjoy it all! Keep up the great work!

[@blackpenredpen]

Eduard Van Beeck thank you!!!!!

[@JotaFaD]

Another very similar way to get to the same result, but without using ln: i^i = ? But, i = 0 + 1i = cos(t) + sin(t)i t = pi/2 (or pi/2 + 2*pi*k) solves the equation. So, i = e^it i = e^(i*pi/2) i^i = [e^(i*pi/2)]^i i^i = e^[(i^2)*pi/2] i^i = e^(-pi/2)

[@xgplayer]

if you use de polar form you get the same answer right away: ( e^( i*(pi/2+2pi•n) ) )^i = e^(-pi/2-2pi•n)

[@gmtutoriais6519]

Much simpler

[@dox1755]

LOOOOOK I Have mooore easier than that. Now. e^pi*i=-1 -1 is i^2 so than equal it e^pi*i=i^2 than multiplye the powers by i/2 e^-pi/2=i^i LOOOOOL

[@Mrwiseguy101690]

+Alper Berkin Yazici Slow clap

[@hach1koko]

José Paulo I don't think that's well justified. With that logic, you could also say : e^(i5pi/2)=i (true) e^(-5pi/2)=i^i We would end up with e^(-5pi/2)=e^(-pi/2), which is obviously false...

[@gregg4]

"Hopefully this makes everybody happy." (10:04) This is the internet! It is mathematically impossible to make everybody happy.

[@spiguy]

*physically

[@Hydrastic-bz5qm]

@@spiguy theoretically*

[@shayanmoosavi9139]

@@Hydrastic-bz5qm all of the above*

[@user9287p]

@@shayanmoosavi9139 *Under the assumption that all possibilities are random, I would concur to the previous comment before me.

[@aasyjepale5210]

@@es-rh8oo psychedelically*

[@meme_engineering4521]

I am just getting addicted to this channel

[@boband75]

I like the back stories you provide, and your logic and steps are very easy to follow!! Please keep this channel alive, watch it every day! It’s great for my engineering students too.

[@Timelaser001]

6:50: 'You know this is a real number. So real." XD

[@blackpenredpen]

: )

[@shayanmoosavi9139]

LOL XD but do you know that complex numbers are also as real as other numbers?

[@istudy2194]

@@shayanmoosavi9139 complex numbers aren't real They are just helpers for complicated mathematics

[@istudy2194]

@@shayanmoosavi9139 our school teacher said that when you go for math major, you study about symtots or whatever its called Lines that seem to intersect but don't since they aren't real. (They don't exist) Just as i^i is Real but I isnt, it further explains complex numbers as an expander of mathematics

[@shayanmoosavi9139]

@@istudy2194 the word you're looking for is asymptote. We say the function asymptomaticaly approach a value when it gets really close to that value (its idea is connected to limits). For example the function f(x)=1/x asymptomaticaly approaches 0 as x increases but it'll never reach 0. A similar concept is convergence. This concept is used in infinite sums (aka series). Now let's get to the main point. Numbers are just tools. None of them are real. They're just concepts. That's it. Let me explain with an example. What is "2" exactly? And I don't mean that you show me 2 fingers. Explain to me what 2 is _without_ referring to any physical object. We _invented_ numbers because of necessity. Our most basic need was how to count so we invented natural numbers. Then as we advanced and developed complex (no pun intended) economic systems we needed to keep the records of debt so we invented the negative numbers (ancient civilizations like China used negative numbers for debt). I think you get the idea. As we advance our needs get more complex (pun intended) so we invented complex numbers to help us. They're very helpful. They're used for modeling different phenomena. You'll find them in electrical engineering (they're used for modeling the signals), quantum mechanics (for modeling the wave equation) and almost everywhere else. The conclusion is numbers are very helpful tools and they're just a concept. They're not "real" (pun intended).

[@shoaibmohammed3707]

I have no idea why some people dislike your videos. Honestly, the work you do is amazing. It's very understandable and really nice! Thank you so much!

[@christiansmakingmusic777]

I really enjoy the efforts you make in complex algebra calculations. Many people get beat down with endless calculation but few ever tell them there are calculations that no human can do, so don’t get discouraged. Just increase your focus and attention span over time. I’ve had professors who would assign us twenty 3x3 matrix inverse problems to be done by next week, but couldn’t do one on the board without making ten arithmetic mistakes.

[@victorj9582]

This is how I did it: Rewrite i^i in terms of e and natural log: i^i = e^ln(i^i) Bring the "i" power to the front: = e^(i*ln(i)) Also bring the 1/2 power from the "i" to the front: = e^(i/2*ln(-1)) We know ln(-1) is equal to πi according to Euler's formula; e^πi = -1; therefore, πi = ln(-1) = e^(i/2*πi) Bring πi to the numerator which will result in i^2 which is equal to -1: = e^(-π/2) The end.

[@Rzko]

you use over-complicated things assuming easier things that are enough to have the solution. There is absolutely no need to use the complex log.

[@mike4ty4]

For what it's worth, a few other values of i^i are: 0.00000072494725159879381083665824397412631261... 0.00038820320392676624723252989870142711787... 0.20787957635076190854695561983497877003... (this is the one in the video) 111.31777848985622602684100793298884317... 59609.741492872155884501380729500106645... The all lie along the exponential curve y = e^(-pi/2 + 2pix). But only where x is an integer, does that curve represent a value of i^i.

[@tannerzuleeg1229]

111.31777848985622602684100793298884317 = e^(3pi/2) which means ln(111.31777848985622602684100793298884317)=3pi/2 59609.741492872155884501380729500106645 = e^(7pi/2) which means ln(59609.741492872155884501380729500106645)= 7pi/2 Using radians as exponents on e, you can equate the powers of i if they are imaginary. Obviously x=0 in your example (making it i^i) So when you graph it, I think the first of an infinity of y intercepts would have to be e^(-pi/2). every time you add 2npi (n=integer) you get another intercept. I could be wrong.

[@bl_ninjat9012]

why is the title crying?

[@rohithsai3570]

I'm watching this while I am in a movie theatre.

[@douglasstrother6584]

"5" is a complex number: a knuckle sandwich is lunch.

[@quaji1]

It's possible to simplify the second part and skip everything from 7:01 until 11:10 by stating that "i" in polar coordinates is r=1 and theta=pi/2+2pn (and not just theta=pi/2) and using the same formula.

[@mjdev-i1p]

The nice thing about your videos is that they are very calm and relaxing. And your enthusiasm cancels out the boredom :D

[@esu7116]

i^i=e^(iπ/2)^i=e^(-π/2) done.

[@psibarpsi]

Yeah. That's exactly what I did.

[@estelle_chenxing]

...

[@godeffroydemontmirail3661]

Yes, that's all. Very simple in fact 😁

[@esu7116]

@אהבה יהוה wtf? I wish you could have constructive comment.

[@victorkkariuki]

I really love the accent.. As a bilingual myself, it feels so awesome to have the ability to switch anytime and speak a different language. Come to think of it, I am definitely multilingual

[@samiam9235]

If I've learned anything, it's always have a pokeball ready. Just in case.

[@blackpenredpen]

yes

[@LightYagami-el3ft]

Another way is i^i = x Square both sides (-1)^i = x² Now put e^ipi = -1, we get the final answer as e^(-pi/2)

[@KarelSeeuwen]

You're a credit to the human race. Keep up the good work. [Edit] ++ I just read some of the comments regarding the multiple solutions police incident. The importance of making mistakes cannot be understated. I for one walked away from this video with the message that thinking is far more important than wrote learning. How the hell else are we going to make progress in this world people.

[@chengme]

you have a great personality, plus you are doing great, love it!

[@RealUlrichLeland]

This is probably the only bit of enjoyably vaguely complicated maths that I have ever understood. Good video, well explained. Don't listen to people who say that your accent makes you hard to understand, I found it a lot more comprehensible than many people with English as their first language.

[@fNktn]

You can skip the whole ln transformation part by substituting i = e^ai, with a = pi/2 in this case, to get (e^i*pi/2)^i which is of course equal to e^-pi/2

[@sidali9057]

We can simply use the Moiver's formula i = i. sin(π/2) = e^i(π/2) i^i = [e^i(π/2)]^i = e^(-π/2)

[@philologo8323]

I only barely put my toe into the water of mathematics, and these just encourage me to dive in! Keep up the awesome content!

[@barbiefan3874]

you could write i(base) in exponential form which would be e^(iπ/2) and then rise it to i'th power which would be (e^(iπ/2)^i) and then multiply the exponents: e^(i*iπ/2) = e^(-π/2) i think that would be a lot easier then calculate the ln(i)

[@sobianiaz30]

There was no need to add 2*pi*n in the original angle theta because by definition of coterminal angles Theta+(2*pi*n )=Theta.

[@Imperio_Otomano_the_realest]

it's sort of wrong, but θ needs to be defined as π/2 + 2πn

[@sayanjasu]

Shit gotta be real tough when you gotta whip out a whole new black board in the middle

[@jameeztherandomguy5418]

You dont have to he overcomplicated the explanation 1. e^πi = -1 | Known 2. sqrt(e^πi) = i | Sqrt both sides 3. e^(π/2)i = i | Simplify 4. e^((π/2)i)^i = i^i | Raise both sides to power of i 5. e^(π/2)i*i = i^i | Power rule 6. e^-π/2 = i^i | i*i = -1. Final Answer.

[@ZipplyZane]

And now I just realized that Matt Parker is who introduced me to this channel. I'd been wondering how I found you.

[@ProCoderIO]

I learn about it from 3Blue1Brown

[@abhijitprajapati3764]

i just found the channel randomly

[@adityashivam3066]

You can calculate i^ any power using rotations Multiplying by i mean π/2 rotation So i^n is simply nπ/2 rotation Which would be equal to e^inπ/2 i^i = e^iiπ/2 = 1/√e^π

[@ReubenMason99]

One of the following: i^i, (i^i)^2i is real. Same proof as irrational^irrational

[@xgplayer]

is i irrational?

[@TheUpriseConvention]

Sergio Garcia It's irrational, as you can't put it in the form a/b where a and b are integers; the definition of a rational number.

[@Gold161803]

TheUpriseConvention Gaussian integers are complex numbers z for which Re(z) and Im(z) are both integers. I imagine there's an analogous definition for a Gaussian rational. I've never seen this in practice though, so I can't be sure. Doesn't matter though, since rationality plays no role in the original commenter's argument

[@Cannongabang]

You are right i^i is either complex non real or real If it is real we are done If it is complex non real, (i^i)^(2i) = i^(-2) = -1 so we are done. Cool one. But I'm still not too sure if exponentiation of complex numbers is "well defined"

[@xamzx9281]

Reuben Mason i^i is real, i=e^(i*pi/2), i^i=e^(i*pi*i/2)=e^(-pi/2)=real

[@ashishpastay8711]

This can be done in more easier way... Since, we know that e^(iπ)=(-1) e^(iπ)=i^2 Taking natural log on both sides iπ=2*ln(i) Multiplying (-i) on both sides, (π/2)=ln(¡^-i) Taking exponential on both sides e^(π/2)=i^-i Multiplying (-1) to the power s on both sides (i^i)=e^(-π/2)... Thankyou.

[@vibhanshuvaibhav2168]

This is what I did.. Let i^i = t log(i^i) = log t ilogi = log t Now, e^iπ/2 = i So, log(i) = iπ/2 Then, log t = i^2π/2 = -π/2 Hence, t = e^-π/2 = i^i

[@shayanmoosavi9139]

That's when we already know what is ln(i). He did the same thing but he also explained what is the ln of a complex number. Also I think you made a mistake in the second last line. It should be ln(t)=i.iπ/2=-π/2 log notation is confusing because it's the logarithm with the base 10. If you mean natural logarithm use ln instead. I know some of the math tools like MATLAB use log for natural logarithm and log10 for base 10 logarithm but we use ln in standard notation.

[@emree5962]

How can you keep adding 2pi? Is the formula not valid for -pi

[@bloodyadaku]

Since we know that e^(iπ) = -1 and i = √(-1) couldn't we just substitute the i in the base with √(e^(iπ)) so that i^i = e^(iπ)^(1/2)^i ? And then from there you just multiply the powers in the exponents so you have e^(i*iπ/2) = e^(-π/2)

[@bonbonpony]

My thoughts exactly. But I guess that then we wouldn't have some fun with complex logarithms :q

[@thatwhichislearnt751]

You are using (a^b)^c=a^(bc), which is not true, in general. You learned this rule for reals and positive base, but it fails for many examples with complex numbers. Therefore, multiplying the exponents is an improper argumentation.

[@vincentsmyang]

@@thatwhichislearnt751 Could you give an example?

[@TheDslide]

I found it like this maybe you find this interessting : i^i = e^(i*ln(i)) = e^(i*ln(sqrt-1)) = e^(i*1/2*ln(-1)) = e^(i^2 *pi/2) (∵ ln(-1) =i*pi) = e^(-pi/2). Feel free to point out any wrong steps , im trying to learn *thumbs up*.

[@wilhelmt.muller170]

complex to the power of complex: Exist GER: YOU WILL NEVER BE REAL

[@luxaeterna5281]

I'm not good at maths, but the simple fact that i^i seems a little face crying made me figure a thing or two before watching

[@oualidezzoubeidi1149]

Loved the video! would also work if we just replace i with e^i(pi/2) then basically i^i would become e^(-pi/2)

[@shomen96]

6:18 right here, we have e^[(pi/2)(i^2)] and if we have a power of a power then we can multiply. So it can be written as e^[(pi/2)(2i)] Which gives us e^[(pi)(i)] which is -1

[@KennyMccormicklul]

blackpenredpen: right? me:

[@tomdanub4231]

Lol!

[@knuckles140]

Another simple way: 1) i --> cis(pi/2) --> e^(i*pi/2), arctan(pi/2) --> undefined 2) e^(i^2*pi/2) e^-pi/2

[@harshit3964]

He is holding an additional brain in his hand that is giving him extraordinary powers to solve problems ! 😂

[@geralds361]

Even simpler, just using Euler's identity e^iπ = -1 by direct substitution: i = (-1)^1/2 i^i = (-1)^i/2 = (e^iπ)^i/2 = e^(-π/2)

[@vancecollie6833]

Out of curiosity, whenever you add 2(pi) for each additional rotation, wouldn't the e's exponent eventually approach some sort of limit? In that case, could we find some sort of infinite sum that could give us another definition of i^i since you could theoretically have any amount of infinite rotations when considering this problem? Regardless, great video! Thank you for teaching me something!

[@hefesan]

The limit is just 0, but the infinite sum converges at about 0.00038

[@penghuiyu5538]

The way he gets pi/2 is kind of wrong in Complex Analysis. What I would say is that I would choose the principal branch for Ln(x), then the corresponding theta is pi/2 by default. (You really need to know CA to understand that.)

[@lmmartinez97]

It's such a same that you have to be so careful in order to not offend the internet police and avoid the bullshit they can put out. You transmit passion about what you do and seem to genuinely love math, but some people just have too much free time and they look for every tiny informalities so they can whine about something they probably wouldn't do themselves. I know it's not my responsibility, but I do apologise for it. Please, keep making these videos.

[@Ounaide]

Wait, what are you talking about ? Sounds like I missed something here

[@grogcito]

Luis Miguel Martinez he named the imaginary axis "complex axis" on his sin(z)=2 video

[@Hepad_]

I agree with you but many math people would consider his answer false or at least incomplete if he didn't talk about the 2pi modulo

[@RomeForWar]

My prof in University wouldn't let you pass an exam if you ignored multiple solutions to a problem. Some people can be annoyingly pedantic, but it's also true that when doing maths you should always be as complete in your proof / answer as you can be.

[@ulilulable]

Yeah, while I agree with Luis in most cases regarding the damnable internet PC thought police offense stealing nonsense, when (and possibly only when) it comes to mathematics and logic, strictness is essential. I would think and hope that blackpenredpen also knows this and won't get discouraged by mathematical corrections. :)

[@nickcunningham6344]

The tricks you use to reach a solution are so clever they feel illegal. I love it

[@mattgillespie6457]

Do you have a major in Mathematics? Also great video. I'm planning to do engineering with a possible math minor for fun/ semi-practical uses. This channel has helped me find a passion:)

[@tcocaine]

Same. very interesting problems indeed, and as a norwegian high school student, I'm learning MASSIVELY from it. Learning imaginary numbers before even being taught it, is benefitial.

[@Ounaide]

He is a math teacher

[@tcocaine]

+Ouanide That's not an answer to his question.

[@Ounaide]

TheDucklets I don't recall ever talking to you.

[@tcocaine]

So what? I can still correct you.

[@yante7]

i^i = ? i polar form (dist 1, ang pi/2) = e^i(pi/2) thus: i^i = (e^i(pi/2))^i = e^(i*i)pi/2 = e^(-pi/2) technically, ang is +2(pi)n for n ∈ ℤ, so: i^i = e^(-pi/2 + 2(pi)n)

[@michaelc.4321]

I read the thumbnail as 1^(-i)

[@estebancanizales3303]

You're great ive never seen these mathematics and you help me piece it all together

[@leoneschle3112]

Isn't it crazy, if you think about it? You have one expression, i^i, but it has infinitely many values! It's like a Super-Schrödinger's cat!

[@2FaceGames]

I actually got the same in a much simpler way, simply using euler's identity: e^(pi*i)=-1 to solve ln(i), we gotta solve for x in e^x=i sqrt(e^(pi*i))=sqrt(-1) (e^(pi*i))^(1/2)=i e^(pi/2*i)=i therefore ln(i)=pi/2*i

[@holymotherduck3636]

complex^complexe = real ? mmmhhh i don't know, this seems kinda complex to me...

[@uberswine]

Well, this is the most epic redemption ark I've ever seen! Thank you for your videos, I watch them with awe in my heart

[@Miguelista1000]

Roses are red Violets are blue There's aways an asian Better than you

[@DashRevoTV]

*always

[@borisdorofeev5602]

If there's always an Asian Better than you Life doesnt matter Commit Sepuku

[@goodplacetostart9099]

Violets are blue ! Great

[@2tri749]

Then just FIND YOUR PASSION AND WORK HARD ON IT. Success will follow you! :)

[@phorinzyx2866]

2Tri r/woosh

[@chikitabanane9086]

i swear that you're a genius... My brain has justed been over used at the end

[@Sid-ix5qr]

10:17 Where's my Math Teacher?

[@OptimusPhillip]

By Euler's formula, we know that i can be written as e^(pi*i/2). Thus, i^i = (e^(pi*i/2))^i. Raising an exponential to a power is equivalent to multiplying the exponent, so i^i = e^(pi*i*i/2). i*i = -1, so i^i = e^(-pi/2), which is a real number. I'm still paused at the start, so I don't know if this is mentioned in the video, but the description mentions i^i^i, so I'll give that a try. i^i^i = e^(-pi/2)^i. Again, multiply the i into the exponent, and you get e^(-pi*i/2). Back to Euler's formula, this (e^(pi*i/2))^-1, or i^-1, or 1/i. 1 = -1*-1, so 1/i = (-1*-1)/i, which by the associative property is equivalent to -1*(-1/i). -1/i = i, so this is -1*i, or -i.

[@Sovic91]

Me, after seeing the thumbnail: "I don't need sleep. I need answers"

[@steffssnapshots]

xDD

[@ricardomilos7419]

Great video! I have a question: Is it correct to say that: 1^π = cos(2*n*π^2) + i*sin(2*n*π^2) ?

[@AndrewErwin73]

I love it... "this will cause a lot of arguments in the comment section..."

[@amurss8249]

You can also use polar form: i=e^(iπ/2) so i^i= (e^(iπ/2))^i =e^(i*iπ/2)=e^(-π/2)

[@oenrn]

So, something imaginary, when given imaginary power, becomes real? Cool!

[@abdullahpatel2612]

Wow nice way to sum it up

[@NathanSMS26]

This is a very complicated way to solve the problem. If you convert i into a different representation of it in the argand plane (like going from Cartesian to polar coordinates) you get i = exp(i*pi/2) then raising this to the i power lets you change i^i=exp(pi/2*i^2)=exp(-pi/2). If you’re familiar with different forms of complex numbers then this is mental math

[@John----Smith]

great teacher, and your english is good.

[@blackpenredpen]

Thank you.

[@qwicy]

"Pause and try it first." lol

[@findystonerush9339]

lol

[@AhmeddIbrahim]

Can someone explain how he got pi/2 for the angle at 5:50

[@kristofersokk1580]

Ahmed Ibrahim theta is the angle, for "i" it is 90 degrees, aka pi/2 radians

[@scitwi9164]

We measure the angles with regard to the direction of the unit (the number `1`), which is assumed to be at the angle `0` radians. Then you measure the rotation of that unit, counter-clockwise. E.g. the number `-1` is at the angle 180° or `π` radians to the unit. The imaginary unit `i` is half-way there, because it is a unit perpendicular to `1`, so it is at the angle `π/2` to the unit (or 90° if you prefer degrees).

[@rastrisfrustreslosgomez544]

Radianes are the unit of choice when dealing with the complex plane. 360° degrees (a full revolution) equals 2Pi, from there it´s just algebra

[@weeseonghew2]

in radian form π is basically a equivalent of 180 degree, hence, π/2 is basically 90 degrees

[@goodplacetostart9099]

In imaginary axes angles change e^πi=e^180i. ....(1) π=180(in imaginary axes) π/2=90 You can check equation (1) in Wolfram Alpha Computational Intelligence

[@GlobalWarmingSkeptic]

Pi is like that friend that pops in at the most random times just to say hello.

[@chatherinehu3804]

I have a easier way to solve it you can express in the r e (angle) form

[@johnny_eth]

Any complex number with modulo 1 to the power of i produces a real number equal to e^-θ or if you want to be pedantic e^(-θ+2nπ) for integer n.

[@6359]

"2 pi n or 2 n pi, up to you" I always write n 2 pi, I feel excluded

[@giovannipelissero1886]

In Italy we write +2kpi, I know: k is the Satan's son.

[@karryy01]

In vietnam we write k2π.

[@robert_wigh]

Yeah, we write exactly the same in Sweden: n*2π

[@malak9759]

I thought about it this way and as the result doesn't match the one suggested in the video, I don't know what went wrong in my approach: i^i= e^i.ln(i) =e^((1/4)i.ln(i^4) =e^((1/4)i.ln(1)) =e^0=1

[@huangkuantun]

what did i see at the beginning???

[@blackpenredpen]

: )

[@jge27455]

Maybe you should explain why we can write ln(i). Ln is define on ]0;+infinity[ to R. e(i.theta) is a definition/notation.

[@ДмитроПрищепа-д3я]

Just define ln(x) as a number/numbers, that e^ln(x)=x for any complex x.

[@Craznar]

Well, if 3 is real, the i is imaginary ... for it to be complex it should have non zero real and imaginary components :)

[@BloodyxScy]

Well, since the real numbers are just a subset of the complex numbers, all real numbers are both real and complex. why should it have non-zero imaginary and real components?

[@franzluggin398]

If you look at the last video blackpenredpen uploaded, you see it's "(irrational)^(irrational)=rational?". If we look at this video as a continuation of that last one, then I think we can infer the line of thought that led to this one. What is my point here? Irrationals are all the numbers that are real, but not rational. We know there exists a rational base of the real numbers, but I doubt it is possible to write down explicitly, as it is an uncountable base. Therefore, we simply divide the reals into rationals and irrationals, even though we could write them as a direct sum of the rational numbers and the rational base of the real numbers: we just don't bother. Therefore, the last video answered the question whether it is possible to satisfy an equation of the form: (non-rational)^(non-rational)=rational. This video, then, answers the question whether it is possible to satisfy an equation of the form (non-real)^(non-real)=real. Unless I misinterpreted your point, it boils down to "why should you be allowed to use purely imaginary numbers if you exclude purely real numbers?". The answer is: because we look at all the complex numbers that are explicitly not real, in analogy to the last video's real numbers that are explicitly not rational. The fact that we can easily split the complex numbers into a real and imaginary part because their real base is so easy has no bearing. Just as it did not in the last video. Sorry for making this longer than it probably needed to be, but I hope I made my point clear.

[@Craznar]

'if 3 is not complex, but real' then it follows that 'i is not complex, but imaginary' ... there was a conditional in my original comment.

[@thomasg6830]

Christopher Burke Agree. A solution that is not located on any axis of the Gaussian number plane would be interresting.

[@alxjones]

Christopher Burke Let z = e^(a+bi) with a and b nonzero real numbers and b NOT a multiple of pi/2. Then take w = a-bi. So, z and w are complex numbers with nonzero real AND imaginary parts, with z^w = [e^(a+bi)]^(a-bi) = e^[(a+bi)(a-bi)] = e^(a^2+b^2) which is a real number.

[@davidbrisbane7206]

No pigeons were injured in the filming of this video.

[@marcs9451]

I really wanted to know How much is i^e

[@guktefngrshoo7465]

call me daddy that’s alt right imagery you got as a profile pic. Get help, please. I love you

[@SpeakMouthWords]

a kind of ugly complex number somewhere between -1 and -i on the unit circle in complex space

[@mdyusufzaman9356]

This is so BEAUTIFUL something that doesn't even exist has connection with something that exists

[@liliacfury]

Wow! I'm pretty much a 13 year old and can kinda follow whats going on here. I love how you teach!

[@blackpenredpen]

Yay!!!

[@federicovolpe3389]

Yash 2223 I’m 14 and I can follow, having watched a lot of videos and having read a bit over the argument.

[@shayanmoosavi9139]

@@yashuppot3214 the idiot one is you who think that children can't learn complicated concepts. If it's explained correctly then even a six year old will get it.

[@togbot3984]

Here is another way of calculating ln(i) as: e^(iπ)=-1 Take square root both sides e^(iπ/2)=i Take natural log on both sides ln(e^(iπ/2))= ln(i) iπ/2=ln(i)

[@oldfire3107]

Simpler proof: e^(i*pi/2)= cos(pi/2) + i*sin(pi/2) e^(i*pi/2) = i e^(I*pi/2*I) = i^i

[@Mrwiseguy101690]

+JT He used Euler's formula to derive an alternate definition of i. i = cos(pi/2) + isin(pi/2). This is an elementary definition that is easy to derive if you understand Euler's formula and the complex plane. It is most certainly a proof.

[@franckdebruijn3530]

I loved the video (as I like all of them). This one kept me wondering though that a simple power-operation can lead to multiple results. In the Real domain, this does not happen, or does it? We do not say that 9^(1/2) is 3 or -3, but is really 3. It always gets me confused. Since, when solving quadratic equations and applying the square-root operation then we do want to keep track of the possibility that the square-root of a number can also be the negative counterpart ...

[@crunchamuncha]

As n gets bigger, i^i approaches 0. How is that possible? Could you explain why it happens?

[@bonbonpony]

Because the exponent is negative, so this is basically 1 over something. And if that something (the denominator) gets bigger, then `1` is being divided into more and more pieces, which gets smaller and smaller, along with the entire fraction, until they vanish at 0.

[@crunchamuncha]

Bon Bon yeah, I understand that. But my question is how is it possible for i^i to have different values for different values of n.

[@bonbonpony]

+crunchamuncha My friend will be making a series of videos on complex numbers soon, and this question will be explained there too. I'll let you know. If you have any other questions regarding complex numbers, something that you always wanted to know, or that bothered you, or that was hard for you to understand, feel free to ask it here, I'll send those questions to my friend so that he could explain them too in his videos and make them more useful to people ;)

[@jidma]

yeah I have that same question: if for any integer k i^i = e^(-pi/2 + 2*k*pi) then for k=0 : i^i = e^(-pi/2) and for k=1: i^i = e^(-pi/2 + 2*pi) which means: e^(-pi/2) = e^(-pi/2 + 2*pi) e^(-pi/2) = e^(-pi/2).e^(2*pi) 1 = e^(2*pi) and that is not true... how did that happen?

[@jidma]

+Bon Bon I received a notification about a reply from you but I can't find your reply here.. weired

[@mojolotz]

Mathematicians seeing "i over i": "I can fill some blackboards with this easy".

[@ΜΙΧΑΗΛΚΑΤΤΗΣ]

so the number i^i has many values?......

[@blackpenredpen]

ΜΙΧΑΗΛ ΚΑΤΤΗΣ yes!

[@wurttmapper2200]

ΜΙΧΑΗΛ ΚΑΤΤΗΣ Yeah, is like sqrt of 2 or sininv of 90. Nice greek, Michael Kates

[@BigDBrian]

no, but ln(i) does.

[@franzluggin398]

It is a number that you get from solving an equation. An equation that happens to have infinitely many answers. You have that in the reals as well. sqrt(4), for example, is usually assigned the value 2, but it could as well be (-2), since (-2)^2=4 also solves the equation that the sqrt function seeks to solve. Then there is integration, which has _uncountably_ many solutions (+C, C in Reals). And in that case, you probably already know how you can make the answer unique: Just add another condition, in the case of integration a starting value, in the case of i^i the restriction that theta has to be in [0,2 pi[, for example. Just keep in mind that there are other solutions, in case you need them sometime.

[@bonbonpony]

Not the number. The operation. Exponentiation can have more than one answer if the exponent is not a real integer (nth roots are an example of that, because they have fractional exponents). `i^i` is not a number, it is an operation (exponentiation), so it can have more than one answer, since the exponent is not a real integer. Each of these answers is a single number on its own.

[@techsinc]

Much easier if you note exp(i*pi/2) = i. Then i^i = exp(i*i*pi/2) = exp(-pi/2) = 1/exp(pi/2). No need to invoke ln(i), etc.

[@1_1bman]

try irrational^rational = rational^irrational

[@MuffinsAPlenty]

That's a fun one! How about this: sqrt(3)^4 = 2^(log2(9))

[@profdrsumitbanerjee2641]

This sum you can do in one line also. Because i=cos(pi/2)+isin(pi/2)=e^i(pi/2).Hence i^i=e^(-pi/2)=0.20788 (approx).

[@morganmitchell4017]

I've been wondering for a while. What age group do you teach? I understand most your videos, but not the second order differential equations or the more complicated series

[@maxguichard4337]

I think he just does what he finds interesting/ what is recommended.

[@shayanmoosavi9139]

Second order differential equation is taught in university. We're learning them right now. In order to know what is a diffrential equation you should first learn calculus. You should know derivatives and integrals. An n'th order differential equation is : F(x,y,y',y'',...,y^(n))=0 Which means an expression of x, y which is a function of x, y' which is the derivative of y with respect to x, y'' which is the second derivative of y with respect to x,...,y^(n) which is the n' th derivative of y with respect to x. (n should *only* be in parentheses so as not to get confused with powers). An example is y+y''=0. I really don't want to get into details because I don't know your mathematics background. Hope that helped.