i-th root of i 

Do you enjoy complex numbers and up for a challenge? If so, check out x^x=i ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-vCdChDmMYL0.html

The most surprising thing about this is I understood more than half of what he explained

*Learns about imaginary numbers* RU-vid: What's 1/i? What's e^i? What's i^i? What's sqrt(i)? What's log(i)? What's i-th root of i? What's sin i? What's sin-1(2)? What's i factorial? What's integral of x^i?

0:32 "I don't like to be on the bottom, I like to be on the top" Me: *scrolls down to comments to see if anyone made a joke about that*

"i" understand this a whole lot better now.

Do the e-th root of i and the i-th root of e

“I dont like to be on the top. I like to be on the bottom” - blackpenredpen

Or you could just i = e^(iπ/2) ith root both sides i√i = e^(π/2)

Wow, complex roots of imaginary numbers are real......math is so weird!

As mentioned before, I love this channel. Found it randomly, but so glad I did. Currently a grad student taking complex analysis (as an elective, took it 2years ago). All of your complex videos are just supplementary/review for me, but they're still very satisfying to watch. Got any plans for singularities?

I just stumbled upon your videos while procrastinating. You're a really warm and enthusiastic talker, keep it up sir! :)

I'm in love with mathematics now

black mic red shirt yay!

The reason why i^i = e^(-pi/2) is because i^i = e^(log(i)i) = e^((pi/2)i * i) = e^(-pi / 2). There are infinite answers because the log(z) has infinite answers; as it is defined as ln |z| + arg(z)i, where arg(z) is the angle (which can be, in this case, pi/2, 5pi/2, 7pi/2, ...).

You're explanations are simple and crisp, really helping me learn mathematics better

This guy rekindles our faith in the fact that things that seem to be very tough, are actually very easy, we just need to take one right step, and everything else becomes so easy!

Drink game: Take a shot after he says the letter "i"

You are a math god

You could actually just start with e^(ipi)=-1. Take the square root of both sides: e^(ipi/2)=i. Then take the ith root of both sides. e^(pi/2)= i^(1/i).

I'm even more amazed that you can formulate these math questions with answers.

Who thinks new BPRP is more interesting and better? *raises hand*

e, pi, and i: The Zeus Poseidon and Hades of math

Another way to do this is using De Moivre’s theorem, where i on the complex plane would be pi/2, so theta is pi/2/i, so e^i theta would simplify to e^pi/2

I did some other method : Consider i^(1/i) and get i = (-1)^(1/2) and -1=e^iπ so finally : i=e^(iπ/2) so in the end we have i^(π/2) as the "i"s cancel out !

This video is amazing. Thank you for the break down!

I don’t even know how to begin to do this

0:33 that’s what she said😂😂

That was the question that i had not been able to solve in my exam thanks

Euler’s number and pi, name a more iconic duo

That ball that you always keep in your hand makes you look like an ood from DW xD

I'm going to use this information when I file my taxes.

but the real question is... what is i-th root of i based on Wolfram Alpha?

0:30 the quote of all time

YAY! This was interesting!

0:33 of course you do. Your entire channel is about doing hard work. Lol

nicely done.... the use of multi color to force focus reminds me of a math professor in the 1960's at the University of Maine.

Your voice is a lullaby 😭❤️

Sir there can also be my method: Let i^1/i=x Take ln both sides.. Now 1/i×ln(i)=ln(x)-----(1) Now as we know e^iπ=-1 And(-1)=i^2 e^iπ=i^2 then take ln both sides then (iπ)=2ln(i) Which equals (iπ)/2=ln(i)------(2) Now put (2) in (1)= {ln (i)=iπ/2} 1/i×iπ/2=ln(x)==== π/2=ln(x) so e^π/2=x I hope this can also be a solution...☺

Imagine if imaginary roots of other imaginary things could be real. Like trees.

i blinked and he got pi out of seemingly nowhere

Wait ur a teacher!?

I still find it odd that "i" with absolute magnitude of "1" can ever twist itself into something outside of that, as in > 1.

There are other solutions as well: (e^(pi/2+2*n*pi))^i = i for integers n.

00:32 Lost my concentration there

such an interesting edge case!

" I don't like to be on the bottom, i like to be on the top " (0:31) - blackpenredpen 2017

"What we do now? We do our usual business" if math was that easyyy

alternate method: i^(1/i)=x i=x^i now we know e^(i . pi/2) =i therefore i=e^(i . pi/2) = x^i so, x=e^(pi/2)

I just wasted 2 minutes being confused by a problem I never even knew existed and still don't understand

“I dont like to be on the bottom, i like to be on the top” - pen guy

0:30 "I don't like to be on the bottom. I like to be on the top."-Blackpenredpen, 2017

These just make me so cheerful..... I have to limit myself to one a day ( and then I'll start again from the beginning because I'm still forgetting stuff like when I was a kid. )

damn I didn't know about imaginary numbers more than how to write them and how to show them on complex plain. yet after watching several other videos of yours I managed to get this one by myself.

I love it when blackpenredpen features blackpenredpen.

Why so complicated though? Q: i^(1/i)=? Solution Rewriting the BASE i on the polar form, by analyzing the complex plane we get that r=1 and θ=π/2+2n, n∈ℤ. Therefore, i=cos(π/2+2n)+isin(π/2+2n). By Euler's formula this can be written as i=e^(i(π/2+2n)). Keeping the EXPONENT on it's original form, we can conclude that i=(e^(i(π/2+2n)))^(1/i)=e^(π/2+2n) by simplification A: i^(1/i)=e^(π/2+2n) By plugging in n=0 we get the answer i^(1/i)=e^(π/2), as presented in this video.

You can check its true, if you raise both sides to the power i, you get i = e^(i*pi/2) which is the complex polar form of a vector with argument pi/2 and magnitude 1, which is of course just i on the complex plane. The family of solutions are associated with all arguments that equal pi/2 mod 2pi, so any integer number of times around the units circle landing back at i.

You are the most brilliant math person I have ever come across. I am in awe.

And 5 years later: Just go straight for Eulers identity: i'th root of i = i^(1/i)=e^(πi/2i)=e^π/2 From 5 steps to 3 steps

I accidentally created a generalized formula for the i-th root of any complex number with a radius of 1 in school which consists of i-th root(z) = e^a, a is equal to the angle of the complex number in polar form. The first version of my formula is the i-th root(z) = i-th root(r) • i-th root(e^(i•a)), a is still the angle of the number in polar form (only works with the radius of 1).

When I saw the video I took a few minutes to think about it for myself and then come back to see if I was right and I was. Thanks I like keeping my mind sharp with little math riddles like this one.

'THATS IT' PROCEEDS TO VANISH IN THIN AIR

I don't even think about this kind of stuff. It's crazy that some dude was like, oh there's an i'th root of I, let's solve what it is...

This is so fantastic and beautiful. Thank you for sharing!!

You should probably also update this version with e^(pi/2+2*pi*n)

This guy : make a complex math things Me : 1/1-

Black pen blue pen would make it easier for my eyes to read. Thank you so much for your videos.

Bprp: "Now what? Do the usual business." Me (being a 2 year old): Shit myself.

hush, hush... i to i.. too shy, shy..

By the fundamental theorem of algebra, there are exactly n roots of nth power. So, there are exactly i roots of i :)

"Let me do a REAL quick video for you" I see what you did there

(X^2-5x+5)^(x^2+7x+12)=1 Find all real values of x. There will be 6 values of x. Love from India.

"technically, i^i has infinitely many answers" - that is crazy, considering that i is a number. That's like saying 2^2 or e^e has infinitely many answers! I love your videos about calculations involving i even though I could never solve them on my own. But your explanations are always clear and step by step. Thanks!

*Me trying to sleep* My brain: Hey dude, what could be i√i? Starts to do mental math, got it it's e^π/2, now time to sleep. My brain: are you sure? This is how I end here.

So you have this chanel too. Subscribed to this too ❤️❤️🔥🔥😂😂👍👍.

What would an ith root(x)graph look like?

If I have learned anything from this channel it is that if you something strange to i, e and/or pi are going to pop out of the equation.

That shirt is crazy!! Also, I love your vids ! 😄

Omg, u did it so easily. The way that I've done was making: Y=i^(1/i) ln(Y)=ln(i)/i then, i = e^(iπ/2) ln(Y)=ln(e)×iπ/2i ln(Y)=π/2 Y=e^(π/2)

That moment when you try to checkmate a grandmaster and he just literally flips the board. Great stuff man, I didn't this was a question that bothered me since 3 minutes ago but thanks for the answer anyways.

Damn math rooting an imaginary number with another imaginary number’. You get a real number 👏

Could also just write e^(pi i/2) = i and take the root of both sides.

The ad is trying to kill the youtuber 🤣

Well, _i_ root of _i_ is pretty easy to calculate. The _a_ root of _b_ can be rewritten as _a_ ^(1/ _b_ ) So, _i_ root of _i_ = _i_ ^( 1/ _i_ ) Now, 1/ _i_ = -i. You can figure that out geometrically / visually with ease. _a_ ^ ( - _b_ ) = 1 / ( _a_ ^ _b_ ) So, _i_ root of _i_ = _i_ ^( 1/ _i_ ) = _i_ ^ (- _i_ ) = 1/ ( _i_ ^ _i_ ) To calculate _i_ ^ _i_ , we can use one of euler's formulas. For starters, _a_ ^ _b_ = _e_ ^( ln( _a_ ) _b_ ), thus _i_ ^ _i_ = _e_ ^( ln( _i_ ) _i_ ) ln( _a_ + _bi_ ) = ln( length( _a_ + _bi_ ) ) + angle( _a_ + _bi_ ) _i_ , so _e_ ^( ln( _i_ ) _i_ ) = _e_ ^( ( ln( length( _i_ ) ) + angle( _i_ ) _i_ ) _i_ ) = _e_ ^( ( ln( 1 ) + angle( _i_ ) _i_ ) _i_ ) = _e_ ^( ( 0 + angle( _i_ ) _i_ ) _i_ ) = _e_ ^( ( angle( _i_ ) _i_ ) _i_ ) = _e_ ^( ( 1/2 _pi i_ ) _i_ ) = _e_ ^( 1/2 _pi i i_ ) = _e_ ^( -1/2 _pi_ ) becuase _i i_ = -1 Now, _e_ ^( -1/2 _pi_ ) All of that says _i_ ^ _i_ = _e_ ^( -1/2 _pi_ ) We still have to find 1/that. 1/( _a_ ) = _a_ ^(-1), so 1/( _e_ ^( -1/2 _pi_ ) ) = _e_ ^( -1/2 _pi_ ) ^(-1). Stacked exponents multiply, so _e_ ^( -1/2 _pi_ * -1 ). The negative cancel, resulting in: _i_ root of _i_ = _e_ ^( 1/2 _pi_ ). The last part here can be calculated. In conclusion, the _i_ root of _i_ = _e_ ^( 1/2 _pi_ ) ~~ 4.8105

Another way is to get this result is to convert i into it’s polar form, you get there a lot quicker

You could easily motivate this by analogy with a square root. Like square root of 4 can be found by finding a number a such that: a^2=4 ==>a=2. Similarly, the ith root of i means find an a such that: a^i=i ==> a=e^(pi/2)

I *really* think you need to explain the multivalued nature of complex exponentiation. It is really beautiful math but it seems like people resort notational tricks which work sometimes, but not always. And it doesn’t foster understanding

Could one also write i^(1/i) as (e^(pi/2)i)^(1/i), because e^(pi/2) = i according to Euler's formula, allow the i and 1/i to cancel out, and take the remaining e^(pi/2) as a final answer?

The whole point of the question was to explain that e term, and you bring that from within the thin air, solving the sums fast doesnt mean you miss the real part of the solution

The thumbnail has tortured me after a tough maths test. Stop

So much love given to - i = e^(iπ2(k+¼)) - in all of its forms and so little given to - i = -e^(iπ2(k-¼)) i^(1/i) = -e^(-π/2 + 2kπ) or e^(π/2 + 2kπ) You can't just add multiples of 2π to the positive result and declare victory.

Why did you multiply by i over i? You could have substituted e to the i pi/2 as first step. And then cancel out the i and get the result in 2 steps. :)

wow i thought it will be complicated but then its just simple by knowing what is i^i and the rest are just very easy

Of cooourse, (e^(pi÷2))^i is just equal to e^(i(pi÷2)) which is i in polar form.

Explaining this Our dude here became closer to being a pirate

raising to the power like that isn't allowed for imaginary numbers

The greatest crossover in history.

Sir, i have a question, We know i = √-1 If we have 1/i, can we write it as, 1/√-1, and then √1/√-1 => √(1/-1) => √(-1/1) => √-1, basically 1/i => i, can we write this? If not then why?

Me seeing the thumbnail: *horror noises* Kids in the intro: YAY!

My mind is blown and my curiosity satisfied

Shortest blackpenredpen video ever