Seriously helpful videos. Incredibly basic applications but you review the process well enough that it is easy to expand to more difficult problems. You are getting me through my course in complex analysis.
Thank you so much! Super helpful. I am in my second year and my complex analysis exam is in 3 weeks. One of the hardest modules for me so this really helped me understand. :)
for the second example if we have 1/(3-z). why cant we convert that to 1/3(1-z/3), why do we need to have (z-1) in the fraction and why is the restriction |2/z-1| < 1, where did that come from ?
+Nick R hey guy, here is the trick, he said if |z-1|>2.......for a series to converge we must have |z-z.|3 that means z. is zero then we have to turn |z|>3 to |z|
+Nick R I also find this (Ex. 2) very confusing. The question, for me, is rather why we (at 4:26 ) choose to look at it in powers of (z-1) valid for |z-1|>2 at all? Nothing was specified about this on the board as in the previous exercise. This is all very new to me but the problem to solve might have been specified a bit unclear so I might have missed some info about Exactly what was asked of us.
+kiwanoish you are right by saying that it wasn't specified on board. but he just made mention of it that we gonna look at it at |Z-1|>2 note that z is a complex number and Z-1 is the distance from the origin of a circle centered at 1 ... this implies for any z to be within the radius of convergence then |Z-1| must be less than 1... according to Taylor series but for |Z-1|>2 that means we are looking at the region outside the circle of radius 1...then we have to use Laurent series. but we have to scale it to look like |Z-1|
Thank you, it was just that as this is a bit new I was unsure if he mean't "let's solve this problem for |z-1|>2" or if this was some trick that I did not understand the purpose of to solve the problem. It's more clear now since I watched some more clips. Thanks!
Great job on the videos. On this one (the lecture part), the pan of the audio is heavily to the left. Its really awkward to hear with headphones. The intro is stereo, but the talking is odd. Just a thought.
function represented by a Taylor series must be analytic at the center, Laurent series expands the idea of Taylor series, but Laurent series can represent functions which are not analytic at the center
thank you for sharing this great video of Laurent series. But sir, would you be so kind to make your penmanship a bit clear? I find difficulty in reading your writing. Thank you sir.
my finals is in 1.5 hours and you are one heck of a lifesaver.
Great series of vids. I'm studying for my final and this series really helps. Thank you!
Seriously helpful videos. Incredibly basic applications but you review the process well enough that it is easy to expand to more difficult problems. You are getting me through my course in complex analysis.
Thank you so much! Super helpful. I am in my second year and my complex analysis exam is in 3 weeks. One of the hardest modules for me so this really helped me understand. :)
Many thanks from Poland sir, this helped me a lot!
Thank you, this video help me to night for my exams
these videos have helped me immensely! thank you so much :)
EXCELLENT LECTURE -WHICH COUNTRY ? -AMARJIT -INDIA
Hi. Can i have permission to process the audio (boost volume, compress, etc) and re-upload or send it back to you?
شكرا دكتور 🙏❤️
Thanks from Ukraine
for the second example
if we have 1/(3-z). why cant we convert that to 1/3(1-z/3), why do we need to have (z-1) in the fraction
and why is the restriction |2/z-1| < 1, where did that come from ?
+Nick R hey guy, here is the trick, he said if |z-1|>2.......for a series to converge we must have |z-z.|3 that means z. is zero then we have to turn |z|>3 to |z|
+Nick R I also find this (Ex. 2) very confusing. The question, for me, is rather why we (at 4:26 ) choose to look at it in powers of (z-1) valid for |z-1|>2 at all? Nothing was specified about this on the board as in the previous exercise. This is all very new to me but the problem to solve might have been specified a bit unclear so I might have missed some info about Exactly what was asked of us.
+kiwanoish you are right by saying that it wasn't specified on board. but he just made mention of it that we gonna look at it at |Z-1|>2
note that z is a complex number and Z-1 is the distance from the origin of a circle centered at 1 ... this implies for any z to be within the radius of convergence then |Z-1| must be less than 1... according to Taylor series but for |Z-1|>2 that means we are looking at the region outside the circle of radius 1...then we have to use Laurent series. but we have to scale it to look like |Z-1|
Thank you, it was just that as this is a bit new I was unsure if he mean't "let's solve this problem for |z-1|>2" or if this was some trick that I did not understand the purpose of to solve the problem. It's more clear now since I watched some more clips. Thanks!
Great job on the videos. On this one (the lecture part), the pan of the audio is heavily to the left. Its really awkward to hear with headphones. The intro is stereo, but the talking is odd. Just a thought.
I don't get the difference between this and Taylor series. Isn't he doing just taylor series in all of these examples?
me neither
function represented by a Taylor series must be analytic at the center, Laurent series expands the idea of Taylor series, but Laurent series can represent functions which are not analytic at the center
***** Thanks, I just learned this topic today by practising a few hours and watching solutions
Thank you! very helpful!
thank you for sharing this great video of Laurent series. But sir, would you be so kind to make your penmanship a bit clear? I find difficulty in reading your writing. Thank you sir.
Thank you sir
why do you write the summation from n = 1 and not n = 0?
Because |z|>1
Because it's 1/z multiplied by the (1 + 1/z + (1/z)^2 + ...) so your first term has to be a 1/z, not 1.
course hero
You are moving too fast bruv....