Small nit: At 19:45 cos(z) is missing a negative sign, it should be cos(z) = (e^iz + e^-iz)/2 not cos(z) = (e^iz + e^iz)/2 (which makes sense because that would simplify to e^iz)
For those who don't know, it goes like this: e^iz = cos(z) + isin(z) as per Euler's formula e^-iz = cos(-z) + isin(-z) since cos is even, we have: cos(-z) = cos(z) since sin is uneven, we have: sin(-z) = -sin(z) thus e^-iz = cos(-z) + isin(-z) = cos(z) - isin(z) so, if we want cos(z), all we need to do is start by adding e^iz and e^-iz: e^iz + e^-iz = cos(z) + isin(z) + cos(z) - isin(z) = 2cos(z) (since isin(z) - isin(z) = 0) then divide by 2: (e^iz + e^-iz)/2 = (2cos(z))/2 = cos(z) If you want sin(z), all you have to do is subtract both terms: e^iz - e^-iz = cos(z) + isin(z) - (cos(z) - isin(z)) = cos(z) + isin(z) - cos(z) + isin(z) = 2isin(z) Same here, divide by 2i this time: (e^iz - e^-iz)/2i = (2isin(z))/2i = sin(z) Thus we have: cos(z) = (e^iz + e^-iz)/2 sin(z) = (e^iz - e^-iz)/2i
Thank you for taking time to include a large number of functions in a short presentation. Meaningful big picture with details. I will have to review several times, but very helpful. Thanks.
Just beautiful stuff. Crystal clear explanation and motivation. Captivating introduction to Complex Functions. I eagerly look forward to the complex logarithmic function and using complex analytic functions to solve partial differential equations that involve Laplace & Inverse Laplace Transforms.
Thank you so much professor, I just found this video and this topic when I have the course of signals processing this semester which started just 2 weeks ago. I am so grateful for this, please keep uploading sooner...
Is there a forum somewhere to exchange ideas or for questions? I would be interested to know how DMD and HODMD (book of mr Vega and LeClainche) are connected from your perspective? Do you have a vide about it somewhere? Or simply through Koopman operator?
I love that you include the board erasing at the end. Helps to feel accomplished for the journey, and is just simply cute to see!
Small nit: At 19:45 cos(z) is missing a negative sign, it should be cos(z) = (e^iz + e^-iz)/2 not cos(z) = (e^iz + e^iz)/2 (which makes sense because that would simplify to e^iz)
True
Yeah, I started arguing with the video :)
@@Urgleflogue 😅😅😂
For those who don't know, it goes like this:
e^iz = cos(z) + isin(z) as per Euler's formula
e^-iz = cos(-z) + isin(-z)
since cos is even, we have: cos(-z) = cos(z)
since sin is uneven, we have: sin(-z) = -sin(z)
thus e^-iz = cos(-z) + isin(-z) = cos(z) - isin(z)
so, if we want cos(z), all we need to do is start by adding e^iz and e^-iz:
e^iz + e^-iz = cos(z) + isin(z) + cos(z) - isin(z) = 2cos(z) (since isin(z) - isin(z) = 0)
then divide by 2:
(e^iz + e^-iz)/2 = (2cos(z))/2 = cos(z)
If you want sin(z), all you have to do is subtract both terms:
e^iz - e^-iz = cos(z) + isin(z) - (cos(z) - isin(z)) = cos(z) + isin(z) - cos(z) + isin(z) = 2isin(z)
Same here, divide by 2i this time:
(e^iz - e^-iz)/2i = (2isin(z))/2i = sin(z)
Thus we have:
cos(z) = (e^iz + e^-iz)/2
sin(z) = (e^iz - e^-iz)/2i
@@Bard_Gamingbecareful with odd and uneven, many non even functions are not odd
I am locking forward to hear the next lecture 👍
Yeah, staying tuned for the complex logarithm
Thank you for taking time to include a large number of functions in a short presentation. Meaningful big picture with details. I will have to review several times, but very helpful. Thanks.
This series very nice, recognizing your work better than arbitrary ebisode thanks a lot prof
Thanks sir to bring all the things at one place for which I was kept searching.
Thanks for the lecture, professor!
Thanks. When you wrote zⁿ = [R...]ⁿ at about time point 01m:50s , did you intend to write another set of [ ] within?
Yes, there lacks an inner pair of brackets: [ R [ cos(theta) + i sin(theta) ] ] ^ n
Well spotted. I was wondering if I was the first to see this. That's why I always read the comments first.
Just beautiful stuff. Crystal clear explanation and motivation. Captivating introduction to Complex Functions. I eagerly look forward to the complex logarithmic function and using complex analytic functions to solve partial differential equations that involve Laplace & Inverse Laplace Transforms.
Thank you so much professor, I just found this video and this topic when I have the course of signals processing this semester which started just 2 weeks ago. I am so grateful for this, please keep uploading sooner...
If you check the playlist you can find all videos ahead of time
@@Eigensteve Yes I found it. Thanks a lot
You are a great teacher... thank you very much for these awesome lectures...❤❤❤.
Is there a forum somewhere to exchange ideas or for questions?
I would be interested to know how DMD and HODMD (book of mr Vega and LeClainche) are connected from your perspective?
Do you have a vide about it somewhere?
Or simply through Koopman operator?
Love from Türkiye ❤
What is called here "Power series" is in my view better described as a complex polynomial.
Great lecture! 😊
Bro, I would like to know how do you do your videos, like, the set up and equipments from scratch
THANK YOU A TON
Thanks
excuse me sir i think their is mistake with z in place of theta as z is a complex no.
like e^z
7:00
He looks like doppelganger to steve jobs except ge is wearing bracelet
I wish he hadn't erased everything at the end of the video.
Is it just me or is he actually writing backwards, like, huh?
he writes it normal way and then image is mirrored horizontally
Don't say uhh ahh okay?