Even in complex numbers, sin(2z) is periodic with period pi, so you pretty much have to get a picture like that. People generally know about sin(2z)=2w when w happens to be a real number between -1/2 and 1/2, and the two sets of dots just go off of the real line when you leave that domain, while still being two sets of dots each with that spacing.

How very Fascinating! Thank you Dr. Peyam! ^.^

Very cool stuff! 👍👍😀

Very cool! I didn't see the simple double angle formula and decided to substitute sin(z) for √(1-cos^2(z)). I'm still working on the solution, but it appears that the golden ratio is involved, along with the complex arccos function

Very nice equation, thanks a lot👍

nice as usual enjoying yer being back posting regularly ❤8

I sure do want more math like that! The only place to see this kind of math 🔜

PROF thanks a lot)

"multiples of myself" 😁😆😄 So, from Dr. Peyam's point of view - there are four: 2 pi M. i = two (Peyam an' I) = 4 Peyams. Put all of them to work, Dr. Peyam - three more videos, please.

Nifty! I'm going into math and I should probably memorize all the angle identities so I can solve problems like this better.

Complex is beautiful😊

Please finish the video by writing out all solutions together so that we can compare them. Thank you. Keep up the great work.

@drpeyam

Жыл бұрын

Thank you but I’m confused what you mean

@GreenMeansGOF

Жыл бұрын

@@drpeyam For example, at the end, we have the formula for the case 2 solutions but you erased the case 1 solutions. It would be nice if you rewrote the case 1 solutions so that we have all the solutions written on the board at the same time. I hope that clarifies the confusion.

Yay we have pim😃

I found 8 more valuse: Z=-iln(+-1+-sqrt(+-sqrt(3)i) +-= Plus or minuse

@drpeyam

Жыл бұрын

Too many +/-

I do not quite understand how2piMi works, but this is very fascinating. Thank you.

@anarchy5369

Жыл бұрын

Periodicity of a unit circle where M is just some integer. Every 2pi, you get to the same point you started at

Pi am I? No, pi are you. :-)

@drpeyam

Жыл бұрын

Hahaha

I really like your content. 🙂 I understand everything. Do you work at the university? Big greetings from Poland! 🙂📏📐📈📉👍

@drpeyam

Жыл бұрын

Thank you!!! I work at Brown University :)

What a fun problem! I think you made a small error when drawing the solutions in the complex plane: case 1 is above the real axis, not below. This is because ln(-2+√5) is negative, so -(1/2)*ln(-2+√5) is positive. In fact, -2+√5 = (2+√5)^-1, so -(1/2)*ln(-2+√5) = (1/2)*ln(2+√5). Thus the imaginary part of the case 2 solution is exactly the negative of the imaginary part of the case 2 solution. Isn't that fun?

Mysterious!!!! Please change the title

@drpeyam

Жыл бұрын

Why?

@samosamo4019

Жыл бұрын

@@drpeyam qualifying as mysterious!

@drpeyam

Жыл бұрын

It is mysterious!