booooo

I am SO getting one!

Good I thought was the only one

That’s almost a diss but my smart friend has helped me more than my greatest professors just because I do the homework with my smart friend.

I actually almost liked this better, I feel like I'm getting a little more use to seeing i and sin and all that here, like their real and workable, watching 3b1b makes me question a lot of stuff, I love them both equally

@@abird9724 you can be as good as black/red pen if you do lots of exercises, but that alone can't help you reach 3b1b level

In general I think, Grant is great at instilling necessary intuition for problems, but black black pen red pen does more interesting and somewhat unusual topics, ie ln(negatives, a+bi . . . .)

Nice physicist solving

z approximately equals 2 to be rigorous lol

@Hepad More like engineers.

Any explicit number is comparatively small, overall.

Right from harmonic motion

You’re welcome! I am happy to hear this.

How old are you now

@@hyperplane69 the real questions

@@hyperplane69 around 90 probably

@@3c3k80*

thanks!!!

- said no one ever*

@@elymajdou5668 shut up and leave

@@arnavanand8037 he left ages ago

Indeed it was..

hmm... sorry... lol~

Jonas Lecerof yes, but look at the colour of his top O_o

Jonas Lecerof and it’s not even a pen it’s a chalk

Black board red shirt

ঈট

I heard it in the word _complex._

*z*

Well he used z for a complex number, which is usually composed of x + iy. So essentially that z is an ex-x. :)

+Tanith Rosenbaum I usually see a+bi, but it doesn't really matter since variable names are the least important thing in the entirety of mathematics, being a choice of style with no influence on the answer so long as you don't conflate anything.

Well, with a+bi, the joke wouldn't have worked :)

blackpenredpen you should learn how to write dotted lines like prof. Walter Lewin

blackpenredpen u

San Samman best comment ever

No

wtf

✅

Shut up, that was funny. You're a dad now

Damn the amount of continuous puns is enough to change even a woman into a dad

If you enjoyed this, consider taking complex analysis in university. It's filled with this :)

I'm in the second semester of 10th grade and this video was relatively easy to understand.

@@viktorramstrom3744 not much of a flex, he explains it really well

Keep learning math, grab some practice books (calc is fun imo), start practicing! Mathematics is a very powerful tool to learn

PokemonBronze wow!!!! How wonderful!!! I am very happy to hear this!!! Thank you for sharing and this definitely brightens my day!

@@blackpenredpen this is a jee adv question that u solved u r a legend

Then you must have never solved a advanced level sum.

@@govindbarwa1806 exactly this is at most mains level, and still comparatively easy

: )

thats like saying, z = " insert correct answer"

lol the fun part is arcsin‘s domain is -1， 1

Edward AA yes , in the real numbers there is no answer (ie. undefined). Need to be in the complex number domain to have an answer and, as shown, it’s tricky.

Inverse sine domain is [-1,1] :) but nice one. You almost got me there😂😂

This is very nice!!

Very elegant, love it! 👌🏼

USE -i PLEASE

or use i^-1

or i^i^2

@@GamingWithJumbo WHY? or maybe i^i^(π/2 - i*ln( 2+sqrt(3) ))

Good thing I'm not a mathematician

I like it, I just wish I could follow it #lost

In general, sin(z)=n If z=π/2 -i•log[n±√n²-1] And Cos(z)=n If z=i•log[n±√n²-1] Because sin(π/2-x)=Cosx

I hope I don't make a mistake but I think there is no proof of the existence of euler's formula in the complex world, so can we admit it?

@@romantchamalatdinov3640 This is a viable solution if all you are doing is finding some numbers you can plug in that satisfy the equation, but it does open up more questions than it answers. As a former engineering math professor let me tell you this solution is hyperbolic if you pardon my pun. If you look at sinh you may see better what he is trying to do here. This question is not explainable in 20 minutes and a lot of information is lost since one needs to understand that i is orthogonal to the REALS and this evolves into a helix.(and this is a hyperbolic explanation on my part since we need to consider tangent space and sub-tangent spaces etc.). . The topic is worth exploring in more depth in another video and I would hope Dr. Chow explains this is pretty much just a parlour trick with this level of explanation in his description. This is better understood in group theory. If you have a background in Group theory, specifically Lie theory and how when exponentiated, the Lie Group, this short paper may interest you. www.academia.edu/74997571/Geometric_Thoughts_with_Guest_Algebra_i_i_

Like the comment that says "12:34 that r is still stuck inside of that ln"

hehehehe!

Replying to a 4 year old comment with no reply other than bprp..... Ok

Wyatt Sullivan nice!!!

that does look very similar to inverse hyperbolics

this is actually quite amazing!!!

I was fascinated enough by this video to try doing this by myself in 12th grade :") along with the relevant solutions for arctan and arccos. Btw, since (n+sqrt(n²-1))×(n-sqrt(n²-1))=1, you can turn the second part of the RHS into +/- ln(n+sqrt(n²-1))

I can give you a brief rundown of how to switch where the plus/minus is: 2+sqrt(3) and 2-sqrt(3) are recipricals of each other, as can be shown by multiplying them together and getting 1 as the product. ln(a)=-ln(1/a) as ln(1/a) can be rewritten as ln(1)-ln(a) and ln(1) is 0 (or alternatively by rewriting ln(1/a) with a negative exponent and bringing down the exponent). Therefore ln(2+/-sqrt(3))=+/-ln(2+sqrt(3))=+/-ln(2-sqrt(3)).

Lisa Greene You legend

@@leahgreene5880 nice man. I like it😎

@@leahgreene5880 this is awesome! Does this frequently work? I never would have noticed they were reciprocals

@@Reallycoolguy1369 This only works in some specific cases :) When you have a number x + sqrt(y), then x^2 - y has to be equal to 0 for them being reciprocals. In this case, 4 - 3 = 1.

Leone Tassinari thanks! Be sure to see the description since I did make two small mistakes.

I saw the |z| = a^2 + b^2 error, but it didn't affect the proof at all. Anyway I must have missed the second Edit: oh I see, the imaginary axis. That's cool man, nothing major. Sometimes I make the same mistake myself. Anyway you've widened my orizons. I guess tomorrow I'll be looking into complex trigonometry :D

The only thing that’s impossible is division by 0. Everything else is possible, even if it isn’t real.

Sinx ning qiymati -1《Sinx《1

@@paulchapman8023 i mean, division by zero *is possible* its actually non determined, because there are infinite solutions For 10/2 = 5 we can think that 2×5=10 For 10/0 = y there is no number that makes 0×y = 10 so its impossible For 0/0 = z there are infinite answers that fulfill z×0=0 There are other ways to showing this but I think this is the nicest and least complex

YUP!

Whitechalkwhiteboldchalk

BlackblackboardWhitechalk

@@MT-in3tp black²boardwhitechalk

@@blackpenredpen how old are you?

Genius

Genius = i·Genus Imaginary taxonomic category?

LOLLLL

Avana hahaha

It's exactly what I told to myself at this moment of this video lmao

@@blackpenredpen hey are you from india (Andhra) ???????

Also make 12:34 a meme "that r is still stuck inside of that ln"

Thanks!

What hate

LOLLLL

"I have another one!" that made me laugh so hard

sameee!!

same

Valhalla

Dark magic!

Вам повезло, дружище. Китайский акцент реально напрягает

but you won't be able to see that much smaller mic thus no one will ever comment on my mic again... (My Mic likes the attention : )

He wants to be an Ood.

Hahaha! 😁⚡⚡

But that wouldn't be fun. Would it?

Why will he want a small Microphone?

Donez Horton-Bailey lolll nice!!

otis neill I was wondering about that too!

otis neill yes

Yes, and this has now been added to the description. Good catch, otis! And good correction, blackpenredpen!

wait... r cant = z so easy! especially since z has a complex component... unless absolute value assumes the Pythagorean formula on the ocefficients to get a single real number

the absolute value of a complex number (z) is defined to be the "distance" (r) away from the origin of the co ordinate system of all combinations of real and imaginary numbers (R x I). if z = a + bi then a is the real component of z and b is the imaginary component; or a is how far along the real axis z is and b is how far along the imaginary axis z is. if the modulus of z (|z|) is defined to be the distance away from the origin, we can draw a right triangle out of the values a, b and the distance r; where r is the hypotenuse of the right triangle. so if we wanted to know the absolute value of z and we knew the components of it were a and b we could use the Pythagorean theorem to work out how far away from the origin it is since we know the hypotenuse squared (r^2) is the sum of the square of the other two sides then we can take the square root of both sides to find that r = sqrt(a^2 + b^2). this would lead us to the conclusion that is z = a + bi then |z| = sqrt(a^2+b^2); if we define |z| to be the distance away from the origin. i hope that helped :)

Lol Ig you are in love with nerds

This needs 10,000 likes

mueez adam thank you!!!

Picard's theorem says: If a non-constant function has a taylor series with an infinite radius of convergence, that function must output every complex number with at most 1 exception. e^z outputs every value except 0, and sin(z) outputs every value with no exceptions.

Yes, in fact you can extend **any** existing function into the negative values and the complex plane

You could already extend logarithms to include negative numbers using badic logarithm rules. Log a (-b) = Ln (-1 * b)/Ln (a) = (Ln (-1) + Ln (b))/Ln (a) Know if you somehow knew e^ipi = -1, you know that ln (-1) = ipi Therefore: Log a (b) = (Ln (b) + pi*i)/(ln (a) However were not done yet. If you have some experience with the unit circle and eulers formula, you know that e^ixpi = -1 where x is an odd number. Therefore, log a (-b) has infinitly many complex solutions

karolak kolo will like to see extension of factorial into the negative values and the complex plane

@@sergey1519 look up Gamma Function

There isn’t a unique argument for a number since the complex exponential is periodic. Since “the” complex natural logarithm depends on the choice of an argument function, there isn’t a unique choice for the natural logarithm either. Every choice has a ray from the origin on which it is not defined. The “traditional” choice is to have arguments in the interval (-pi, pi), which leads to continuity issues along the negative real axis. You can change the interval, move the problem ray somewhere else, and calculate a natural logarithm on the negative real axis. Whether or not there’s any gain to doing this, I have no clue.

“That can’t possibly be real!”

Ruben i like your profile pic. Diabolic tutor

yes pls

Which one?

Cédric P. it is equivalent to saying -ln (2-sqrt (3)) = ln (2+sqrt (3)), which is equivalent to 1/(2-sqrt (3)) = 2+sqrt (3). if you multiply the left expression by (2+sqrt (3))/(2+sqrt (3)) you'll see its true.

probably yes , because another comment simplifies this by a vast amount using cosx=root of 1-sin^2x and substituting sinx=2 cosx=root3i

Thank you!!

True, he's not entirely careful about a lot of the properties still holding (the video would probably be twice as long if he did!). But you can check these properties using the definitions he gave for the functions. To pull the same trick as a math professor would, those are good exercises for the viewers ;)

I agree, I watched the video again carefully and this new definition is derived from old one, so everything is fine!

Arek Krolak you guys are virgins

@@highdino2181 minecraft

@@MuffinsAPlenty : Not sure he can use the Euler equation to prove it, since that is a circular proof then. To prove it, you would need to write sin and cos as an infinite series and show the equivalency.

that's great! I am glad that you like my choice of problems!

What do u think of finding @ in this A^@=-A it'll be interesting

Log_A(A^@)=log_A(-A) @(A) =log_A(-A)=log_A(Ai^2) @(A)=log_A(Ai^2)= log_A(A) +log_A(i^2) =1+2log_A(i) But log_A(i) =(Π/(2lnA))i @(A) =1+(Π/lnA)i :)

lol!

For blackpenredpen, yes. Anyone else, no.

Thanks for reminding haha

hehehe

josh davis yup!!!!

Euclid of Kekistan Yes. I have that around 18:45

Back again

That's not hard, you just use partial integration formula, integral of (u*dv)=u*v-integral of (v*du), with u being x^2 and v being the exponential in the first use, and u being x, and v being e^-x in the second use of the formula.

Dario Barišić i used integration by parts with the DI method and i got the answer. Thanks anyways :)