Mam, in question no 21(at 58:40), as -1 and +1 are on the real axis why didn't you take pi*i instead of 2*pi*i for finding residue? Thank you mam great lectures.
Because that points lies inside the curve. For the points which lies inside the curve, we take 2*pi*i ( sum of residue). pi*i (sum of residue) we consider for half circle. If your doubt remains, see the video of cauchy integral & real improper integral.
@@arindammaity4065If point lies inside the circle or semicircle we take 2*pi*i , & if point lies on the real axis of semicircle i.e. on boundary we take pi*i. Hope you understand. If doubt remains see the recommended videos.
in question no 17, why option b)? Answer may be option a) coz only positive values of n are considered.z=±√n only z=±i√n correspond to negative values of n Please correct me if I am wrong
Both the values of n are allowed because for the simplest case if we have sinx = 0 then x =n.pi. where n is integer can have positive and negative values both values follows the result. That's what we said in the question. Hope you understand.
No,, option b is correct,, because z=0 is due to 1/z so it is a simple pole,,,if you still have any doubts,,,see the videos of singular points. Hope that will help..
1:07:15 Even z = 0 is also a simple pole Why didn't you apply the method which you applied for other singular points for the point z = 0? The answer is that the method is valid only if the numerator is not equal to zero at the point considered which you didn't mention in your video on calculating residues as well
question no. 29 is asked for finding the value of function at z=(2-i) not the value integral so the answer d will be correct and question no. 28 the integral is analytic so value will be 0
Q.29 Yes,, we did find the value of f(z) at the given point because as you can see that value of f(z) is given in the form of integral in question so we did find that integral.
Generally ,we use that method for higher order pole,,,but when sine/cosine terms are given in function then we prefer to solve with expansion. And when algebraic function is given in denominator,,,then we use that method.
Mam ek humble request hai agar possible ho to plss kuch subjects ke short revision ke videos upload kr dijiye feb ke exam me kafi help ho jaegi mp me kafi confidence aagya hai thank you for your hard work🙇🙇🙇
z=+R,-R are real points lies on real axis ,,,and for the points on real axis we consider pi*i* sum of residue. If you still have doubt,,,you should visit our video on real improper integrals.
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Nice video ase ase sab topics p bnado video
Thank you so much mamm...... this is very helpful.....❤❤❤
Many Blessings form my side 💙
Thank you for these awesome videos
Thanku very much,,please do share with your friends if you find the contents useful
Mam, in question no 21(at 58:40), as -1 and +1 are on the real axis why didn't you take pi*i instead of 2*pi*i for finding residue?
Thank you mam great lectures.
Because that points lies inside the curve. For the points which lies inside the curve, we take 2*pi*i ( sum of residue).
pi*i (sum of residue) we consider for half circle.
If your doubt remains, see the video of cauchy integral & real improper integral.
Thanku,,, please do share with your friends if you find the contents useful.
@@vyingphysics I understood mam Thank you
@@vyingphysics But in gate solutions(2019) video you did 2πi for the semicircle .
@@arindammaity4065If point lies inside the circle or semicircle we take 2*pi*i , & if point lies on the real axis of semicircle i.e. on boundary we take pi*i. Hope you understand. If doubt remains see the recommended videos.
All previous yaers question base videos are unique 🙏
Thanks 😊
(Yaers) -years 🙈
Good explanation maam. Expecting more csir pyq solved topics from your side. May God Bless
Thanku....will upload soon
mam for que 16. Can use convert into 2* int o to infinite pi/n*cosec*pi/n
in question no 17, why option b)?
Answer may be option a) coz only positive values of n are considered.z=±√n only
z=±i√n correspond to negative values of n
Please correct me if I am wrong
Both the values of n are allowed because for the simplest case if we have sinx = 0 then x =n.pi. where n is integer can have positive and negative values both values follows the result. That's what we said in the question. Hope you understand.
Ma'am jo branch point wala question h us mai d option zeyada acceptable h? .. correct me if I'm wrong
P.s 37:09
No,, option b is correct,, because z=0 is due to 1/z so it is a simple pole,,,if you still have any doubts,,,see the videos of singular points.
Hope that will help..
@@vyingphysics yes yes i got the point thank you and you are oosm💕
Mam aap NET physics ka Whole syllabus cover kr lo na
Will take some time, but we will definitely do
1:07:15
Even z = 0 is also a simple pole
Why didn't you apply the method which you applied for other singular points for the point z = 0?
The answer is that the method is valid only if the numerator is not equal to zero at the point considered which you didn't mention in your video on calculating residues as well
I have completed all the methods/ concepts in complete complex analysis Playlist. And I recommend to watch them first.
A ton of thanks♥♥. Thank you so very much😭😭😭. Pls keep doing more concept videos also. ❤❤
You're very welcome. We will make videos...but Don't cry.
Thanks a lot ma'am, it means a lot to people like me. Keep going❤
It's my pleasure
1:26:41 recidue nikalne k baad bhi kyu recidue zero ho gaya samjh nehi aya?? Kaya koi rule hai removable singularity k liye??
Mam at the time 44:44 hmme kse pta chla ki hmare points upper quadrant m lie kr rhe h
Thank you ma'am 😊
Welcome
Really helpful series
Nice explanation
Thanku 😊
Mam apne video ki starting me kaha ki mene last video analytics ke questions explain Kiya tha konsi video hai mam vo?
Check Playlist of complex analysis 2nd video from 30:45, you will get that question.
question no. 29 is asked for finding the value of function
at z=(2-i) not the value integral so the answer d will be correct and question no. 28 the integral is analytic so value will be 0
Q.29 Yes,, we did find the value of f(z) at the given point because as you can see that value of f(z) is given in the form of integral in question so we did find that integral.
Mam brunchcut ,brunch point bhi thoda para Dena ,one lecture please
Superb mam
Thanku,
some valuable suggestions are most welcome.
Please share with your friends.
@@vyingphysics aap kaha se ho
Appki phd ho gyi kya?
Last question me agar residue at n order wale se kare to same result aaega kya??
Generally ,we use that method for higher order pole,,,but when sine/cosine terms are given in function then we prefer to solve with expansion. And when algebraic function is given in denominator,,,then we use that method.
And mam you are amazing
Thanku 😊
Mam ek humble request hai agar possible ho to plss kuch subjects ke short revision ke videos upload kr dijiye feb ke exam me kafi help ho jaegi mp me kafi confidence aagya hai thank you for your hard work🙇🙇🙇
Thanks for kind words,
Given the time constraints, I can only work on PYQs for this exam, but will definitely cover that too over time
great !
Que 16 how to find out which quadrant an exponential function belong to?
Exponential function is written as cos(pi/4) + isin(pi/4) which belongs to first quadrant. Hope you understand.
Tnx mam
maam in question number 5 isnt it 2*pi*i(sum of residue)?
z=+R,-R are real points lies on real axis ,,,and for the points on real axis we consider pi*i* sum of residue.
If you still have doubt,,,you should visit our video on real improper integrals.
@@vyingphysics Thank you maam i got it.
Ma'am... It's a humble request that the white board and the written words are not very clear.... So focus it
All the videos are recorded and uploaded at full HD, try changing the video quality, and anyways will improve on video quality as well
mam in sum 28 z=pi/4 2nd order pole kse hua
Please provide matterial for questions based on Green function...
"Except previous year problems of Green function"
Z k power -1 is analytic or not? How
Mam in ques no. 16.. 1/4 × 1/✓2 (...) Ye kaise hua
Complex ka level bhut high kr diya hei net walo ne 😢
But if you have seen complex full series ,, i don't think you'll miss any question.
Plz mam
Thankyou so much ma'am