Solving x^4=1

[@bprpmathbasics]

2nd way: 3:35

[@hassanawdi3793]

Its so hard for me to solve 😅

[@sanjukumari7684]

e^i×pie = x can also be a solution?

[@Nikioko]

@@sanjukumari7684 e^πi = -1, and (-1)^4 =1

[@trumpetzone0]

x=1 1^4= 1

[@NuzranRahat]

Its really easy when you watch the "x^3=8" video were he shows you all the solutions to that equation. But this one showed me that there are more ways you can solve these equations. Thank you ❤❤❤

[@ferretyluv]

I saw your comment and I just automatically knew the answer as 2 because I know the cube root of 8 is 2. So I’m confused how that’s relevant.

[@thisjosiah]

@@ferretyluv You have no idea what's possible

[@nasdfigol]

​@ferretyluv since the root is to the 3rd, there are 3 solutions meaning there are 2 more solutuons

[@mgancarzjr]

​​@@ferretyluvlook up roots of unity. They're complex number solutions but pretty easy to solve if you can convert degrees to radians and then understand that e^(i * pi) is equivalent pi radians in the complex plane.

[@mgancarzjr]

​​​​@@ferretyluvso for example, x^3 = 8 has a solution at 2 but also 120 degrees rotated into the complex plane and 120 degrees rotated beyond that (360 degrees / 3 = 120 degrees, so unique solutions at 0, 120, and 240 degrees) 2 * e^(0 * i * pi) = 2 * 1 2 * e^(2 * i * pi / 3) 2 * e^(4 * i * pi / 3)

[@jayJEEt.08]

1:00 those markers switch😳

[@samukaze5810]

He's the master of blue pens and red pens

[@marcogiai-coletti354]

Your second method explains the concept really well.

[@timeonly1401]

2:59 OMG!! The FIRST TIME I've ever seen BPRP flub a marker color change!! It'll go onto my list of "What were you doing when such-&-such event happened?". 😂😂❤

[@rotemlv]

Another way is to make a visit to trig land and use De Moivre's tools. Probably the most annoying way to write on paper but it is my favorite, as it helps you see the geometric properties of the roots of a complex number.

[@koennako2195]

He mentions this at the end of the video. He says something along the lines of "convert to polar form" which is the same as using De Moivre's Theorem.

[@ultrio325]

Pro tip: Always write all polar complex numbers in terms of Euler's formula, trust me it'll make literally everything easier

[@emconstrucao.2235]

Another way of solving it would be to set y=x², and then y²=1, y=±1, x²=±1, x=±√±1. Finally x=1, -1, i and -i

[@Nikioko]

Quick way using Euler's formula: x⁴ = 1 In the complex plane, the four solutions form a regular tetragon (= square), inscribed by a circle with the radius ⁴√1 and the centre at the origin: x = ⁴√1 ⋅ [cos(n ⋅ 360°/4) + i ⋅ sin(n ⋅ 360°/4)] n = 1→4 x₁ = 1 ⋅ [cos(90°) + i ⋅ sin(90°)] = 0 + i ⋅ 1 = i x₂ = 1 ⋅ [cos(180°) + i ⋅ sin(180°)] = −1 + i ⋅ 0 = −1 x₃ = 1 ⋅ [cos(270°) + i ⋅ sin(270°)] = 0 + i ⋅ (−1) = −i x₄ = 1 ⋅ [cos(360°) + i ⋅ sin(360°)] = 1 + i ⋅ 0 = 1 Long way using algebra: x⁴ − 1 = 0 Apply third binomial formula: (x² + 1) ⋅ (x² − 1) = 0 Apply the identity i² = −1 on the first bracket: (x² − i²) ⋅ (x² − 1) = 0 Apply the third binomial formula twice again to get the fully factored form: (x + i) ⋅ (x − i) ⋅ (x + 1) ⋅ (x − 1) = 0 According to the rule of the zero product, the whole product is zero if one of the factors is zero. So we get: x₁ = −i ∨ x₂ = i ∨ x₃ = −1 ∨ x₄ = 1 𝕃ₓ = {−1, 1, −i, i}

[@zachansen8293]

Why aren't there infinite answers?

[@Nikioko]

@@zachansen8293 Because an nth degree polynomial has n solutions. You can factor out each solution and get the factored form of the sort (x − a) (x − b) (x − c) ... = 0. According to the rule of the zero product, the whole product becomes zero when one of the factors is zero. That is the case for x = a, x = b, x = c, etc. If you have a 3rd degree polynomial with the factored form (x − a) ⋅ (x − b) ⋅ (x − c) = 0, then the normal form is x³ − (a+b+c) ⋅ x² + (ab+bc+ac) ⋅ x − abc = 0 You see that in the factored form, every factor is a binomial with x. So, in the multiplication of these factors, the highest power of x is the number of factors. And that's why there are not infinite answers.

[@zachansen8293]

@@Nikioko is e^ii3pi not a solution? 5pi?

[@yiutungwong315]

Good 💯👍

[@mokouf3]

​@@zachansen8293 Using polar form, we have x = e^(nπi/2) which may give you an impression of having infinite many answers. In fact, you need to remember that e^(2kπi) = 1, if you set n = 4k+1 in x = e^(nπi/2), you have x = e^((4k+1)πi/2) = e^((2k+k/2)πi) = e^(2kπi) * e^(kπi/2) = e^(kπi/2)

[@tankman1301]

I revised complex numbers a few weeks ago, but you completely caught me off-guard

[@obinator9065]

if you want to troll someone solve this using the general quartic equation

[@joeljain10]

quartic equation Blue comment spotted

[@nathanoher4865]

@@joeljain10so it’s not just me, neat

[@bossstarling217alt2]

​@@joeljain10Blue comment?

[@notoriousd.i.g.87]

You're evil lol.

[@zerid0]

Another way (that works well with values other than 4) is to notice that |x^4|=|x|^4=1 So |x|=1. Therefore we can write x in the form x=e^(iy) x^4=1 means e^(4iy)=1 Which means 4y is a multiple of 2π so y is either 0, π/2, π or 3π/2 modulo 2π which is exactly 1, i, -1 and -i

[@takix2007]

Write "x⁴ = 1" => "|x⁴| = |1|, and arg(x⁴) = arg(1) [2.pi]". You also write x as a.exp(i.theta), then x⁴ = a⁴.exp(4.i.theta) a⁴ = 1 with a in R+ (by definition of a complex number modulus) => a =1. 4.theta = 0 [2.pi] => theta = 0 [pi/2].

[@Yootlander]

He really is crazy with the marker swap

[@Chordrider]

He is fast but still easy to follow. Virtuoso use of the whiteboard. Just enough talking. I don't need to learn any math but I watch anyway. Good fun!

[@davidwillmore]

0:40 Let me guess, +/- i, right?

[@eclipse5708]

Wrong. It's +/-i or +/-1. There are 4 solutions, not 2.

[@joshuajoestar407]

@@eclipse5708pretty sure that’s why there’s a time stamp.

[@kavinesh_the_legend]

1,-1,i,-i

[@silense4644]

j, -j, k, -k, e₄, -e₄, …

[@kavinesh_the_legend]

@@silense4644 I understand j,-j,k,-k but I don't know what are the last two

[@silense4644]

@@kavinesh_the_legend just an extension of hypercomplex numbers. e4 to e7 is all the other imaginary numbers in an octonion, e8 to e15 in a sedenion etc

[@kavinesh_the_legend]

@@silense4644 I ain't paid enough to understand this💀💀

[@MorgothDarkLord83]

​@@kavinesh_the_legend No one is "paid" to understand anything

[@toasty-fy3]

Absolutely AMAZING video, I used it to explain the concept to my students and they understood it perfectly! Thank you so much, looking forward to more amazing videos!

[@DonTheRealMan]

Mathematicians: *stumped by negative square roots* Rafael Bombelli: Just calls √(-1) "i" and moves on... I love how based mathematicians are🤣

[@reizinhodojogo3956]

he ran out of real numbers and had to imagine new numbers /j

[@thisjosiah]

@@reizinhodojogo3956 That is kind of actually true though

[@highviewbarbell]

@@thisjosiah ssssort of yeah, i guess. He found a place in math where definitions were missing and extended the number system into another dimension

[@lukasjetu9776]

​@@reizinhodojogo3956 /j...azz

[@pritamdas5439]

Leonhard euler used 'i' for the first time to represent √-1 But he was not the first to study imaginary numbers

[@siener]

I think the simplest, most general and most useful way to solve this is to write 1 in it's polar form and then take the natural logarithm on both sides, and the rest is easy. Then you can replace the "4" with a "n" to get the general form for the roots of unity. The general form is also pretty easy to visualize: There are always n solutions, evenly spaced on the unit circle.

[@takix2007]

I'm not sure you are allowed to take the natural logarithm of a complex number without defining it beforehand.

[@takix2007]

Also, you don't need ln to do that. You write "x⁴ = 1" => "|x⁴| = |1|, and arg(x⁴) = arg(1) [2.pi]". You also write x as a.exp(i.theta), then x⁴ = a⁴.exp(4.i.theta) a⁴ = 1 with a in R+ (by definition of a complex number modulus) => a =1. 4.theta = 0 [2.pi] => theta = 0 [pi/2]. Longer, but "safer" as you do not take a potentially unallowed shortcut.

[@okaro6595]

I did it geometrically what angles multiplied by 4 give 2pi*n: 0, pi/2, pi, 3pi/2 so you get 1, i, -1, and -i.

[@davidwillmore]

Same!

[@magnusmalmborn8665]

The polar method generalizes to all powers, so maybe do a video with it?

[@Todo_2912]

This is the first video i didn't watch. I took the problem, solved it and skipped to the end. My solution looks a lot like the first solution and is correct^^

[@system3378]

What about x = sqrt(1) and x = -sqrt(1) ?

[@rdspam]

Roots of unity - e^((2*pi*k)/n)i. So e^(1/2,1,3/2,2)*pi*i, or i, -1, -i, 1

[@thej3799]

You are so smart! Also you are an amazing teacher and presenter. It humbles me how much math I do not understand. This channel is always making me think.

[@kururhai4531]

unrelated to video but I'm glad I found your channel currently our class completed calculus 1 and is teaching calculus 2, your explanation of questions really helped me and cleared up previous queries about some integrals and all doubts

[@Lightning78821]

A more complex way of doing this: r^(1/n)=r(cis(Θ)). The 4 theta values would be 0,90,180 and 270. Input and solve

[@El_Presidente_5337]

This dude just randomly showed up in my recommendations when I began going to university and started to fresh uo my math.

[@BeeDaWorker]

Show the polar method pls because y not Also I'm about to learn it soon so it'd be nice to see that stuff applied in a simpler question like this

[@RickyRatte]

Yeah yeah, this guy is good at maths, that's cool and all, but what's really impressive here are the marker handling skills

[@LolaSteininger]

I saw the thumbnail and felt so smart, answering “1”, until I watched the video.

[@mokouf3]

3rd way, roots of unity: x⁴ = e^(2nπi) x = e^(nπi/2) = ±1 or ±i

[@fernandojackson7207]

Akchually, the factorization a^2-b^2=(a-b)(a+b) only holds for commutative rings. Try it with a couple of non-diagonal matrices. Of course, the Reals, Complexes are commutative rings, but it may be a good idea to qualify the equality from time to time.

[@tchevrier]

you can also use the unit circle for complex. numbers.

[@sie_khoentjoeng4886]

As we know, power with even nunber has positive and negative roits. And also, power with multipication of 4 of i (4,8,12,..) has imaginary roots, since i^4 is 1 (also i^8=1, i^12=1,...) --- So, X^4=1 has two real number, √1 and -√1 or 1 and -1, and imagery number i and -i. Total solutions are X = {1 -1, i, -i.}

[@snaukball8764]

Another way is to imagine the complex plane, and just divide 360° into 4, which is 90°, then find which points lie on the unit circle at angles 0, 90, 180, and 270 degrees.

[@justafellowgamer-c5w]

I couldn't wait to grow up and take your classes

[@acekingbones]

Or you could just recall the power properties of i, where it alternates between i, -1, -i, and 1

[@nicef4919]

Please find all values of 4^x=x^64 I got two 256 and 1.0229

[@OptimusPhillip]

My attempt: If x^4=1, then x^2=1^(1/2) 1^(1/2)=±1 If x^2=±1, then x=1^(1/2) or (-1)^(1/2) (-1)^(1/2)=±i Therefore, x=-1, 1, -i, or i.

[@joseeoliviero6078]

I think most people would have taken (x^2 + 1) equals +-sq rt of -1 which is also +- i. But I really like the method you used. Thanks

[@AtomicSteel]

You are a life saver

[@PawełDąbrowski-n3w]

X⁴-1⁴=0 (X²)²-(1²)² = 0 (x²-1)(x²+1) = 0 From the first parenthesis we get: X=±1 From second: X²+1=0 X=±√-1 X=±i So we have 4 answers.

[@HargunCuberOG]

Nice one dude 👍👍

[@NavalSharma-r9z]

Where were you

[@yiutungwong315]

X = 1 and -1 (Real Number) X = i and -i (Complex Number)

[@ralphbehar]

Is no one gonna talk about the BOXES of markers in the bottom left????

[@nastykerb34]

Underrated comment

[@Twi_543]

That the bottom rigth but cool ig

[@DriftinVr]

x^4-1^4 (x^2)^2-(1^2)^2 (x^2-1^2)(x^2+1^2) (x+1)(x-1)(x^2+1) x^2=-1 x=+-i x=1,-1,i,-i

[@frankoptis]

I miss the formal stuff. x e R, x = {1, -1}; x e C => x = .... or something like that

[@nxcnibiru]

"There's no more real integers that would turn into 1 with a non-0 power so it must be something imaginary; I think that fits yea x⁴ -> (x²)² -> (-1)² -> 1"

[@Garfield_Minecraft]

|i| = 1

[@sigmagamer111]

u should try doing singapore's GCE o level elementary and additional math papers

[@thisjosiah]

e^(n(iπ/2)) *where n=any integer

[@rohangt1]

x^4=a => x=√a, -√a, √ai, -√ai

[@streambotnt2095]

super helpful!

[@cooltaylor1015]

x=1 solved from thumbnail. Yes it's trivial. It may not be the only solution. But it's RIGHT THERE

[@j.t.z.s.]

(x^2-1) (x^2+1)=0 (x^2-1)=0 or (x^2+1)=0 x^2=1 or -1 x=±1 or ±i

[@General12th]

So good!

[@kirkkohnen5050]

Polar coordinates make finding all n of nth roots trivially easy.

[@7Carryon]

When we have to solve an equation, the teacher have to say in which set. R or C? But the teacher knows what have taught to his students, so never saying the set. That is understood! By the way, sqrt(x^4) = 1 => abs (x^2) =1 => x^2=±1

[@starleaf-luna]

the 2nd way feels like when I'm taking a math test and I've done sth wrong, but I try to maybe weasel out of it and get an answer.

[@MagicProG]

usually, if you want complex decisions you should write z^4 = 1

[@RyanLewis-Johnson-wq6xs]

X^4=1 X=±1 X=±i

[@Brid727]

my way: x^4 = 1 x^2 = +/- 1 case 1: x^2 = 1 x=+/-1 case 2: x^2 = -1 x=+/-i x=+/-1, +/-i ezpz

[@TonyFisher-lo8hh]

The original piroblem is incomplete,. It does not specify the number domain. If real, two distinct solutions; ifs complex then four.

[@matu949]

x⁴ = 1 x = 1, because 1^any (positive) number is always 1

[@hodayfa000h]

Doesn't multiplying by i rotate 90°? So multiply it 4 times get 360° so one, and the same thing but in the negative, same location of 1.

[@yaseenelhosseiny]

X=plus-minus1, plus-minus I (sqrt(-1))

[@_Rainbooow]

x = 1 so easy. literally could do this in a second

[@wabbagabba]

great video 🎉🎉

[@eclipse5708]

x^2 = +- 1, then x = i, -1, -i, or 1. Done.

[@lool8421]

tbh it's kinda obvious if you know about complex numbers (-i)^4 = (i)^4 * (-1)^4 = 1 * 1 = 1 alr wrote down 3 answers there and 1^4 is just trivial

[@juanj.smithrobertob.sarcos1347]

I am still looking for the APPLICATION of (-1)^1/2, that is why I am SOL(Shouting out loud) for recess.😁😆😅😁😆

[@m.h.6470]

Solution: x⁴ = 1 |-1 x⁴ - 1 = 0 (x²)² - 1² = 0 (x² - 1)(x² + 1) = 0 x² - 1 = 0 (x - 1)(x + 1) = 0 x₁ - 1 = 0 |+1 x₁ = 1 x₂ + 1 = 0 |-1 x₂ = - 1 x² + 1 = 0 |-1 x² = -1 |√ |x| = √-1 |x| = i |x₃| = i x₃ = i |x₄| = i -x₄ = i |*-1 x₄ = -i

[@jomariraphaellmangahas1991]

The real answer is x = iⁿ, where n is element of integer

[@AidenHardy596]

Before watching Pretty sure the solutions are 1, -1, i, and -i

[@joeljain10]

Why i feel like I'm having Deja vu watching this?

[@JimmyMatis-h9y]

hmmm now x⁴ = -1 .... so 4th power equations aren't just "sharp parabolas"... 🤔 ty professor.

[@NLGeebee]

Great first solution on the left. On the right side, you should first substitute i² = -1 before applying ±√

[@anonymouscheesepie3768]

(x^2 - 1)(x^2 + 1) = 0 x = +-1, +-i

[@Creepmond]

That's nice, but what about x^8=1?

[@davidwillmore]

(i/2)^n for all integer n from 1 to 8?

[@samamethyst997]

ill try to do this without watching the video so we gotta prove x^4 = 1 take the square to the other side making it a square root so x= fourth root of one therefore converting that to x = 1 because root of 1 or fourth root of 1 will result into 1 therefore the answer being x = 1

[@Fengwunna100]

Why did you use abc and x? How did alpabet come?

[@DevMan2342]

Do you not know what variables are?

[@Fengwunna100]

@DevMan2342 i dont know

[@DevMan2342]

@@Fengwunna100 why are you on a math video if you don’t know basic math. Go to school first

[@theflyingcrane1008]

Bruh I’m majoring in law idk why I’m watching this but it’s cool af

[@sid6076]

Now I get x does equal 1 but it can also equal -1 and i and -i

[@Caysen52TheSecond]

any format of 1 is correct including i and -i.

[@rajatdogra96]

+-1 and +-i 😂😂

[@idk39381]

hi can u do the 8 hours series back (need indices,surd,log question)

[@zachansen8293]

wouldn't the polar form show that you hadn't found *ALL* the solutions to the equation since there are infinite ones? Huh, WA also shows only 4 solutions, but it sure seems that e^(i(2n+1)pi) nEZ are all solutions

[@bravcoder]

looking at it, I think the answers are going to be, +1, -1, +i, -i

[@Zianagi]

My uneducated brain wondering on what is going:

[@nolimitzs00]

x⁴+1=0 would be little bit more interesting

[@shirinpatel9339]

x^4=1 1^4=1

[@tom-kz9pb]

It could see the 1 and -1 solutions, in just a glance, but always forget about the pesky "i" complex solutions that I know have to also be there.

[@OnePlusOneEqualsOnePlusOne]

Sooooooooooooooooooooooooo, X is still equal to 1?

[@Marandahir]

Shouldn’t it just be x = i ?

[@lwhxe]

Why x = (-i)? Since (-i)^4 = -1? How is this possible?

[@lwhxe]

Oops, I'm stupid... -i means other way round...

[@yiutungwong315]

(-i)^2 = (-1) (-1)^2 = 1 So (-i)^4 = 1

[@lwhxe]

@@yiutungwong315 Yeah, I got it just a sec later...

[@JohnVance]

In the apocalypse, this guy will be the only one left with dry erase markers.

[@jeffocks793]

I really enjoyed that. I want more! (Ps I can't see a subscribe button 😮)

[@Wienco303]

Solution: 1

[@Wienco303]

Solutions: 1 -1 i -i

[@DooMNewLechia]

I wonder if x can be equal to e^pi*i

[@lwhxe]

It can! It's just -1.