Problem Solving | Complex number basics.

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We look at some basic tools involving complex numbers for problem solving contests.
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Опубликовано:

25 июн 2024

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Комментарии : 62

@madladphy2570 3 года назад

Are you planning a complex analysis course on the channel? Would absolutely love it

@MichaelPennMath 3 года назад

We will have to see. I may request to teach complex analysis for an upcoming semester and then the answer would be yes!

@Invalid571 3 года назад

@@MichaelPennMath Please do, I'd also like to see a complex analysis course from you. ☺

@flipsareflippingfun2971 3 года назад

Yea, I’ve seen Peyam doing alot of that stuff, maybe you could do some sort of integral series where you show an integral being done through complex analysis and the typical real way

@TheRockMorton 3 года назад

Most excellent complexity instruction! i love i

@alperenkoken 3 года назад

Could you make problem solving videos about geometry bashing(vector,bary,inversion...)?

@captainsnake8515 3 года назад

30-minute video + michael penn + complex numbers!? Hey Ferb, I know what we're gonna do today

@TechToppers 3 года назад

Old stuff... Did you watch the movie?

@MarieAnne. 3 года назад

At around 32:00, when equating −P(1) = ζQ(1) + ζ²R(1) = ζ³Q(1) + ζR(1), we should end up with Q(1) + ζR(1) = ζ²Q(1) + R(1). Solving for R(1) gives R(1)(1-ζ) = Q(1)(1-ζ²) or *R(1) = Q(1)(1+ζ)* [not R(1) = Q(1)] Plugging into previous equation Q(1) + ζR(1) = ζQ(1) + ζ³R(1) we get: Q(1) + Q(1)ζ(1+ζ) = ζQ(1) + Q(1)ζ³(1+ζ) Q(1)(1+ζ+ζ²) = Q(1)(ζ+ζ³+ζ⁴) Q(1)(1+ζ²−ζ³−ζ⁴) = 0 Now the same argument can be made, but with a different expression. Since 1+ζ²−ζ³−ζ⁴ ≠ 0, then *Q(1) = 0* → Q(1)(1+ζ) = *R(1) = 0* We can then proceed to show that P(1) = 0 as shown in the video.

@megauser8512 3 года назад

Yeah, I saw the mistake that he made too, but I just read your corrections, and you are right, with no mistakes.

@quantabot1165 3 года назад

Yes!! I would love to see a complex analysis playlist....it would be just amazing

@giacomovicentini3495 3 года назад

I second that! I had planned for a loooong time to start learning complex analysis, and I see this as an amazing opportunity!

@virajagr 3 года назад

32:05 it should be zeta² Q(1) not zeta Q(1) although answer is still same

@mikeschieffer2644 3 года назад

How does this change things?

@mikeschieffer2644 3 года назад

So then R(1) = Q(1)(zeta + 1)

@inigovera-fajardousategui3246 3 года назад

That's it

@mikeschieffer2644 3 года назад

If R(1) = Q(1)(zeta+1) instead of R(1)=Q(1) could you show me how the answer ends up being the same? Thanks!

@virajagr 3 года назад

@@mikeschieffer2644 it's because the equations are homogeneous equations in terms of Q(1) and R(1). Homogeneous equations are equations which doesn't have any constant value. If you got n homogeneous equations in n variables, (0,0,...0) is always a solution to it called the trivial solution. Infinite solutions can exist if the equations are linearly dependent. You can learn more about this in Linear algebra.

@goodplacetostop2973 3 года назад

34:10 Hello Michael, hello everyone, what’s up? I have one suggestion for Mr Penn AND one homework for viewers. Suggestion is from IMO 1997 short list. Do there exist functions f and g (both R -> R) such that : 1. f(g(x))=x^2 and g(f(x))=x^3 for all x in R ? 2. f(g(x))=x^2 and g(f(x))=x^4 for all x in R ? Homework : Choose three random real numbers X1, X2 and X3 from the interval [0;1], independently of each other. 1. What is the probability that X1 is stricly bigger than X2 + X3? 2. In general, what is the probability that the largest of the three random numbers is strictly greater than the sum of the other two? 3. Bonus (the source only gives the answer but not the whole explanation) : for X1, X2... Xn chosen in the same conditions (n bigger or equal than 2), what is the probability than X1 > X2 + ... + Xn

@virajagr 3 года назад

1. To choose (x1,x2,x3) randomly in [0;1], we can choose a random point in the unit cube, with one vertex at origin, and the sides parallel to axis. Then we construct the plane x = y+z. It will cut the cube in 2 portions. One of them will satisfy x > y+x and the other x < y + z. The former has a volume 1/6 and volume of cube is 1 so probability is 1/6 2. Look at X1 > x2 + x3 and x2> x1+x3 First implies x1>x2 and second implies x2>x1 which means the set of triplets for first one is exclusive to the set of triplets for the second one. Similarly we can do for the third inequality as well. Since they all are mutually exclusive, probability would just add up i.e. 1/6+1/6+1/6 = 1/2 In general it would be 1/(n-1)! 3. X1>0 has probability 1 = 1/1! X1>x2 gives y=x line through a unit square with vertices (0,0),(1,0),(0,1),(1,1). It divides in half so probability = 1/2 = 1/2! X1> x2+x3 is already done and 1/6 = 1/3! Hence generalization should be probability = 1/n! Maybe we can do the same geometrical interpretation in Rⁿ and the nth volume of the favourable region would be 1/n!. We can prove this using induction.

@iridium8562 3 года назад

@@virajagr how could you prove that the volume of that region equals 1/6?

@MichaelPennMath 3 года назад

thanks for the suggestion. I'll give it a go!

@virajagr 3 года назад

@@iridium8562 it would be a tetrahedron with base area = 1/2 and height = 1. V of tetrahedron = Ah/3

@maxxis4035 3 года назад

I've just started working with complex numbers, great video!

@hellosquirrel7271 3 года назад

I wanted a video like this since so long ! Thank you so much :3

@abdlazizlairgi9690 3 года назад

Best teacher ever I've learned many things from you, thanks

@2funky4u88 3 года назад

would be epic if you covered contour integration and some theory behind that

@noumaneelgaou1624 3 года назад

We can in simple method we have polynomials P(1)+XQ(1)+X^(2)R(1) has 3 different root then this polynomials is nule then P(1)=Q(1)=R(1)=0

@mangai3599 3 года назад

Oh, at 18:18 that was really 👍 👌 👍 👌 👍 👌!

@mangai3599 3 года назад

At 5:55 I think scaling is more used word for that!!

@mijmijrm 3 года назад

what i think is cute about complex number is: it's simple to create integer hypotenuse rt triangles. Eg. (2j1)^2 = 3j4

@devildracofrostborn1340 3 года назад

amazing knowledge in all videos

@cauchy2012 3 года назад

Just amazing I was need this video so much thank u you are the best

@bamonroe Год назад

This is great thanks! I was looking for some examples of patterns that look unsolvable with real numbers, but become solvable with complex numbers, like the sum of squares problem you explain starting at 18:30.

@dimitrisandroid6483 3 года назад

More problem solving videos with basic theory please!

@grzechu9751 3 года назад

good video, you should do a video about vieta jumping

@djvalentedochp 3 года назад

Nice work

@psy7669 3 года назад

amazing

@mathsaddict7029 3 года назад

what is the easiest way of contacting you , professor?

@nuclo7586 3 года назад

I dont get it. P(1)=0 implies automatically P(X)=(X-1)A(X) where A is a polynomial, if P is a polinomial. Isnt that trivial ?

@quantabot1165 3 года назад

84K,

@fashnek 3 года назад

7 midroll ads before 9:00 :(

@rikhalder5708 3 года назад

Please tell what's difference between norm and modulus

@MichaelPennMath 3 года назад

No difference, it is just the setting. norm usually is reserved for vector spaces and modulus for the complex numbers. I think this is fairly universal.

@rikhalder5708 3 года назад

@@MichaelPennMath thank you

@rikhalder5708 3 года назад

@@MichaelPennMath but why magnitude of a vector is represented by modulus. The magnitude of vector should be represented by norm notation. I am very confused.

@roberttelarket4934 3 года назад

The powers that be need to eliminate the words imaginary numbers and complex numbers once and for all! Their presentation should be by dyads as taught correctly in modern abstract algebra.

@mangai3599 3 года назад

At 34:11 🤒😵🥺🥺🥺 wait ✋ I thought that Proffessor will also cover some complex numbers in geometry.