"being a physicist I bound to do something to upset the mathematician" got me 😂😂 Thank you for this, I've been struggling to understand my lecturer for complex analysis and your video helps me tremendously!
Complex analysis was one of my favourite courses at the university of the west indies, cave hill campus. I have a few more ' what if' questions, not only in the field of Complex analysis, but in other areas also 2:10
EDIT: This was indeed a stupid question on my part!! I forgot a basic fact about fractions 🤣. No need to answer this question, but I'll leave this comment up here, just in case someone has a mental relapse I did! If you want a good laugh, feel free to read my unedited question below 🤤 This may sound like a stupid question, but can f(z) be a polynomial? My reason for asking is this: Suppose we are integrating over a closed contour that does NOT include the point z=0 but does include a complex z_0 (where z_0 is not zero). The solution to this integral, assuming the integrand has the form f(z)/(z-z_0), is 2πif(z_0). But now, what happens if I rewrite the integrand as (f(z) + z_0)/(z - z_0 + z_0). All I've done was shift both the numerator and the denominator by z_0. z_0 is just a complex number, and not the integration variable, so I think this shift should be allowed. If I define a new function, say g(z) = f(z) + z_0, then the integrand is now g(z)/z. Remember that we have a contour that does not surround z=0. Therefore, the integral should equal 0, but according to the integral we started with, it should equal 2πi*f(z_0). This solution does not agree with the true solution provided by this integration rule (or identity or whatever it's technically called...😅). What am I misunderstanding about the limitations of solving an integral like this?
Could you explain why you just multiply 2pi*i with f(z_zero)? dont you have to take the integral and then multiply the result of the integral by 1/(2pi*i) ? Also, say z_zero in the integral was 4, could you only find f(4), since the formula has it as f(z_zero) equals the whole thing, or could you say find f(5) ?
The way the cauchy formula is stated by you, suggests to the learner that you are trying to evaluate f(z) nought. However you rearranged the formula so that you could evaluate f(z). This is a common mistake of teachers and lecturers. There is a way to overcome overcome
"being a physicist I bound to do something to upset the mathematician" got me 😂😂
Thank you for this, I've been struggling to understand my lecturer for complex analysis and your video helps me tremendously!
Complex analysis exam later today and this video might just have saved my life ty Nick!
Glad I could help!
This dude has 2.64k subs?How? He is that good. Straight to the point but understandable. Bravo Nick!
I tried many videos in few youtube channels to understand this. But you are the best💜️
Thank you so much 😀
Good work, I felt complex analysis was not as complex! Thanks!
This is bound to make the mathematician happy!
A really big thank you, I understood it because of your exemple, something my professor doesn't do!!!
Happy to help!
Complex analysis was one of my favourite courses at the university of the west indies, cave hill campus. I have a few more ' what if' questions, not only in the field of Complex analysis, but in other areas also 2:10
Beautifully explained, Thank you.
Well explained Nick! 🔥
Very good video, thank you very much!!!
Glad you liked it!
Besides these really weird pis which neither do look like lowercase pis nor like capital ones, its a very good video. Cheers!
This was a great video . The only thing I thought could be better were the examples number and complexity could've increased .
or there could be a part 2 to this video for that ! your explanation was quick and simple 😊
Great content bruv
Thanks man this is really helpful
This is so helpful. Thank you man!
WONDERFUL !
What happens if the simple closed curve is not positively oriented? 2:10
Thank you sir 😌
EDIT: This was indeed a stupid question on my part!! I forgot a basic fact about fractions 🤣. No need to answer this question, but I'll leave this comment up here, just in case someone has a mental relapse I did! If you want a good laugh, feel free to read my unedited question below 🤤
This may sound like a stupid question, but can f(z) be a polynomial?
My reason for asking is this:
Suppose we are integrating over a closed contour that does NOT include the point z=0 but does include a complex z_0 (where z_0 is not zero). The solution to this integral, assuming the integrand has the form f(z)/(z-z_0), is 2πif(z_0). But now, what happens if I rewrite the integrand as (f(z) + z_0)/(z - z_0 + z_0). All I've done was shift both the numerator and the denominator by z_0. z_0 is just a complex number, and not the integration variable, so I think this shift should be allowed. If I define a new function, say g(z) = f(z) + z_0, then the integrand is now g(z)/z. Remember that we have a contour that does not surround z=0. Therefore, the integral should equal 0, but according to the integral we started with, it should equal 2πi*f(z_0).
This solution does not agree with the true solution provided by this integration rule (or identity or whatever it's technically called...😅). What am I misunderstanding about the limitations of solving an integral like this?
Well Expained
Thanks
Thank you sir!
You are welcome!
Wonderful
Thank you so much dude
No problem
Thanks Nick
Could you explain why you just multiply 2pi*i with f(z_zero)? dont you have to take the integral and then multiply the result of the integral by 1/(2pi*i) ?
Also, say z_zero in the integral was 4, could you only find f(4), since the formula has it as f(z_zero) equals the whole thing, or could you say find f(5) ?
To overcome the problem. But i will leave you to work that out 9:27
Nice video!! one question tho, at 8:12 why does f(z0) = the derivative of f(z)?
Because f(z)=exp(z) the derivative is itself: f’(z)=exp(z)
Can n be generalized to all real number? I face fractions all the times.... (E.g. n = 5/2)
Thank you !
i don`t understanded please show more example
Why is it n!/2*pi*i in the formula, but when doing the solution you say 2*pi*i/n! ?
Please 🙏 show solved examples
Hey man, can we speak on private somehow ? i really need help understanding somthing.
Dankeee
Why (2pi.i)/1
I'm not immediately sure. I'm sure a derivation of the Cauchy integral formula would probably explain why.
The way the cauchy formula is stated by you, suggests to the learner that you are trying to evaluate f(z) nought. However you rearranged the formula so that you could evaluate f(z). This is a common mistake of teachers and lecturers. There is a way to overcome overcome
👌👌
Stop Back music
I find the examples too simple , somewhat more complicated ones would show more of this very useful formula
Thanks for the feedback. I'll try to add some more complicated examples in future videos.
who the hell is coshy?
Learn French!
very very helpful! thank you so much
Thank you brother