i-th root of -1

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Күн бұрын

Learn more complex numbers from Brilliant: 👉 brilliant.org/blackpenredpen/ (20% off with this link!)
We know the principal square root of -1 is the imaginary unit i but what if we have the i-th root of -1? We will do a quick review on converting a complex number from the standard form a+bi to its polar form re^(i*theta). Then we will see the surprising answer that the i-th root of -1 will actually give us.
Related videos:
the proof of the Euler's identity (with the Taylor series): • Euler’s identity proof...
i-th root of i: • i-th root of i
🛍 Shop Euler's Identity t-shirt (as seen in the video): blackpenredpen.creator-spring...
10% off with the code "WELCOME10"
0:00 what's the i-th root of -1
5:30 check out Brilliant
6:08 challenging question (-1)^pi
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Пікірлер: 452

@blackpenredpen Жыл бұрын

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

@ATIKURRH Жыл бұрын

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 Жыл бұрын

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

@bryan8038 Жыл бұрын

Formula for √x - √y = √xy

@kafrikotroll8610 Жыл бұрын

What's your opinion on this kzfaq.info/get/bejne/isqJnJZ3rbybcmg.html About his theories? 🤔 Is he right?!🤯

@kafrikotroll8610 Жыл бұрын

What's your opinion on this guy kzfaq.info/get/bejne/grGJlJiWla26g5s.html and his theories is he saying the truth?🤔 Ty VM sifu.

@geboaebo Жыл бұрын

"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 Жыл бұрын

Like pie

@chillx656 10 ай бұрын

360° = 2pi?

@geboaebo 10 ай бұрын

@@chillx656 ye

@fizixx Жыл бұрын

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 Жыл бұрын

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 Жыл бұрын

@@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 Жыл бұрын

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 Жыл бұрын

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 Жыл бұрын

@@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 Жыл бұрын

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.

@ivanxdxd Жыл бұрын

a circle only if pi is normal

@dinosaric4862 Жыл бұрын

How would you graph the roots of a number?

@glitchy9613 Жыл бұрын

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

@klementhajrullaj1222 Жыл бұрын

And (-1)^e ??? 😀😉

@Rhovanion85 Жыл бұрын

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

@BobY52944 Жыл бұрын

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+

@musicmakelightning Жыл бұрын

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!

@j.u.4.n620 Жыл бұрын

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

@Mathologix Жыл бұрын

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

@j.u.4.n620 Жыл бұрын

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

@aaaaaattttttt5596 Жыл бұрын

@@j.u.4.n620 lmfao

@Zombatt Жыл бұрын

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

@adiaphoros6842 Жыл бұрын

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

@michaelwagner6877 Жыл бұрын

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

@marcrofSA Жыл бұрын

I'm from Brazil and I'm really enjoying your content and with the videos having the option of subtitles in Portuguese it's even better. Congratulations on the quality of the classes and I hope you continue with the great work.

@davekensk8 Жыл бұрын

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 Жыл бұрын

Thank you very much!

@getnie6867 7 ай бұрын

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

@user-db4lk7yg3o 6 ай бұрын

@@getnie6867 parenthesis

@slytherinbrian Жыл бұрын

(-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 Жыл бұрын

Yes!

@hervesergegbeto3352 Жыл бұрын

(-1)^π = -e^(π+2k)

@mhm6421 Жыл бұрын

@@hervesergegbeto3352 There is no negative.

@avrahm9029 Жыл бұрын

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

@bernardobuffa2391 Жыл бұрын

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

@pouyamovie3253 Жыл бұрын

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

@kdgcovers7530 Жыл бұрын

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

@blackpenredpen Жыл бұрын

Thank you!

@adarah00 Жыл бұрын

Love your channel bro 😊

@Nutshell_Mathematica Жыл бұрын

Your teaching not just improve my knowledge They really motivate me

@shahzadahmed6160 Жыл бұрын

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

@thenickli Жыл бұрын

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

@mathematician8113 Жыл бұрын

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

@abdulmalek1118 Жыл бұрын

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

@neilgerace355 Жыл бұрын

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 Жыл бұрын

real chads use tau/2 lol

@aflahmm2784 Жыл бұрын

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

@walls_of_skulls6061 Жыл бұрын

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

@GameMaster-pz9pw Жыл бұрын

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

@ultimatous4588 Жыл бұрын

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

@kovaxim Жыл бұрын

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

@existing666 Жыл бұрын

i'm in the same boat XD I wish school was more like this

@fLaMePr0oF Жыл бұрын

(-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

@VictorGallagherCarvings Жыл бұрын

WOW! What a great explanation.

@Gabsleggy Жыл бұрын

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 😂

@szymongrzebyk1964 Жыл бұрын

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

@hearstboy Жыл бұрын

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 Жыл бұрын

The tiny mic is an added bonus :)

@pardontheleft2692 Жыл бұрын

this was very neat, thank you

@stevenschilizzi4104 Жыл бұрын

Brilliant, as usual!

@jorgesoberon6866 Жыл бұрын

Bloody amazing!!!

@TheMemesofDestruction Жыл бұрын

Fantastic! Thank you!

@enisheadpay Жыл бұрын

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.

@scraps7624 Жыл бұрын

Your marker technique is just incredible lol

@5ilver42 Жыл бұрын

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 Жыл бұрын

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.

@leom8051 Жыл бұрын

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 😆

@ihatethesensors Жыл бұрын

I like that shirt with Euclid's formula with the Pi emphasized in red.

@mrajsatyam Жыл бұрын

You are my best teacher 🐱

@tutorchristabel Жыл бұрын

very informative, i love it

@pinedelgado4743 Жыл бұрын

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

@herbie_the_hillbillie_goat Жыл бұрын

Love your comment. 😁

@HeyKevinYT Жыл бұрын

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

@pinedelgado4743 Жыл бұрын

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! :) :) :) :)

@diamondnether90 8 ай бұрын

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

@aidkik580 Жыл бұрын

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

@Math0821 Жыл бұрын

I like your Film,very good

@0xABADCAFE Жыл бұрын

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

@donovanmahan2901 Жыл бұрын

-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.

@leonardobarrera2816 Жыл бұрын

Awesome!!

@srividhyamoorthy761 Жыл бұрын

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

@srividhyamoorthy761 Жыл бұрын

Pl read my other comment @blackpenredpen

@rajaekhafif2457 Жыл бұрын

Thanks u so much

@-_Nuke_- Жыл бұрын

That was neat! :D

@dmitrykostikov3909 Жыл бұрын

Really beautiful

@atlantiz4120 Жыл бұрын

I understood everything without knowing English amazing!

@kosterix123 Жыл бұрын

Nice to see. Very useful in daily life lol. But e^pi is really hard to approximate.

@galus_anonimus Жыл бұрын

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

@Rekowcski Жыл бұрын

i loved this

@aguyontheinternet8436 Жыл бұрын

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

@bryanlangley5337 3 ай бұрын

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!"

@tupublicoful Жыл бұрын

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

@johnoconner706 Жыл бұрын

good stuff.

@General12th Жыл бұрын

Very cool!

@igarciaasua9 Жыл бұрын

(-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)

@cmilkau Жыл бұрын

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?

@rihankota2021 Жыл бұрын

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

@WinterSoldier42 Жыл бұрын

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

@cesarrivas2295 Жыл бұрын

Please do the 100 trig questions 🙏

@Henrix1998 Жыл бұрын

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 Жыл бұрын

Yes. That’s equivalent to solving a^b=b^a and I have a video for that kzfaq.info/get/bejne/hq9hgch42bDTZ6c.html

@aryanprasad2406 Жыл бұрын

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

@liyisu Жыл бұрын

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

@jderick17 Жыл бұрын

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 Жыл бұрын

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 Жыл бұрын

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 Жыл бұрын

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)

@gheffz Жыл бұрын

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

@gheffz Жыл бұрын

I will try (-1)^Pi later!

@joseponce37 Жыл бұрын

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

@lechatrelou6393 Жыл бұрын

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

@alibekturashev6251 Жыл бұрын

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

@barthennin6088 Жыл бұрын

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 Жыл бұрын

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

@kaveenshankar7740 Жыл бұрын

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

@YahyaKhashaba Жыл бұрын

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

@faizakhtar5329 Жыл бұрын

Please solve n^i. n is any natural number

@388C4CGREEN Жыл бұрын

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

@oferzilberman5049 Жыл бұрын

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

@joshuazeidner8419 Жыл бұрын

pretty cool

@SavageGreywolf Жыл бұрын

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

@sberacatalin2250 Жыл бұрын

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

@Peter_1986 Жыл бұрын

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

@ryokucha_9101 Жыл бұрын

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.

@iljas275 Жыл бұрын

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

@SpringySpring04 Жыл бұрын

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

@joyli9893 Жыл бұрын

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

@joyli9893 Жыл бұрын

Wow! Somebody liked my comment already!

@superlinux Жыл бұрын

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

@leadiodide8243 Жыл бұрын

1:57 BASED

@rotemlv Жыл бұрын

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.

@HeartlessConservativ Жыл бұрын

Did this one in my head, lol. Just because of the famous e^(i*pi)=-1, so so the i-root of these gives is e^pi (or e^(pi+2pi*z)

@KrSaPoww Жыл бұрын

"Wait, is this all Eulers Identity?" "Always has been 🔫"

@kenbrashear210 Жыл бұрын

Where did you get the identities for you picture/poster on the wall?

@blackpenredpen Жыл бұрын

It’s my merch!. You can check my merch store.

@Imran-Shah Жыл бұрын

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?