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I'm glad that you found the analogy useful.

In the last part you have (2-1)(e^iu+1)=0. So either the first or the second parenthesis is zero. Obviously (2-1) is not zero, so (e^iu+1) must be zero which is the euler's identity.

@getcomplex That's ✅

💛

@@getcomplex 2(e^(iu) + 1) - 1(e^(iu) + 1) = 0 || e^(iu) + 1 = 0 ~> e^(iu) = - 1,, -1 + 1 = 0 || 2 - 1 = 0 ~> 2 = 1,, 2 - 2 = 0 ,, 1 - 1 = 0 || (2 - 1)(e^iu + 1)=0 ~> 2e^iu - e^iu + 2 - 1 = 0 ~> e^iu + 1 = 0 || u = pi

Thanks. Glad you liked it 💛

💛💛💛

Happy to hear it's been useful. Send me an email. You can find it in the descriptions of the channel.

Thank you. Happy you liked it 😍😉

I'll drop the background music in my upcoming videos. I agree that it might be annoying. I was just experimenting and although the figures have gone up but I find the background music distracting myself, too. Thanks for sharing your opinion.

@@getcomplex Many thanks - I have just become a subscriber.

Happy to hear that. I hope you watch and enjoy the upcoming videos 💛

Thanks bro 💛

Glad you enjoyed it

Bro has just started making videos and is happy you are watching them 😄💛

Maybe even a series of videos as a playlist. Nice suggestion. I'll do it.

Glad you liked it 💛

Wow that's such a kind compliment 😍 I'm happy you liked it Arda.

It actually is 💛

Hey Nazish. I use powerpoint and its animations. With a voiceover on the slides being screen recorded.

Thanks a lot. Happy you liked it

You are welcome. 😍

It really is!

Glad it was helpful!

Hey I just wanted to let you know that I enjoyed your video. At the end of the day it's all semantics and pedantic. People looooove being technically correct especially on the internet. Don't let yourself being dragged down. I know where you were comming from. You can look at it from different perspectives. If you look at certain aspects like scalar multiplication and Complex Number Multiplication of course that is different. if you want to definie a multiplication in R² like (a,b) * (c,d) = ( ac - bd, ad + bc ) then you can show it's isomorphic to the complex multiplication. But I agree. The way Vector Spaces and Complex Numbers are introduced in school they are different beasts. Keep up the good work ^-^

​@SimchenThank you so much for your heartwarming comment. I really appreciate it. I'm working on this channel beside my work which is not even related to physics or math. My job takes me about 50-60 hours a week and this channel is where I talk about my passion. Despite all the different comments I got on this video, I should admit that I learned a lot and what I am truly trying to show here is my effort towards what I love and this is the journey I have just begun amd will continue for the rest of my life. In fact, a kind of documentation of my journey. Again, thanks for your comment. Really means a lot to me.

I wholeheartedly accept your criticism and admit that my explanations might have caused some confusion, especially in the part I was talking about multiplications. As a physics graduate, I have some knowledge about the topics you mentioned, but I couldn't fully explain what I meant which show I should be more careful about making my next videos. In this video I was trying to show that complex numbers are not JUST like vectors and I clearly didn't talk about other forms of multiplication. I'm going to create another video and include some good comments under this video (including yours for the beginning of the video to roast myself 😄) and explain more about what I had misleadingly presented. And I totally agree with you on buying books and reading from the source itself. I have tried to stick to these sources so that I don't cause any confusion, but sometimes I miss things. Some of the comments under this video have been a wake-up call to me in order to make better videos in the future. Sooo, the bottom line is that I take full responsibility for my videos and try to make better videos which are clear and thorough in the future. Thanks for your time

By that part of the video I meant that we don't have the algebraic multiplication for vectors. But it seems that my explanation is a bit misleading. In this video I was trying to justify that complex numbers are not just vectors, but more.

@@getcomplex should check out geometric algebra... has a way of multiplying vectors which doesn't reduce down to just the dot product but that is a component of it. And it unifies complex and vector algebras into one algebra. One product to multiply vectors or imaginary numbers or even things like quaternions

Thanks for the suggestion. Yeah I checked and learned geometric algebra and it is really nice. I have made another video talking about multiplication of complex numbers which you can find in the caption of the video and in that I have made it more clear and also shown how we can find the area of a polygon using complex multiplication. I really appreciate your time for letting me know about these useful information you have. I'll definitely study quaternions later.

Yeah you are right, but seem to be very angry. It's not worth it my friend😄. It's just a bad video as you say and by reading your comment I might have learned something new, so thanks for your information, but not your tone.

Glad you liked it.

...under addition.

Level million plus, there are no vectors any more, just branes😂😅😮

🤓🤓🤓

😄👍

Yeah I see. I actually don't know what modern academia says, but I'm trying to learn some science stuff and make videos about them. I've learned a lot from some nice people commenting under my videos and I find this process an amazing learning experience. And it is helping me to even make better videos and improve.

Thanks for your explanation.

Glad it was helpful!

Thanks, will do!

So why do these complex numbers work in quantum physics? Firstly, there is rotation in quantum physics: spin. ,the second is the wave function There are places in the square where the function is zero, and this is similar to waves canceling out each other. If a^2+ab+ba+b^2 has axb=- bxa, that is, if you do vector multiplication, a^2+b^2 You get it. Geometric algebra can do this with bivectors. This is a model. Now we will do the same thing with another model. (1+1i)(1-1i)= 1.1+-1i+1i-(1i.1i)=1.1+0i-(-1)=1+0+1 😉 These are all models, not reality, they are tools to model reality.. If it doesn't work for you, you can find a different model and throw away the old one.

Now why should we rotate with "i" when we already have a model that allows rotation in 2 dimensions? Now imagine that you multiply the (a+b) vector with the (a-b) vector in this model. (a+b) (a-b)=(a.a)-(b.b)= a^2-b^2. If we do this on the complex plane (a+bi) (a-bi)=a^2+b^2 We get it. Moreover, this exactly mimics the anticommutative property ordered by quantum physics. We ensure the order by using complex conjugation in 2 dimensions and doing a juggling trick. (a+bi)(a- bi) =a^2- bi+ bi- (bi)^2 I choose a=1, b=1: 1-i+i-(i)^2=1+0i+1. We have provided the quantum order, but the problem is this: (a^2-b^2) and a^2+b^2 are not the same. This is not the same as Hamilton's model, which rotates a,b pairs in 2 dimensions. For this reason, rotation with "i" There must be a rotation operation in 3 dimensions. More precisely, the process of reflecting vectors on two planes positioned at 90 degree angles is a rotation action. Like the electric and magnetic fields... we think of the electric field as horizontal and the magnetic field as vertical and at the same time turn clockwise with "-i". You will rotate it counterclockwise with "i".😉😎👍 We defined axb=- (bxa) in the vector product (you can also interpret this as rotating fields as bivectors, your interpretation is your model) You are modeling the same thing in the + bi- bi=0 operation. Remember the right hand rule, the thumb is the result vector and points to the 3rd dimension. (axb) = turning right, thumb up. (- bxa) turning left, thumb down. now (+ 1) thumb up,(-1) thumb down. (a+bi)(a- bi) =a^2- bi+bi- (bi)^2 ...do we see what we are doing with - bi and + bi? We do the labeling in the third dimension with +i and -i. What a big trick😀😅

Nicely mentioned 😄 As far as I know, the dot poduct is like a map that maps two vectors to a scalar which is a real number. The number 3 is that scalar.

As far as I know it's a kind of vector multiplication. Maybe I should have also talked about cross product which is another kind of vector multiplication. To be clear, even cross product, which is another kind of vector multiplication is different from a complex number multiplication and the answer includes a combination of unit vectors that when looked at as one single vector is perpendicular to the planne in which our two first vectors exist. My main point in this video is to show that complex numbers form a field but vectors live in a vector space. And I can safely say that I tried my best to explain that and can also promise you that I'll try to get better at explaining things on this channel. Thanks for sharing your opinion and if you think I might have made a mistake in explaining the concept please tell me so that it is clearer to me.

I don't think we can safely say that without restricting what you're referring to as a vector. The dot product is a special case of the inner product (in this case taking two vectors to form a scalar), yielding an inner product space. All inner product spaces are vector spaces. The dot product is just a special (however common) case. To summarize, dot products don't exist without vector spaces (e.g. algebras) forming them. I am unsure why you're suggesting the idea of tossing out dot products as vector products to be a comfortable one.

I reccomend the videos on geometric algebra by mathnoma. He does a great job of talking about vector multiplication as the sum of the dot product and the wedge product (aka cross product). This flows seamlessly into complex numbers and into the law of sines and cosines. Really interesting videos and I wonder why this wasn't taught in my university math and physics degree. They are really excellent videos.

@@getcomplexAs others have pointed out, the biggest problem with this video is defining multiplication as the dot product. A much more natural definition of multiplication of vectors is the geometric product (the inner product+the exterior product). If you do this, and you have a Cl(1,1,0) Clifford algebra, then complex numbers become isomorphic to vectors and it makes more sense to say they are the “same”. (The 1,1,0 refers to the fact that one of your bases have to square to negative one while the other squares to positive one, which is also what makes complex numbers useful, but is not unique to them)

Yeah I got it. I just used the dot product as an example but it has made the video confusing. I'm making another video to show how complex numbers are related to dot and cross products so that it makes more sense. I don't know anything about Clifford algebra but I'll study it. Thank you so much for explaining.

Hi I draw what I want to show on powerpoint and add some replay animations. Then record screen the presentation and speak on it. When you draw something in power point, it gives you an option in animations called replay which draws your strokes on the screen.

@@getcomplex So these dynamic math videos are made on PowerPoint ?

Yes. Just power point animations.

And sometimes I use photoshop and camtasia too. But mostly power point animations

@@getcomplex I wanna publish some new math concepts on RUclips. Is it a good idea or I should do a paper for a journal. I learned English recently for that. I'm francophone from Mali. Please answer me

We can do what you said, but it would still be different from a field because of the multiplication. As you said the inner product is an additional structure and not mandatory for a vector space, it's not a built-in structure. However, in the structure of a field we have a built-in multiplication and that's where we can say complex numbers are different from vectors. It was the point I was trying to make.

A vector space with basis {1,i} and field ℝ is indeed ℂ, but that sells ℂ a bit short. What makes the complex numbers special and not just ℝ² is the multiplication. The complex numbers are equipped with a product which makes them an algebra. An algebra over a field F is an F-vector space equipped with a bilinear product. A bilinear product is a product that multiplies two vectors to get another vector (and follows some other rules). ℝ³ equipped with the cross product is an example of an algebra. ℝ² equipped with the bilinear product × such that (a,b)×(x,y) = (ax-by,ay+bx) is another algebra and is basically what the complex numbers are. But the complex numbers aren't just an algebra, either. They have an identity element under this multiplication, and all the nonzero complex numbers have inverses under this multiplication, and this multiplication is associative and commutative. All this together with the fact that (ℂ,+) is a group and that the product is distributive over addition means that (ℂ,+,×) is a commutative division ring, which is also called a field. So ℂ is a vector space, but it's also so much more.

Amazing explanation. Thank you so much. I'm going to dig deeper into these concepts like sets, groups, maps, rings and fields. And maybe share what I learned as a new post. Really appreviate your time wriring a thorough explanation 🙏

@@getcomplex In the video, you simply showed that if someone thought the product of vectors is the same as the dot product, they are mistaken."

@@thes7274473 Could you recommend a source to learn this ? This sounds super interesting but for someone like me totally unreachable with my current knowledge.

Nice one 😁😍

😍😍😍

Thanks

Yeah you are right. It is my bad. Thanks for mentioning 💛

Oh that's my bad. Thanks for mentioning.

Glad it was helpful!

Happy to hear that. I'll try to elaborate on each notion with examples in my next videos. Thanks for watching

I'm afraid I don't know

Of course. I'll be solving more cases in 3D

Glad you liked it Leona. 🥰