same :D

Angel's Of Revelation , tf

We all

This is better than homework

Way better

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

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

Imao

You could say he was a complex person

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

Got the same thing when tried to solve it ;)

Pi/2 radian is already i vector.

It's the first thing that struck me

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

Me too :)

This is so accurate!

IAAGO ARIEL SCHWOELK LOBO relatable

Same here 😂😂

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

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

Lol

bruhther

@@damuddohonson2282 wtf

*I (capital i)

@@damuddohonson2282 bro…

it is!

@@blackpenredpen Isn't it?

Redchalkwhitechalk

@Fred The Llama yes

lmao

Wow. Nice!!!!!!

@@blackpenredpen Thanks In love your Videos

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

Nice hahaha

@@justarandomdudelol7702 why?

You'd love the MIT open course ware videos.

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)

Much simpler

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

+Alper Berkin Yazici Slow clap

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...

: )

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

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

@@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

@@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).

Eduard Van Beeck thank you!!!!!

*physically

@@spiguy theoretically*

@@Hydrastic-bz5qm all of the above*

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

@@es-rh8oo psychedelically*

yes

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.

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.

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

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

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

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.

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. :)

*always

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

Violets are blue ! Great

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

2Tri r/woosh

Yeah. That's exactly what I did.

...

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

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

Lol!

I learn about it from 3Blue1Brown

i just found the channel randomly

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.

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

is i irrational?

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.

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

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"

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

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.

He is a math teacher

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

TheDucklets I don't recall ever talking to you.

So what? I can still correct you.

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

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

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.

@@thatwhichislearnt751 Could you give an example?

xDD

Wow nice way to sum it up

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.

Thank you.

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

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).

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

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

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

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

In vietnam we write k2π.

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

Yay!!!

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

@@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.

He's Chinese so your joke doesn't work😂😂😂 But nice try though you made my day. Thanks :)

I wanted to make a Kouhai joke. But it's all still imaginary.

sin(π)

Another senpai is from complex world.

: )

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

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.

ΜΙΧΑΗΛ ΚΑΤΤΗΣ yes!

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

no, but ln(i) does.

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.

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.

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

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

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

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?

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.

'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.

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

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.

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.

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.

+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 ;)

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?

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

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

lol

+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.