Euler's Formula - Numberphile

[@elgalas]

We need a video explaining why the polar representation of a complex number includes "e". It's a wonderful bit by Euler.

[@bonob0123]

yes this is exactly the part I have no intuition for. The rest after that is just substitution/algebra

[@SampMan87]

I’m glad I’m not the only one. Without the background of why complex numbers can be represented as re^i(theta), that just feels like a leap. From the diagram it’s easy to see where r and theta come from, but why is there an e^i in there?

[@MusicFanatical1]

He plucked it out of nowhere and ended up where he started. Like assuming 1+1=2 then being surprised to find 0=O. Not a well structured video at all.

[@liamleider4807]

It comes from the Taylor series representations of e^x, sin(x), and cos(x). If you start from the Taylor series of e^x (which you can easily find by a google search) and instead of x, you plug in ix, you will notice that you get the sum of the cos(x) and sin(x) series representations (with a multiple of i next to sin(x)). From there we conclude that e^(ix) = cos(x) + i*sin(x). He should have shown this proof instead of what’s shown in the video, i agree. Nevertheless, great video!

[@eoinlanier5508]

e to any power x represents continuous growth. i represents a 90° (π because we're actually working in radians) rotation. So e^ix is continuous growth at 90°. If you start at a point and move at a continuous 90° (π), you draw a circle. e^iπ is just plugging in π for x. You grow continuously for a distance of π, which is the same as moving π around a circle, lands at -1 as per remembering the basic info of the unit circle. So e^iπ=-1 e^iπ+1=0.

[@AtricosHU]

This is not actually how the formula is derived. You've just *used* Euler's formula to *derive* Euler's formula. Which is great, at least you found no inconsistencies in maths, but it doesn't explain where the z = r * e^(i * theta) comes from. Why is this equality true? Why can you represent z using the exponential of i times the angle? This *comes* from Euler's formula in the first place, which can be proven looking at the Taylor series of sin, cos, and exp, respectively.

[@Pjx1989]

Came here to say the same. The first 5 minutes of the video do not make sense

[@zunaidparker]

100%. The polar form given in the video skips over WHY you can use exp(i*theta) to represent the angle. It's a really bad oversight from a seasoned channel.

[@YawnGod]

Cosmic Horror.

[@AssemblyWizard]

You would also need to prove that exp,cos,sin are analytic at 0, otherwise their Taylor series are useless

[@kubtastic]

Igen

[@avhuf]

it's more intuitive without the zero, then you can see that exp(i*pi) is actually a rotation over pi starting from 1, and thus landing on -1.

[@MasterChakra7]

It's actually even more intuitive with τ, since e(iτ) = 1, thus one complete rotation.

[@iveharzing]

@@MasterChakra7 I like the idea of using a constant that is 2*pi instead of pi, but tau is used way too often for other applications, like integrating over a time interval.

[@Apalapan97]

@@MasterChakra7 found the Tauist.

[@InterFelix]

This is the way I intuitively understood Euler's identity immediately when he first drew the graphical representation of the polar form of imaginary numbers. It's quite beautiful.

[@wallstreetoneil]

it also de-mystifies 'imaginary' numbers and makes them intuitively understandable as Rotations. Humans innately understand rotations, and it's one of the great contributions Euler made - he saw that the number one, could be rotated above the plane, around the origin, and land on negative one - and thus the square root of negative one existed but was just above the plane and we need a rotation to get there.

[@markcrites7060]

I didn't hear him explicitly state this, but for all of this to work theta must be in radians. Later on in the video he starts mixing degrees and radians. This is all common sense to a mathematician but it may not be to a student early in their math career.

[@wild-radio7373]

Ughhh! Radians ♡ Those sneaky units :)

[@rbad6215]

a lot of this is in radians, especially the identity.

[@piotrarturklos]

When a mathematician says "angle", they mean radians because they are fundamental. Degrees are just arbitrary units that people use to satisfy the monkey part of their brains that wants a neat whole number to divide the full angle.

[@ingGS]

Yes, this is actually a recurring joke for Engineering students. Confusing degrees with radians have even resulted in failure of structures.

[@markcrites7060]

@@ingGS I am an engineer. And anytime anything involving r, theta notation (polar coordinates, complex numbers, phasor notation, etc) was discussed in school it was drilled into our heads that theta must be in radians.

[@Bodyknock]

2:08 Tom says "and the polar form is z = reᶦᵒ ..." and then goes on to say how that implies Euler's Formula. But in order to use z = reᶦᵒ as the polar form you're already implicitly assuming that eᶦᵒ = cosθ + isinθ so it's a bit of circular reasoning. It's not at all obvious without having already proven Euler's formula that if z = reᶦˣ that x is actually the angle θ in the polar form r(cosθ + isinθ). I think the proof he was actually intending to use starts off by saying "Assume z = eᶦˣ . (We're not making any assumption other than that it's a complex number, we know nothing about x or its relation to the polar form so far. The only thing we know is that it's a complex number.) Let z = r(cosθ + isinθ) be its polar form. So eᶦˣ = r(cosθ + isinθ) and r and θ are functions of x. Now we want to find a closed form for x in terms of r and θ..." and go from there.

[@ricardorix73]

yeah this was my first thought.

[@Dancingdude-ci6wz]

I was wondering this same thing, thanks for pointing it out

[@tjfrench5285]

I was stuck on the same thing, does anyone have resources to a conceptual derivation of this identity?

[@rtpoe]

"circular" reasoning..... LOL

[@Minty_MH]

@@tjfrench5285 One way to see this is to look at the Taylor series of each of these functions and "extend" their definitions to the complex plane by allowing complex numbers to be plugged into each Taylor series. Then the formula falls out. Another way is (which is actually the same) is to say that we want to extend the definition of derivatives and functions currently defined over the reals into the complex plane. The most sensible way to do this is to impose that these things behave as we expect when the complex values are evaluated along the real line. Take any real valued function whose derivative is proportional to itself and evaluates to f0 at t=0. f'(t) = a f(t), f(0) = f_0 Then evaluating this function at a purely imaginary t and taking the second derivative gets via chain rule that f''(it) = a^2 (i^2) f(it) = -a^2 f(it). Of course, this will also be true component-wise in this function's cartesian representation f(it) = (R(t), I(t)) = R(t) + i I(t) where R and I are real-valued functions that represent the real and imaginary part of f(it) respectively. So d^2/dt^2 (R(t), I(t)) = (R''(t), I''(t) = -a^2 (R(t), I(t)). One might recognize this as the equation for circular motion. If we interpret t as time, this is saying that the acceleration always points inwards from where I'm at towards the origin. Algebraically, we get that the solutions to each component of this differential equation are the trig functions, cos(ax) and sin(ax), so the full solution in each component will be some linear combination of sines and cosines determined by the initial conditions. The initial condition is imposed on us by wanting f(0i) = f(0) = f_0, just like on the reals. Therefore, f(it) = f_0 (cos(at), sin(at)). So, if I take a function over the reals whose derivative is proportional to itself and I extend its definition to the complex plane in the most sensible way, I get that a purely imaginary argument is always circular motion. The frequency of that rotation is determined only by the proportionality constant from the derivative, and the radius of that circle is determined by the initial conditions. Go back now to the differential equation for f. The unique solution to this differential equation is f(t) = f0 exp(at). If we set a=1, t could be interpreted as the polar angle in the cartesian representation of that point. The base of the exponential map, e, is special because it is the unique base of the exponential where the derivative IS itself. That is why the base e is vital for Euler's formula.

[@harper5128]

personal favourite part of euler's definitions of trigonometric functions is how neatly they relate standard functions to hyperbolic functions, whereas only considering the reals they look nothing like each other - neat example is that sin(z) rotated pi/2 around the plane is the same graph as i*sinh(z)

[@cobalt3142]

That's kind of true, what you probably meant to say is sin(iz) = i sinh(z) (sin z is the same as sinh z rotated by pi/2). I totally agree with everything else you said though

[@harper5128]

oversight on my part lol thanks, should specify im still learning ca

[@landsgevaer]

They may not look like each other indeed, but they obey plenty of similar relationships that someone who knows only reals would be able to appreciate; e.g.: cos^2 x + sin^2 x = 1 vs. cosh^2 x - sinh^2 x = 1 cos'' = -cos and sin'' = -sin vs cosh'' = cos and sinh'' = sinh cos 2x = cos^2 x - sin^2 x and sin 2x = 2 cos x sin x vs cosh 2x = cosh^2 x + sinh^2 x and sinh 2x = 2 cosh x sinh x tan' = 1 + tan^2 vs tanh' = 1 - tanh^2 cos^2 x = (1 + cos 2x) / 2 and sin^2 x = (1 - cos 2x) / 2 vs cosh^2 x = (1 + cosh 2x) / 2 and sinh^2 x = -(1 - cosh 2x) / 2

[@shayeri7208]

@@harper5128 will you olease share some tips from where to begin

[@harper5128]

@@shayeri7208 i started off mainly with dr richard borcherds' lecture series on introductory complex analysis and needham's visual complex analysis textbook, the latter is very compact and mostly focused on intuition and application

[@johnny_eth]

The coolest feature of the polar representation is that it makes calculating roots or powers super super easy. Just divide or multiply the angle, and don't forget the 2πn term.

[@DavidLindes]

or τn. :)

[@jmcsquared18]

1:51 This was a bit of a cheat, because that's not really polar form. That every complex number can be written as z=r⋅e^i𝜃 assumes Euler's formula, as well as begs the question of what we could possibly mean by e^i𝜃 in the first place (i.e., raising something to an imaginary power). The actual polar form of a complex number is z=r⋅(cos(𝜃)+isin(𝜃)) which is easily shown by your geometric argument with cosines and sines in triangles. The theorem behind Euler's formula is that the trigonometric part in parentheses cos(𝜃)+isin(𝜃) is equal to the exponential function e^i𝜃 once it's been suitably defined. Of course, you show that using the power series expansions later on, but the presentation is quite out of order, which could confuse viewers.

[@jonathanross6260]

Yup, came here to say this.

[@EebstertheGreat]

Later on, he also makes the curious assertion that e^(iπ/2) = i is an "alternative definition of i." It's not even a characterization of i, since e^(π/2 z) = z for all z = -Wₙ(-π/2)/(π/2), n ∈ *Z.* So for instance, z = -i and z = 1.02132... - 4.86835... i also satisfy the equation.

[@guillermonasa]

e^ix does not mean e raised to ix. It is a silly way of writting exp(ix) where exp(x) is a sum of infinite terms 1+ x + x^2/2 + x^3/3! +…+ x^n/n!. exp(x) has very particular properties, exp(0) =1, exp(1)=e, exp(a+b)=exp(a)*exp(b), and therefore exp(3)=exp(1+1+1)=exp(1)*exp(1)*exp(1)=e^3 . But if you analyze exp(x) when x is ix it has the has same intereting properties but it is silly to write it as e^ix,

[@jmcsquared18]

@@EebstertheGreat yes I was also skeptical about that. That's the product log, right? Honestly, I only resubscribed to Numberphile because Grant gave an awesome presentation here a while back. But this channel really lacks in taking care with their presentations.

[@artsmith1347]

@@jonathanross6260 Me, too. Also, unlike the rectangular form, the polar form is a unique representation of a point on the complex plane only if the value of theta is restricted to the range 0

[@andmefikri7555]

I want to say thank you! Thank you from the bottom of my heart. I am a first year university student with math major, and so far, my experience have been discouraging me away from math. I kept thinking that I made a mistake, that I made the wrong choice, and that mathematics isn't for me. This video has reminded me of my love and passion of math. I took the bargain, math isn't easy, but it rewards your curiosity and perseverance.

[@TomRocksMaths]

[@mestar12345]

Magic moment happens at 1.55. He's explaining basics of complex numbers, and then he just right into using "e to the power of i something", which makes no sense. You need to justify using that magic at that point.

[@drewhead118]

I think it's derived from euler's formula, at that, making the logic here circular

[@joshrowe9653]

When simply defining addition of two integers takes hundreds of pages of dense proofs, this might not be reasonable to demand. To justify the magic could turn this into a three hour, boring video.

[@SiMeGamer]

@@joshrowe9653 "boring" is subjective. I'd personally a more formal proof rather than "magic". As magical as math is, it's still rigorous. I can accept the weird x+iy form despite not understanding why that is acceptable, but that's simply because we define it that way. We have reasoning and it is shown. It's axiomatic. But this whole ordeal with r*(e^(i*theta)) is just given out of nowhere with no semblance of an explanation. The only thing explained is the r and the theta which is nice but the e and the i aren't explained at all. And this is no axiomatic. This is just missing. Why isn't it r^2+sqrt(3/theta)? Might as well try that since it's just "magic". Math is rigorous and we invent axioms in order to move forward with missing fundamentals. Euler's formula is based in algebra and uses constants that we somewhat understand. There is no reason for requiring a new axiom here. You need to derive it, not poof it from thin air. I watched the entire video in hopes that it is brought up but it wasn't. So all it did was make the whole video kinda pointless because it is was based on a false premise the entire time. I like math, but that doesn't mean I know all the fancy stuff. Comments here mention how beautiful Taylor's series is or smth and idk what that is and would've loved to see it. Even if it is skimmed through as many other Numberphile videos are. It's like we are skipping part of the proof, we are skipping the proof in its entirety. There is no magic here. Imagine a magician tells you to pick a card, you do it, and then he looks at the card and prepares the deck and then does his trick to reveal your card. There is no magic there. It's preset for success rather than any sort of discovery. I recommend you watch almost any other Numberphile video. You'll see how if they skip something, they at least mention what they skip in order to show the arguably more interesting and fun crux of the topic. For math minded people, it's likely very frustrating that there is this assertion of a weird complex formula that comes out of nowhere and never explained. Nobody does math this way. You don't just randomly create a weird looking formula out of an infinite amount of possibilities and just hope it's the right one for whatever arbitrary revelation it would lead to. This isn't human and math is very, very human (you can take that in whatever interpretation you prefer).

[@mestar12345]

@@joshrowe9653 "When simply defining addition of two integers takes hundreds of pages of dense proofs," It does not. Axioms in mathematics are completely man made, you can choose your own. There is nothing better when you start with "sets" and then do your 100 pages. You can just start with numbers.

[@maxwellcody6457]

You can derive it by finding the MacLaurin series for e^x and then substituting x for ix. Then rearrange things. I'm sure he showed it the other way around to not assume the audience knows what a MacLaurin series is. That way the content is more accessible to the viewers. You could most likely find a more rigorous explanation on youtube if you want one.

[@joaobaptista4610]

How you represented complex numbers in polar form using exponential without allready knowing Euler's formula? It looks like circular reasoning to me.

[@satyakonala]

Watch veritasiums video on it

[@EebstertheGreat]

@@SaintCergue That's still incomplete reasoning. You are solving that differential equation "by inspection" by already knowing that d/dz exp(wz) = w exp(wz) for complex w and z. We know that's true for real numbers, but to show it is true for complex numbers, we need a definition for the exponential of a complex number. There are several ways to do this, but they all rely on some initial legwork. For instance, you can define exp z in terms of the power series representation, but to do so, you first have to prove that that series converges for all complex z, and then you have to show that the power rule applies to series of complex numbers, etc. Or you could just straight-up define exp as the unique function from C to C for which exp'(z) = exp(z) and exp(0) = 1. But to do that, you first have to prove that such a function exists and is unique. You can't really prove Euler's formula in a short video that assumes the audience knows nothing about complex numbers, because there is too much complex analysis leading up to it to fit in.

[@AssemblyWizard]

It is, there wasn't a proof in the video

[@Mrjcraft00]

To me, the way to come to this conclusion easiest is by using some calculus with Taylor Series for sin(x), cos(x) and e^x. The sum of sin and cos looks like e^x, except that some of the signs don’t match up in some places, specifically the second, third, sixth, seventh, and so on repeating every 4. If you know the cyclic nature of i, it might make sense to try e^ix. This will give you i every other term, but those are the sin terms anyway so you can separate them, factor the i’s out of the odd terms, substitute, and get Euler’s formula.

[@tomcass240]

@@satyakonala yeah veritasiums video was much better than this garbage.

[@ijuhat19]

I prefer to think of Euler's formula as the *definition* of the cosine and sine functions. That is, you start by defining complex numbers, then define the exponential function (on the complex domain) via its power series, and then define cosine and sine to be the real and imaginary parts of this function. The really cool part is that you can then define pi analytically, by defining pi/2 to be the smallest positive root of cosine. Everything else falls out nicely after that. The way things are presented in this video suffers from circular reasoning in my view.

[@tomkerruish2982]

For another cool way to define trig functions, check out Eisenstein.

[@jasonhargis5598]

Thanks for this!

[@littyfam5136]

The only problem I have is proving the re^iθ. Where does e come from here? I can prove it using the Taylor expansions for sin, cos, and e^x, but without that I don’t know of any way to prove re^iθ

[@angeldude101]

@@littyfam5136 e^x is pretty much defined by the properties of d/dx e^x = e^x and e^a+b = (e^a)(e^b). For e^ix, d/dx e^ix = ie^ix, so whatever e^ix is, it has to have its derivative perpendicular to its value and with the same magnitude. That function is the one that draws a circle.

[@romanburtnyk]

@@littyfam5136 Problem is that z=re^ix cannot be proved. It is a definition which has it properties, described here is video. So I like idea of above commenter, create some abstract numbers, explore them and find some funny things

[@Israel2.3.2]

Euler's work is so much fun. A, B, C, ..., N, ... for any sequence whatever, index notation is rare which forces a sort of conceptual elegance in the arguments.

[@bjornmu]

This is all cool, but you don't explain how e "magically" enters the formula in the first place. How come re^(i phi) equals x +iy?

[@ricardomanzano1270]

Thought the same

[@souleymaneadellah1176]

Yeah I was waiting for an explanation to that because thats actually whats fascinating about the equation

[@shivamsahu1371]

I suspect there might be a simpler way to it, but the way I know the formula is from something called "Taylor Series" which is basically a "Tool" to make any function to a polynomial function. And it's beautiful. If you're new to this concept check out 3b1b video on the topic.

[@copperfield42]

ikr, that is why I prefer the way I was teach this by studying taylor series, you get your series for exp(x), cos(x) and sin(x), and then with that you calculate exp(i*x) and alas you get to cos(x)+i*sin(x), and with that you go from x+iy to the polar form

[@dottormaelstrom]

The reason is that one usually first writes down the infinite series for cos sin (their "Taylor expansions") and then the infinite series for e^ix (its "Laurent expansion", which is actually just the Taylor series of e^x but using the polynomial properties of i, ie i*i = -1 and so on) and THEN notices that euler's identity holds for the series, so e^ix is DEFINED in terms of sin and cos, it's not just a magical property falling from the sky. This is the way I was taught at uni, if there are other approaches to introduce the complex exponential I'd be glad to hear them, but basically these are all equivalent reasonings, one can choose one property over another to characterize it and write down the formula.

[@Pfhorrest]

the e^iπ = - 1 version is much more elegant, because it illustrates what the statement is actually saying: rotating by half a turn (π radians) is equivalent to taking the negative.

[@NoActuallyGo-KCUF-Yourself]

"more elegant" no, but more conceptually informative.

[@genessab]

Wait you can’t start your proof with the polar form of complex numbers, does that not presuppose Euler’s formula?

[@robertleduc5553]

He really just pulled the polar form out of nowhere

[@shubhkarmansingh4385]

Ikr

[@jenspersson_lund]

Yes, the video is pulling that out of the air. It should be done the other way around: Expand cos and sin with Taylor series, plug in iθ and then "magically discover" that if you subtract sin from cos discover that the polar form can be written with e. But it makes the video narative less compelling :-)

[@kilianvounckx9904]

It all depends. If you define exp(i*theta) to be that polar form, you can derive everything from that and proof it also happens to be euler's number raised to that exponent.

[@0x4849]

It does, but I think that is allowed and it doesn't matter if you use knowledge, as long as you later prove it. This isn't the original way Euler discovered this, but just a proof. I think Euler discovered this with Taylor series.

[@calmkat9032]

This whole Euler Formula to Trigonometry thing is also my favorite thing in math! The connection between a Taylor Series in the specific case of e^x being the same as cos(x) * isin(x), and how that solves so many problems in trig, immediately solved so many hard-to-grasp concepts in calculus and physics that plagued me throughout all of high school.

[@andrewrussack8647]

It makes electrical engineering SO much easier!

[@happyestus6688]

I love it when Tom's doing the explanation, he's always so enthusiastic! Very contagious energy

[@TomRocksMaths]

[@MoempfLP]

0:49 i is not the square root of -1. The square root of -1 is undefined. i is defined such that i^2 equals -1.

[@SpencerTwiddy]

11:07 - big missed opportunity to draw the cycle with "i" at the top and "-1" to the left, matching their positions on the complex plane

[@protectedmethod9724]

I was thinking the same thing. A useful visualization tool I heard is that multiplying Z by "i" will just rotate it 90 degress counterclockwise on the complex plane, and multiplying by "-i" will do the opposite direction

[@frankheyder2222]

That leaves the question: Is Numberphile an infinite series? I surly hope so.

[@bernhardkrickl3567]

Well, there once was PBS Infinite Series, but it ended nevertheless :(

[@iteerrex8166]

It’s not only gorgeous maths, but it’s also very important in quantum mechanics.

[@tweedyburd007]

and differential equations

[@jessehammer123]

Wow, there’s never been a Numberphile about this? Color me surprised (but excited too!).

[@tuneboyz5634]

Wow surprised little baby 😮

[@AlisterCountel]

Well, you may have thought there was one, but all the sines show it was imaginary. And even this one, well, it’s a bit Complex.

[@jessehammer123]

@@AlisterCountel I’m not sure you’re right. Can anyone cosine this guy?

[@vigilantcosmicpenguin8721]

I suppose it's just so significant they didn't think to make a video explaining it. It'd be like defining the number one.

[@vindj2391]

Everyone commenting about the maths while I'm just admiring that handwriting

[@timm1328]

you did this exactly backwards. Euler derived his formula from the Taylor series expansion of exp(), sin(), and cos().

[@luketurner314]

A great way to memorize the powers of i and -i is to imagine* the unit circle on the complex plane, specifically the points where it intersects the axes (-1, 1, -i, i); since angles are measured from the positive x-axis, consider moving counterclockwise around the circle as positive and clockwise as negative; then powers of i correspond to starting at i and rotating counterclockwise to -1 (and so on); and powers of -i correspond to starting at -i and rotating clockwise to -1 (and so on). Generalizing from powers to multiplying: multiplying by i corresponds to rotating counterclockwise and multiplying by -i corresponds to rotating clockwise. * pun: imagining about imaginary numbers

[@PierpaoloBuzza]

Being a Tau (where T = 2P) enthusiast, I also love e^iT = 1

[@artsmith1347]

An alternate form: e^(i * T) + 1 * (-1) = 0 It uses six important constants: (e), (i), (T), (1), (-1), and (0).

[@artsmith1347]

On further thought, one could also include the first prime number by using pi: e^(i * 2 * pi) + 1 * (-1) = 0 It uses seven important constants: (e), (i), (2), (pi), (1), (-1), and (0). But it becomes numerology at some point. These are more compact: e^i*pi = -1 e^iT = 1 -- this is the most compact form.

[@lakshyarana67]

I love how all these numberphile geniuses are always so much excited and smiling.... Throughout the video

[@DavidLindes]

Minor constructive critique (though I doubt it's changeable now, just something to keep in mind in case you ever have a reason to re-visit this): at 10:54, I'd suggest doing this cycle going counter-clockwise, as then -1 would be in the relative position of where x=-1 would fall (or Re(z)=-1, if you prefer that notation), and 1 would be where x=1 would fall... whereas the current way, you've got -1 where x=1 would be, and 1 where x=-1 would be. Thanks for listening! P.S. At 12:39, I'd do that sequence clockwise, starting from the bottom.

[@Impatient_Ape]

Tom could have also showed how we can get the trig identities for sine and cosine of sums of 2 angles. That is... e^i(a+b) = e^ia * e^ib. Expand the terms on each side using Euler's identity, and then set the real parts of both sides equal to get cosine of a sum of angles and set the complex parts of both sides equal to get the sine of a sum of angles.

[@douglaswebster3084]

Not just the sum of two angles but the sum of n angles using binomial expansion. And for the sum of two angles it's a lot easier to derive than trying to remember how to set up all those triangles!

[@bethhentges]

I have usually seen it in the other direction. Start with the Taylor Series of e^x, cos(x), and sin(x), and use them to prove cos(x)+isin(x)=e^(ix).

[@gregoryfenn1462]

At 5:45 we show some properies of sin and cos, e.g. cos(x) = cos(-x). "Cos of positive theta is cos of minus theta" you said. But these are only real-numbers, you can't just apply it to complex numbers without a new proof that those properties still hold in a new domain!

[@jakobr_]

(-z)^2 = z^2 in the reals and in the complex numbers: (-z)^2 = ((-1)(z))^2 = (-1)^2 z^2 = 1 z^2 = z^2. Observe that every instance of z in the maclaurin series representation of cosine is of the form (z^2)^n where n is a positive integer. QED It’s really a shame that this video starts with all this stuff that depends on the power series expansions and doesn’t get to them until halfway through the video.

[@RichardJBarbalace]

At 10:53, it would make more sense to have the cycle go counterclockwise instead of clockwise to match the actual locations of i, -1, -i, and 1 on the standard 2D graph. Multiplying by i is simply rotating the space counterclockwise by a quarter turn (pi/2). This is why using the turn constant of tau = 2pi makes a full rotation. Similarly, multiplying by -i is simply rotating the space clockwise.

[@RichardJBarbalace]

This is why e^(i*tau) = 1, which is an even cooler identity.

[@kremigmitsahne7197]

His enthusiasm makes him the best teacher

[@arcuscotangens]

When I was tutoring a class in university, I apparently got really excited explaining Euler's formula. At the end of the course, the students gifted me a t-shirt with it printed on.

[@TomRocksMaths]

I want this shirt!

[@arcuscotangens]

@@TomRocksMaths What if I told you that on the front there is a print of Blastoise?

[@marriedbeaN]

We need a tattoo tour from Tom

[@tomkerruish2982]

Oh, that's on Numberphile's OnlyFans account.

[@obiwanpez]

Perfectly timed! I'm teaching Taylor Polynomials on Thursday!

[@mcb187]

This has been my favorite equation ever since I learned about it from *Person of Interest*. The charicter said almost the exact same thing, that the 5 most important and recognized constants in math are all put together, in one very beautiful, elegant, equation.

[@yunglynda1326]

he has such an adorable personality 😊

[@TomRocksMaths]

[@johnchessant3012]

You can also prove Euler's formula by showing that e^ix (cos x - i sin x) is a constant function; just verify that its derivative is 0 using the product rule.

[@anmolkessani4872]

this actually one of the clearest video i have seen on this topic for a while

[@simonlaplace9790]

This video was great, would you consider doing a video on the Laplace or Fourier transform

[@crashmancer]

Agree: if Dr Crawford explained the Fourier transform with the same aplomb, I might finally understand it.

[@protectedmethod9724]

are you only interested in laplace because it is your last name? haha

[@star_ms]

Your last name is Laplace

[@josejrtuti]

Best explanation of Euler’s formula ever! Thanks!

[@TomRocksMaths]

[@CamMackay96]

I love love love this area of mathematics, gave a presentation all about Euler's formula during my undergraduate course at uni. Will I know absolutely everything said in this video? Yes. Will I watch it all anyway? Also yes!

[@TomRocksMaths]

[@v3spirit]

11:05 i never thought it would bother me so much that -1 is on the right side

[@keb7066]

coolest math professor award goes to this guy

[@yashrawat9409]

Learning to solve AC circuits with complex numbers instead of going haywire is just one instance of how powerfull of a tool they can be !

[@jerrysstories711]

I must have forgotten and rederived the cos(a+b) and sin(a+b) formulas 10^2 times in my life so far. That's my fav use for Eulers formula!

[@DeGuerre]

The thing I love doing with Euler's formula is deriving trigonometric identities. Suppose, for example, you want to know an expanded form for sin(A+B). cos(A+B) + i sin(A+B) = e^(i(A+B)) = e^(iA) e^(iB) = (cos A + i sin A) (cos B + i sin B) = (cos A cos B - sin A sin B) + i (sin A cos B + cos A sin B) Match up real and imaginary parts, and you have your identity.

[@Salien1999]

Lovely video on the subject. It works a bit backwards from how I would've explained it, but then again if I made the video it would've been pretty boring. The infinite series portion serves as the closest thing to an intuitive proof of Eulers formula I've ever seen. The graphical explanation requires that you understand that circles on the imaginary plane are described with an exponential function, which is counterintuitive if you don't have Eulers formula in the first place. Also one thing to point out for anyone who is going to go into engineering or otherwise will have to take a differential equations course: you will use this formula ALL THE TIME in that course.

[@jeremyhansen9197]

This video is backwards. What they proved is that the polar form r*f(theta)=r cos(theta)+I r sin(theta). What they need to prove is f(theta)= e^I theta. The way to do this is by expanding cos, sin, and e, but this video uses the expansion of e to prove cos and sin.

[@z4zuse]

10:56 anti-clockwise would have been nicer

[@lucasl4644]

If people are wondering about an application, we use complex numbers a lot in analysis of AC circuits.

[@karraguer]

Hi! Very nice video, however the definitions used here are quite circular. My very best favourite visual evidence of the nature of Euler’s identity is the computation of (1+\frac{i\pi}{N})^N for a increasing N (when the limit tends to infinity, the exact value is -1).

[@KipIngram]

I think the best way to motivate complex number theory is to use a geometric algebra based approach. Define not only the unit vectors (call them xhat and yhat rather than i and j, since we want i for something else later) but also the unit bivector xhat*yhat. In 2D that's the only unit bivector you have. One of the very simple rules of geometric algebra is that vector products anti-commute: v1*v2 = -v2*v1. We also have as a rule that squaring a unit vector yields one: xhat^2 = yhat^2 = 1. Now consider squaring the unit bivector: (xhat*yhat)^2 = xhat*yhat*xhat*yhat Commute the middle two - this negates the result: (xhat*yhat)^2 = - xhat*xhat*yhat*yhat Now just replace the xhat^2 and yhat^2 factors: (xhat*yhat)^2 = -1 Hey - look at that! There is only one unit bivector in 2D and it squares to -1. Let's give it the special name i, and there you go - complex arithmetic comes pouring out. We don't have to DEFINE i^2 = -1; it arises naturally in the context of 2D geometric algebra. This is just one example of the power of the method - in higher dimensions we get quaternion theory as well, and so on. ALL of the physics we do with the normal vector machinery can be done this way as well, and it's cleaner and more satisfying. Finally, note how common it is for us to use the cross product in our usual presentation of physics. That's fine, but strictly speaking cross products only "work" in 3D. You can do all of the same thing using bivectors, and that gracefully works in ALL dimensions. It's a SUPERIOR APPROACH. We should start teaching our kids geometric algebra in early high school. I didn't learn about it until I was in my 50's, and I actually was annoyed at my former teachers for never exposing me to this.

[@themasterofthemansion3809]

It is worth mentioning that equations at 8:40 are near basic equations of hyperbolic functions: cosh(ai) = cos(a) and sinh(ai) = sin(a)i

[@PopeLando]

Tom never told you that his tattoo used to say e^iπ - 1=0. He had to go back to the tattoo shop to get the line added to make his minus into a plus, and it was twice as painful as the whole original tattoo!

[@firehawk128]

Euler's Identity is one of my favourite things in math. It just seems magical.

[@bonob0123]

I guess the confusing part has always been where does the "e to the i theta" come from in the polar form. Like I get theta is the angle but why does raising e to that power times i equate to a rotation by an angle. That is the intuition I still don't see. The rest after that about euler's formula and application at 0 etc is all algebra and makes sense.

[@MusicFanatical1]

He messed up really by starting with something he was trying to prove.

[@denny141196]

It's hard to explain in a comment but I recommend 3b1b's video on the same subject. It's something about the derivative of e^ax

[@bonob0123]

@@denny141196 fantastic reply with useful info! Do you have a link or title of 3b1b's video in order to find it?

[@KipIngram]

I find writing this in the form e^i*pi + 1 = 0, rather than e^i*pi = -1, to just be kind of a cheap trick to shove 0 into the expression so you can claim a fifth constant. For that matter, I think e^i*angle = cos(angle) + i*sin(angle) is far more useful and interesting - that covers everything, not just the one special case.

[@Tivnanmath]

A follow up is needed that shows how the sum/difference/double angle trig identities can be derived from Euler's formula.

[@beningram1811]

12:50 He says it's the same idea except two key things, and goes on to explain it a certain way, but i prefer to think of it differently. Since it's a negative of the first loop, it's doing the same thing but in the opposite direction and starting on the term that is being multiplied. i.e. one of the goes clockwise, the other does the same loop, but anticlockwise.

[@AtricosHU]

It is misleading to introduce i as "the square root of -1", because the complex square root is a multivalued function. The issue is, -i is also "the square root of -1", but the other square root. That's why introducing i as "i^2 = -1" is how textbooks do it. I understand that the video is supposed to be an introduction to complex numbers, but this is what leads to incorrect understanding of the complex square root.

[@teunvanwezel2282]

@@franmiskovic7630 Yes, but why define -i when you can just define i? :)

[@nHans]

Well, if you want to get pedantic, then i is the _principal_ square root of -1. And since the radical symbol "√" is defined as the _principal_ square root of its radicand, it's perfectly correct to write i = √(-1). Saying i² = -1 is correct but ambiguous for the exact reason that you mentioned: There are at least 2 solutions to that equation. Which is i, and which is -i? And when you consider quaternions, octonions etc., you'll discover that x² = -1 has as many solutions as you care to define, even an infinite number if you so wish. I have bigger problems with this video than calling i the square root of -1 🤣. Did you notice the circular logic he used at the very beginning?

[@AtricosHU]

@@nHans Yes, and I have commented on it :)

[@jayyyzeee6409]

To emphasize symmetry, I suggest expressing the first term of the infinite series of cos(θ) as θ^0 instead of 1 (even powers), since the first term of the infinite series of sin(θ) is θ^1 (odd powers).

[@davidellis1929]

The big question about exponential polar form is "What is the meaning of an imaginary exponent?" Once we derive Euler's formula, we can use it as the definition. The usual derivation of Euler's formula uses the Taylor series expansions of e(x), sin(x) and cos(x). Several comments in this thread give a derivation using differential equations. Is there another derivation that doesn't rely on calculus concepts?

[@pauldaniels3176]

Great video but I think you took a round-about way of deriving the infinite series for cosΘ and sinΘ. After writing down the infinite series for exp(iΘ) just collect the real terms for the series for cosΘ and the imaginary terms (factor out i) for sinΘ. I also prefer the exp(iΘ)+1=0 form because it has e, i, π, 1 and 0. You also came very close to one other weird result: exp(iπ/2)=i If you then raise both sides to the power of i you get i^i = exp(-π/2). In other words, i^i is Real!

[@BooBaddyBig]

I actually missed this lesson when I was at school. The teacher summarized it for me in five minutes when I got back. I was in a class of three. I was the only one that got it. And I was by far the worst student. The other two had thousand yard stares and were still recovering from the first lesson. I think the teacher had laid on the technical difficulty a bit too thick and gone through it too quickly or something, whereas the summary I got was just the right level of difficulty, and even though they sat through the summary I got as well, they still couldn't do the exercises and were totally bamboozled. I thought it was pretty funny. The problem was that the teacher was tending to pitch their lessons at degree level for A level students. The teacher was like: Why does he get it and you don't, you had a whole lesson on this!!! lol

[@anachromium]

You probably got an explanation tailored to your specific needs, while the class got a generic version. This might have been a bigger lesson for your teacher than for you!

[@Omnifarious0]

I noticed most of these relationships myself when I was considering how to hand-code the various math functions in C++ so they could execute at compile time.

[@PMA_ReginaldBoscoG]

Finally someone decided to do these things using a programming language.

[@Omnifarious0]

@@PMA_ReginaldBoscoG - I wrote the exp function. But it gets increasingly inaccurate as the exponent gets larger. I do have a plan for dealing with that, but it requires that the type that's used as the exponent support an absolute value function that returns a real number.

[@magnus0017]

I dunno, 2+2=4 comes up a lot . . . Awesome video, this being my favorite kind of math; I don't have the smarts to figure out or learn this stuff, but someone else figured it out, and someone additionally came along and found beautiful ways to show and play with it.

[@DrTWG]

I studied this a year ago , learned javascript and now make some very beautiful pictures by crunching complex numbers . Believe it or not.

[@doctorscoot]

graphic at 11m10s should be _counterclockwise_ so that -1 is on the left ;)

[@trevorallen3212]

Fun fact if you plugin the formula just right you can get hyperbolic trig functions out of it.

[@unclebrat]

What I have always loved about the mathematics is the stunning logic. I think this is what frightens so many people.

[@nHans]

No, what discourages most people from math is not the logic. It's the ugly, inconsistent notation and the confusing terminology. And most importantly, the boring, uninspired way in which it's taught by hassled teachers who're grossly underpaid and under tremendous pressure to complete the syllabus and improve test scores, while being perpetually short of time and patience.

[@unclebrat]

@@nHans I am so sorry to be this late in my response. While I agree completely with your comment of the problems that teachers have, there are certainly greater problems. I don't feel the notation is inconsistent until you deal with mathematics that transcends the usual boundaries (i.e. Whitehead and Russell) or are formulated on abstract rules. Terminology can be learned in a matter of two weeks. A much larger problem is indicated by the use of "The General Equation" so often given to students in textbooks. There is no derivation, no break-down, nothing to tell the student what is going on. The Taylor expansion is usually shown as a general formula with little explanation. These are in textbooks. Of course, my familiarity of this is over 50 years ago, but I still find it prevalent today.

[@PinoyManUtdFan]

Great video. But the proof all relied on the polar coordinate formula. For the uninitiated we have to show why raising e to the power of theta times i is a rotation of theta degrees in the complex plane. This then makes the video complete. 😀

[@shaunhutchinson4707]

"And this is why Sin(theta) is approximated by just theta...if you're an engineer", and I took that personally. But seriously, apart from questionable orders of explaining things earlier in the video, in a quick tongue-in-cheek comment explained to me more why I spent my third year using the small angle approximation during my vehicle dynamics lectures. Sure, my lecturer showed us how close/the limits of using that approximation was with various values for theta using MATLAB a few times, but never once told us where it comes from or why that's the case, especially when its something as simple as it being the first term of a well known series.

[@artsmith1347]

Because it was "obvious" ... to the PhD professor who had long since forgotten why it was intuitive to him as he taught the class. It seems that a number of math facts/skills were glossed over -- or not covered at all -- at the ABET-accredited university I attended. I have since learned many of those things from KZbin videos. For example, I don't recall anyone making larger use of the quadratic formula, such is in: x^4 + 2*x^2 - 5 = 0 is a "quadratic equation in x^2" -- solve u^2 + 2*u - 5 = 0 with the quadratic formula -- then x^2 = u = ( -2 +/- sqrt(4 + 20)) / 2 -- which can be used to obtain all four roots -- x = +/- 1 +/- sqrt(6). I see KZbinrs do that often enough to make me wonder why it wasn't mentioned in one of the classes I took. Maybe it was, and I missed it?

[@davidianmusic4869]

I soooooo needed that. Big thanX

[@TomRocksMaths]

[@TheMindfulCraftsman]

A follow-up introducing Geometric Algebra as a generalization of Euler's formula would be neat next :)

[@ahobimo732]

I've seen a few videos on Euler's formula, but I think this is my favorite so far. Such a fascinating area of mathematics.

[@TomRocksMaths]

[@RedStinger_0]

I'm taking Intro To Complex Analysis this semester, and this formula was really cool and useful when I first learnt it.

[@garthreid355]

I enjoyed working with those definitions for sine and cosine when I was studying Complex Analysis at the BSc level.

[@peterflom6878]

I've read in several places that complex numbers entered math because of the real solutions to cubic equations, where they are needed as intermediate steps

[@justsomerandomguy8210]

I remember someone once tried to explain this to me, I was 8 at the time.

[@jakeb6977]

cosine approximation in sigma notation: sigma(n=0, ) of ((((-1)^(n))(θ^(2n)))/((2n)!)) sine approximation: sigma(n=0, ) of ((((-1)^(n))(θ^(2n+1)))/((2n+1)!)) using infinite terms, as Tom said, does exactly match the functions, but that's not always the intent of a sigma

[@IndigoWhiskey]

perfectly timed as id just got myself to a question on this one. gonna ve a long deep dive and i havent even started the cliffnotes yet.

[@hugodesrosiers-plaisance3156]

I'm gonna show this in my answers at basic trig exam for my classes when asked to find cosTH.

[@JuanLeyva-fm9tj]

How did he arrive at e ^ (itheta). Where is the proof for that. I understand z = x + iy but how is e part of this. Is that just by definition of a complex number representation.

[@gordonstallings2518]

Series expansions are presented at about minute 15. It's enlightening to graph the series expansions. For example, graphing the first term of Cosine delivers a parabola. The first two terms presents a familiar polynomial curve with four roots. Successive terms build outward. Rather humbling to imagine this going on infinitely!

[@irakyl]

something no math educator has been able to teach me, is an intuitive way to see why raising e to the power of i*x makes you rotate in the complex plane. not even 3blue1brown was able to do it. it's just something you have to accept.

[@tabeh-]

this whole video is somewhat "wrong" actually, e^theta is the exponential form of complex numbers not the "polar". the polar form is the one expressed by the trigonometric functions, but i digress. it's probably not to confuse viewers, still kinda rubbed me the wrong way. to answer your question, it's because of how we define the cos(z),sin(z) and e^z functions in the complex plane. they're actually defined by their taylor series (so this plays nicely when you plug in real values). the interesting part is (and this is actually how Euler's formula arises) is that you can arrange the elements of the e^ix series in a way that the "real" part becomes the cos(x) series and the "imaginary" part becomes the sin(x) series. therefore, e^ix = cos(x) + isin(x). and this is all due to the properties of the imaginary unit, no such result exists in the "real" world. hope this helps a little. EDIT: minor mistakes

[@genessab]

It’s sooo beautiful if you look at the Taylor series for e^ix, sin(x), and cos(x). Not only is that the way Euler derived his formula, but just going through the proof yourself makes the rotations feel *obvious*. Gl!

[@gregoryfenn1462]

I think the idea is that it is an extension of derivatives from real to complex numbers: define f(t) = e^(it). Then f'(t) = ie^(it), by applying the classic real-number differentiation rules of e^(ax) -> ae^(ax). Therefore f'(t) / f(t) = i. But this ratio is kind of weird, it means that f'(t) is i-times f(t), so f(t) is changing but in a "sideways" direction. Think about changing 4 to 12 by multiplying it by 3, you are making 4 move along the real-axis. But the imaginary axis is perpendicular to the real axis, so the change from f(t) to f(t)+f'(t) must be a change perpendicular to f(t) too. And any movement that is perpendicular to the initial velocity is a rotation! The reason is simple: the tangent to a circle is perpendicular to its radius. perpendicular motion is the same as a rotation.

[@cleon_teunissen]

As pointed out in the earlier replies: the connecting factor is the behavior under differention. As we know: the exponential function is its own derivative. d(e^x)/dx=e^x The sine and cosine function also have this unique property that you get the same function back when you differentiate. In the case of the sine and cosine the return to the same function happens in a 4-step cycle. d(sin(x))/dx = cos(x) d(cos(x))/dx = -sin(x) d(-sin(x))/dx = -cos(x) d(-cos(x))/dx = sin(x) Now let me define a numerical entity such that I can reproduce that 4-step property (under differentiation) with the expontential function. Let me give that numerical entity the name 'incremental unit', for short: 'i' i^1 = i i^2 = -1 i^3 = -i i^4 = 1 Now the sine and the cosine can be expressed in terms of the exponential function: sin(x) = (e^ix - e^-ix)/2i cos(x) = (e^ix + e^-ix)/2 The property that you get the same function back under differentiation is one that only a few function have. It's very, very restrictive. There is very, very little room for the exponential function and the sine/cosine functions to be any different from each other. The close relation becomes starkly apparent when you read the Taylor series expansion.

[@mestar12345]

You can try to find the explanation from the book "Road to Reality". It just uses the fact that log is defined such that it turns multiplication into addition, and then when you place imaginary part and real part on a plane, multiplication becomes rotation, and in polar form, rotation is just adding angles. No other magic other than introduction of square root of -1.

[@davidbrosius7518]

You didn't explain why the polar form uses re^iΘ, you've explained why the polar coordinates use r and Θ. Why does e come in to play, and why exponentiation, doesn't that just wrap the value around the circle multiple times?

[@Sad_King_Billy]

If I ever had a teacher as passionate as Tom, I might have gotten into STEM a lot sooner.

[@TomRocksMaths]

[@rosiefay7283]

0:14 This goes halfway round the unit circle and then rightwards by 1. I prefer e^{2\pi i}=1, going all the way round the unit circle.

[@petek1365]

If e^(i*pi) = -1 then it is tempting to assume you can take the natural log of both sides to get Ln(-1) = i * pi and then use this as the definition for negative logs. Unfortunately I seem to recall this is definition is not valid but I can't quite remember why.

[@ignaciorodriguez5636]

I've just started to learn this at university!!

[@mosab643]

The actual explanation in fact goes the other way around. You start with the taylor series of cos(theta), sin(theta) and e^theta and then get to the identity. Otherwise one may ask where the term z=re^i*theta even came from.

[@wazikaxani4879]

We just used this in Electrical circuits. Pahsors, which stands for phase vectors.

[@anywallsocket]

Totally would've benefited from an explanation of the polar complex coordinate -- Euler's number is just presented as working on the circle, and seasoned scholars understand why, but the layman don't.

[@drag0nblight]

The Mandelbrot Set made me love this math, Euler's Formula.

[@I.____.....__...__]

12:55 The signs of the powers of 𝑖 and -𝑖 are the same, except shifted two places, they both go + + - -, they just start at different points in that cycle… just line how _sin_ and _cos_ are the same graph, just shifted.

[@peterslattery9581]

I absolutely love how my first introduction to numberphile and the concept of imaginary numbers was from their video on quartonians Every year I have a little anniversary we’re I watch that episode and every year I understand quartonians more and more Also just wanted to say thank you since Because of your videos I had the confidence to declare a major in Pure Maths at Rutgers NB I am very exited and my freshman year has gone well My proof class is challenging but enjoyable keep making amazing content and my goal would be to be on a numberphile video someday till then adios!

[@maaikevreugdemaker9210]

11:20 I have the opinion that this should be explained first after explain i^2=-1, because it fundamentally grasps why we are dealing with rotation

[@maaikevreugdemaker9210]

Nevertheless this video is excellent as an algebraic intuitive introduction to imaginary numbers