i-th root of -1

[@blackpenredpen]

Learn more complex numbers from Brilliant: 👉 brilliant.org/blackpenredpen/ (20% off with this link!)

[@ATIKURRH]

Hello sir, my name is Atiqur Rahman. I am a trader. I want to build up an AI Algorithm trading software system on trading, I have been trying to solve a math calculation on trading for a long time. I request your cooperation in this regard

[@teamruddy611]

If theta is pi, then the complex number is just negative r.

[@bryan8038]

Formula for √x - √y = √xy

[@kafrikotroll8610]

What's your opinion on this kzbin.info/www/bejne/ipW8mmZ8fbtqeJY About his theories? 🤔 Is he right?!🤯

[@kafrikotroll8610]

What's your opinion on this guy kzbin.info/www/bejne/gny8kmibZayJick and his theories is he saying the truth?🤔 Ty VM sifu.

[@geboaebo]

"This rotation which is 180 degrees, but we're adults now, so let's use Pi" I didn't expect to laugh in a math video

[@joyli9893]

Like pie

[@chillx656]

360° = 2pi?

[@geboaebo]

@@chillx656 ye

[@fizixx]

My first exposure to this was of the form: i^i and the question asked to give the results....a real number. I was fascinated with that ever since. An imaginary number raised to an imaginary number gives a real number result.

[@herbie_the_hillbillie_goat]

That blew my mind too. Then I remembered the product of two negative numbers is positive and i^i kind of lost its charm.

[@quantumsoul3495]

@@herbie_the_hillbillie_goat Or basically complex numbers operations are rotations and scaling in the complex plane so it's not really incredible that some times they fall onto the real axis

[@cmilkau]

Imaginary to imaginary power even *always* yields a real result. Compare to irrational number to irrational power, which can yield an integer, but when picked at random remains irrational.

[@manioqqqq]

1.559605... Let's call it x. Then x^x would be 2, even tho x is trancendental. This thing DOES happen.

[@herbie_the_hillbillie_goat]

@@quantumsoul3495 Exactly! It is pretty fascinating early on when you're first encountering complex numbers. These days when I see something "inexplicable" happing on the real numbers, I look to the complex plane to see where the peculiarity is coming from. Rotations on the plane is also why it's not surprising to see pi everywhere either.

[@aguyontheinternet8436]

The challenge question is a trick question! The principle root of (-1)^pi is a transcendental number in its simplest form! Sure you could give decimal approximations like -0.9026853619 - 0.430301217i, but I think we can all agree that just like pi is usually best left as pi, and e is usually best left as e, (-1)^pi is best left as is. Fun fact, if you graph all possible roots of this number, it makes a circle in the complex plane with radius 1.

[@iiiiii-w8h]

a circle only if pi is normal

[@dinosaric4862]

How would you graph the roots of a number?

[@glitchy9613]

@@dinosaric4862 He means all solutions, like how square root of any number has 2 solutions

[@klementhajrullaj1222]

And (-1)^e ??? 😀😉

[@Rhovanion85]

A transcendental number powered by another transcendental number + 1 equals absolutely nothing. How crazy is that?

[@BobY52944]

I love how you started with the basics of the coordinate systems. So many people jump to the equalities and the basic principles are lost. A+

[@avrahm9029]

I have no idea what this guy is talking about but its addictive.

[@slytherinbrian]

(-1)^pi = cos(pi^2) + i sin(pi^2) is the principal solution, but I think the general form would be (2n+1)pi^2 inside the cos and sin, n in integers, right?

[@blackpenredpen]

Yes!

[@hervesergegbeto3352]

(-1)^π = -e^(π+2k)

[@hellohabibi1]

@@hervesergegbeto3352 There is no negative.

[@davekensk8]

Thank you for posting a lot of videos and engaging with us in the comments, especially knowing the fact that you have over 1 million subscribers is amazing. I've been having a rough time pursuing my college degree because I am so intimidated of taking Calculus 1 but the more I watch your videos it made me feel that it's not that bad after all. If only you could make more videos about getting into calculus as a complete beginner that would be nice! But anyways, You're blessing in this world and thank you for keeping up the grind of all of us!

[@blackpenredpen]

Thank you very much!

[@getnie6867]

@@blackpenredpen 4:21 shouldn`t it be 2npi/i?

[@user-db4lk7yg3o]

@@getnie6867 parenthesis

[@j.u.4.n620]

Challenge 1:)prove leibnitz theorem of geometry by using complex numbers

[@Mathologix]

Is there any libnitz theorem in geometry ? If so then state it

[@j.u.4.n620]

@@Mathologix now I've a challenge for u Find the statement..😅lmaooo

[@aaaaaattttttt5596]

@@j.u.4.n620 lmfao

[@Zombatt]

It's very straightforward and therefore left as an exercise for the reader.

[@adiaphoros6842]

Did you mean “A Geometrical Theorem of Leibniz”? Because that’s the only one I found on Google.

[@Imran-Shah]

2:46 You use the rule to convert from radical form to exponential form with a negative base. Forget about the i-part for a moment. This conversion has to be dealt in a very delicate manner when the base is negative as there are a plethora of examples where this move is unjustified. While I am not disputing your work, this "quick" move doesn't sit well. What justification is there to make this move when a complex power is involved along with a negative base?

[@joyli9893]

Chapters: 0:00 what’s the i-th root of -1 5:30 check out Brilliant 6:08 challenging question (-1)^pi

[@joyli9893]

Wow! Somebody liked my comment already!

[@musicmakelightning]

Coolest thing I have seen all day. I wondered about this but never sat down to try to figure it out myself. Thanks so much. Love your videos!

[@neilgerace355]

1:57 Ah, I'm glad we are all still adults :) Most of the things I have seen on your channel, I think I learned in high school or first year university. Most of those things I forgot until you reminded me in your videos. But some things are new to me, such as the Lambert W function. Either way, I like being educated or reminded :)

[@aguyontheinternet8436]

real chads use tau/2 lol

[@diamondnether90]

(-1)^pi = e^(pi ln -1) = e^(pi * i pi) = cos (pi^2) + i sin (pi^2) This is the primary solution.

[@fLaMePr0oF]

(-1)^π e^iπ=−1 and e^2πki=1 for integer k =(e^iπe^2πki)^π =e^iπ2^(1+2k) =cos(π^2(1+2k))+i sin(π^2(1+2k)) (i.e. infinite complex values for k) Let k=0: (principal) = cos π^2 + i sin π^2 ≈ -0.903... - (0.430...)i

[@thenickli]

OMG knowing Euler's identity and looking at the answer it all made sense! i'th root just means dividing the power by i, so e^pi(i) became e^pi

[@Nutshell_Mathematica]

Your teaching not just improve my knowledge They really motivate me

[@michaelwagner6877]

Great articulation, great explanations for conceptual techniques, and an overall easy to follow presentation!

[@Gabsleggy]

A really fun fact is that, as a french fan, when I'm enjoying math class I write my equations with your voice intonation in my head 😂

[@pouyamovie3253]

First of all love your videos, they are very informative and makes math enjoyable. Math is the biggest "what if" ever 😁

[@walls_of_skulls6061]

I have never seen polar written like that! Nor rectangular be called standard form. I just know polar as r

[@abdulmalek1118]

Hello ! I hope you see my comment I saw this nice question so that I recommend it The question is : solve the system of equations a = exp (a) . cos (b) b = exp (a) . sin (b) It can be nicely solved by using Lambert W function after letting z = a + ib Hope you the best ... your loyal fan from Syria

[@hearstboy]

The math was cool! But I couldn't help but be amazed at the magic he pulled off with his markers! I only ever noticed two markers in his hand, yet he was able to write with 3 colours!

[@williamhu9567]

The tiny mic is an added bonus :)

[@kdgcovers7530]

Congrats on hitting 1 million subs!! love your videos 🙂

[@blackpenredpen]

Thank you!

[@ultimatous4588]

I am in 7th class and 12 years only and you taught me calculus. Thnx

[@Macau_Ball]

same but ur 8th class now😮

[@shahzadahmed6160]

Following this channel when it have only 20k subscribers.... amazing journey.thanku

[@adarah00]

Love your channel bro 😊

[@bernardobuffa2391]

Your are an incredible teacher. Even I got it! thanks!

[@scraps7624]

Your marker technique is just incredible lol

[@GameMaster-pz9pw]

I've been watching your videos for a while and usually there's a lot I don't understand, but I started pre-calculus this year and I'm slowly starting to understand more and more of what you're saying lol

[@aflahmm2784]

My fav channel after learning basic calculu😊💪🏽💪🏽🔥

[@0xABADCAFE]

I was blown away by the sudden introduction of the blue pen.

[@mrajsatyam]

You are my best teacher 🐱

[@kovaxim]

I haven't been studying these things for over 10 years and it's still interesting to see and think about, try to solve even though you have no idea how or even what to do

[@galus_anonimus]

Hi! I was wondering if you could give me some information about the whiteboard you're using. I'm looking for one rn and it would really help me out. Thanks

[@aidkik580]

Dude my vision blurred I got dizzy and started drooling 2 minutes into this and by the end I could smell something burning.....

[@bryanlangley5337]

Many of these videos feel like me playing with my graphing calculator in high school. "Hey, what's the i th root of -1?" "Let's try it!" "What does this even mean?" "I don't know!"

[@donovanmahan2901]

-1^pi could sweep along the complex unit circle, with the principle value being e^(pi^2)i. It can't be the entire complex unit circle as you'd end up with a countably infinite set against the continuum that is the complex unit circle.

[@5ilver42]

the multiple answers connects my brain to those spiral shapes I saw in a video about graphing x^x for all values of k, when plotting the imaginary axis off into a third dimension perpendicular to x and y.

[@5ilver42]

I failed my high school pre-calc courses almost 15 years ago, so it's nice to see that these concepts can make a minor amount of sense to me all of this time later.

[@igarciaasua9]

(-1)^π We know e^(iz)=cos(z) + isin(z) So e^(iπ)=-1 Then we elevate it to π (e^(iπ))^π=(-1)^π So (-1)^π=e^(iπ²) But we know that with complex numbers e^(iz)=e^(i(z+2π)) Then (-1)^π=e^(i(π²+k2π)), k=0,±1,±2... In polar form: radius=1 Angle=π²+2kπ rad, k=0,±1,±2... In binomial form cos(π²)+isin(π²)

[@jderick17]

Isn't there technically only 1 solution to the problem, not infinite? I get that (-1)^(1/i) would have infinite solutions due to the argument being of the form pi + 2n*pi, with n being an integer. However, since the initial statement used the radical symbol, it means the principal value only, right?

[@Syndicalism]

The initial statement (-1)^(1/i) didn't use a branch cut making it the principle root. The "known" statement is there to show the only information expected of us to know for this problem but isn't the initial statement. We can make a branch cut to make the solution finite but without a branch cut there are infinite solutions due to periodicity of the complex exponential.

[@jeffimber7152]

I don't think that i = \sqrt{-1} is really Kosher though! Complex numbers as far as I know do not really have "principal roots" because i is neither positive nor negative, much like the complex number 3 - 2i. I think it would be better to say "there are two solutions to z^2 = -1 and they are +i and -i."

[@lukandrate9866]

I will discover USA but the radical symbol is only intended to be used with a natural power(it is defined so). Otherwise, you should use ^(1/x)

[@JohnDoe-m8i]

I thought you couldn't multiply the powers with complex numbers?

[@Neptune_A_Planet]

I am in grade 11 and I solved it using euler's identity ( e to the power i π +1 = 0 )

[@findystonerush9339]

I'm grade 5 and I know that e^i*pi=-1

[@Neptune_A_Planet]

@@findystonerush9339 how? What's your age

[@cmilkau]

x^a = b (a,b,x complex numbers) in general has multiple solutions for x, sometimes infinitely many (e.g. when a=i like in this case). How is "the" answer for b^(1/a) selected?

[@atlantiz4120]

I understood everything without knowing English amazing!

[@szymongrzebyk1964]

Really nice showcase how everything connects in maths and has impact on each other. One thing about roots in this field of math tho, using trygonomic notation one always get infinite number of results since used agle is periodic. The more interesting thing is that in complex numbers you have actually as many results as the number before root symbol. For example if you have sqrt of 1 it can be 1 or -1, sqrt of -1 is i or -i but 3rd root of -i will give you 3 results (i and 2 others results, (-sqrt(3) + i)/2 and (sqrt(3) - i)/2). I talk about principal values ofc because mentioning infinite amount of results every single time is unnecessary

[@mathematician8113]

Can you do more double integrals in a playlist with changing the variables and polar coordinates 🌚and thx💛

[@YahyaKhashaba]

Please tell me why we can't do this to the exponent: 1/i = 1/SQRT(-1) = SQRT(1)/SQRT(-1) = SQRT(1/-1) = SQRT(-1) = i. So exponent is just i since 1/i = i

[@osamaaslam4923]

May I use any material of your video for my research ? No copyright issue? Kindly inform me.

[@enisheadpay]

I find it so interesting that even though i and -1 both have unit length, taking the ith root can so drastically affect the magnitude (the primary root being ~23.14... and getting closer and closer to infinity as you go up, while going the opposite direction starts at 0.04321... and gets closer and closer to 0). When taking roots using real numbers, taking the 1st root keeps the same magnitude, going more positive increases it towards infinity, going more negative decreases it towards zero, and going into fractional roots shifts it into the complex plane but still the magnitude goes down. When taking imaginary (or complex) roots all that intuition goes out the window... I wonder if a similar operation to the root (or exponentiation in general) can be formulated where taking the ith "root" has the same invariant properties as taking the 1st root of a number. Almost like an abstract inverse or complex conjugate to exponentiation.

[@oferzilberman5049]

I swear mathematicians are just making things up out of boredom at this point

[@tupublicoful]

Do you have a pdf version of the poster you have behind you? Looks very useful for students.

[@constellation1769]

Your videos are good as usual, but what happened to the pokeball in your hand though, i didn't get time to watch for a few months and now its gone

[@getnie6867]

4:21 shouldn`t it be 2npi/i?

[@pinedelgado4743]

Awesome as a possum with a blossom stuff!!! I love this!!! Thanks lots for producing and posting!!! :) :) :)

[@herbie_the_hillbillie_goat]

Love your comment. 😁

[@HeyKevinYT]

the joyful way you commented this is what I imagine Anne Frank (your profile picture) was like before going into hiding!

[@pinedelgado4743]

Thank you both!! ;) ;) Anne Frank is one of my autistic special interests along with mathematics which is why I have a pic Miss Frank as my profile photo and why I enjoy blackpenredpen's videos so much! :) :) :) :)

[@srividhyamoorthy761]

Brooo ur such a magician , how do you figure out these kinda questions

[@srividhyamoorthy761]

Pl read my other comment @blackpenredpen

[@dushkin_will_explain]

Cool! That's why mathematics is so pretty beautiful.

[@barthennin6088]

Interesting! ..Alt soln - Take Euler's Identity e^ipi + 1 = 0... subtract 1 from each side and take the ith root of both sides and get e^pi = ith root of (-1)... but of course that only gives you the primary answer, not the 2pi*n...

[@Syndicalism]

Taking the ith root does give the 2pi*n solution. If you only want a finite solution you must make a branch cut.

[@Henrix1998]

This gave me an idea. Would it be possible to find non-trivial values a and b that satisfy the following two equations: a√x = b and b√x = a (a'th and b'th root). x could be any or specific number

[@blackpenredpen]

Yes. That’s equivalent to solving a^b=b^a and I have a video for that kzbin.info/www/bejne/hnqUf5h9qa-ibdU

[@leom8051]

I’m more used to see the : pi + 2npi as pi(2n+1) but it’s the same ( just personal preferences) Anyway, thank you for these explanation 😆

[@isjosh8064]

What’s the domain for the function f(x) = (-1)^x?

[@rihankota2021]

(-1)^π answer is e^(iπ²) Since -1 is equal to e^iπ From Euler's identity...

[@weggquiz]

very informative, i love it

[@388C4CGREEN]

I love how he says “n” as a variable. “2嗯派i”

[@WinterSoldier42]

Can you make a video on the integral of sin(ln(x))dx ???

[@Dom-kp6ur]

aren't you supposed to also multiply the 2npi by 1/i as well?

[@orenaharoni8763]

6:08 -0.902-0.4303i

[@rotemlv]

Interestingly,, you can also get the general solution using De-Moivre's root formula: -1 = cis(pi) -> -1^(1/i) = cis((pi + 2pi*k)/i) = e^i*((pi+2pi*k)/i) = e^(pi + 2pi*k) Though, I gotta wonder about the range - the formula limits us to a integer k in [0,n), yet here we have infinitely many solutions. I guess since i is imaginary, no integer can be defined as larger or smaller than it. I wonder how to justify that it works, lol.

[@GrandAdmiralMitthrawnuruodo]

Is it correct to say: i = (-1)^(1/2), because if i^2 = -1, but ((-1)^(1/2))*((-1)^(1/2)) = ((-1)*(-1))^(1/2) = 1^(1/2) = 1 but that is not equal to i^2 because that must be -i. Am I getting something wrong or wheres the mistake?

[@alibekturashev6251]

hey do this one pls find a such that a^x and loga(x) have only one solution

[@VictorGallagherCarvings]

WOW! What a great explanation.

[@Math0821]

I like your Film,very good

[@SpringySpring04]

As a joke, I will say that trying to calculate the i-th root of -1 results in: YOU DESTROYED THE FABRIC OF SPACE-TIME

[@kaveenshankar7740]

"It is a real no , but it is also a complex number " i dont know why , but it almost killed me

[@ryokucha_9101]

I'm Japanese junior high school student. This movie is very interesting. It make me like math than before. I want my friends to watch this movie…Oh, I forget that I don't have friends who understands complex number.

[@gheffz]

Beautiful... I didn't even "imagine" to do this... infinitely brilliant! Thank you!

[@gheffz]

I will try (-1)^Pi later!

[@Billonaire_saga1.0]

can you give a reason why (x)^n=1 has n complex soln when n is natural whereas it is not so when n is real

[@mohammadtaha7316]

Hello I want you to please solve this question using tabular method I'm so confused Integration Cosec^3(x)

[@septianhari582]

Does it mean e^pi = e^3pi? If we ln both side, does that mean pi = 3pi? Does that mean....1 = 3?

[@cubicinfinity2]

Is any indeterminate form plus zero equal to itself?

[@superlinux]

AAh!! after thinking about it, this is how you would represent a series or a sequence of real numbers.

[@DanBurgaud]

BlackpenRedPenBluePen . . . YEAH!

[@SavageGreywolf]

Steve just casually proving e^(i*pi)+1=0 in the first two minutes of the video in a way that a junior high schooler could understand

[@pardontheleft2692]

this was very neat, thank you

[@the_untextured]

If a root is a function, then it cannot have infinite results. What you gave is the result to the equation z^i = - 1 Rather than z=i th root of - 1 But still, for the rest, interesting!

[@Bhuvan_MS]

Then the root must yield "i" solutions to the problem. But how can you get "i" solutions? Tell me..

[@the_untextured]

@@Bhuvan_MS the rule is applied for integer powers of z. Not complex.

[@Syndicalism]

A root isn't a function. However, a root with a specified branch cut can be made into a function.

[@the_untextured]

@@Syndicalism yes, a root is a function. What you are thinking about is the inverse *expression* of a power, rather than a function. A function is a one to one relation or a many to one relation, which a root is.

[@aryanprasad2406]

Respected sir, I have a question that, why you didn't go with (2n-1)? Because we want odd integer series starting with 1?

[@Peter_1986]

(-1)^(1/i) = (e^[i⋅π])^(1/i) = e^(π). Actually fairly easy, as long as you remember the complex unit circle.

[@liyisu]

you surely have a lot of black and red pens in stock!!! :)

[@faizakhtar5329]

Please solve n^i. n is any natural number

[@stevenschilizzi4104]

Brilliant, as usual!

[@djttv]

How can we do this double integral: SS(xy/(x^2+y^2)^(3/2))dydx, where x and y are independent. Any ideas or videos on this? My results are not making sense, so I'm doing something wrong. Thanks

[@aguyontheinternet8436]

Now for another question. How would you compute an accurate decimal approximation of such a number

[@thecodingintern3032]

Ok I kinda did something I dont really understand, So what i did was take the sequence of natural numbers as such 1, 2, 3, 4, 5, 6, 7, 8....... and then formed another sequence by taking the first member of the sequence that is 1 and adding the numbers in the sequence to it meaning that first I would add 1 and then 2 and then 3 and so on, which makes 1, 1+1, 1+1+2, 1+1+2+3....... or 1, 2, 4, 7, 11, 16, 22, 29, 37..... then I would repeat the same step to form the next sequence 1, 1+1, 1+1+2, 1+1+2+4, 1+1+2+4+7 .... or 1, 2, 4, 8, 15.... and suppose I repeat this infinitely, the sequence it seems to approach is 2^0, 2^1, 2^2 and so on. essential 2^n why is it so that it approaches this sequence? please explain

[@elekitu2718]

Because 2^{n+1}=(2^n + 2^{n-1} + ... 2^1 + 2^0) + 1, it's easy to prove that if the first terms of the sequence are 1, 2, 4, ..., 2^n, then at the next step the sequence will start with 1, 2, 4, ..., 2^n, 2^{n+1} Since the initial sequence starts with 1, 2, it follows that after n steps it will start with 1, 2, ... 2^{n+1}

[@leadiodide8243]

1:57 BASED

[@lechatrelou6393]

I didn't know we could do this, and it's now giving me noghtmares

[@iljas275]

can you solve the equation y^2-x^3=1 in positive integers?

[@sberacatalin2250]

Exceptional! Excellent! Thanks! 👍😉🙏🇷🇴💎😍👀🙂👍

[@joseponce37]

Hey bprp, can you solve the integral of (e^x²√(x²-1)+1)/x²-√(x²+1)-1 ?

[@cesarrivas2295]

Please do the 100 trig questions 🙏

[@noamlachkar4817]

It does not make sense to use the formula exp(a)^b=exp(a*b) with complex notation, example : exp(2i(pi))^0.5=-1 but also =1.. the square root in this video is false you can get several others paradoxes using this formula