48.10 ....tell us the valiid reason... function is boundes and entire they why it's not bounded ...the theorem says that entire and bounded function are constant
Sir. I have one doubt in 33:56. In option 3, 3 does not belong to the domain, |z| less than 1. Then how can we choose the given function? Can we take any value for z??
32:40 Sir please solve my doubt if function is f( z)=2/(3+z) then how we can say -3 is pole of f while -3 is not in the domain, and second 3 is also does not belong the domain D then how we calculate value of f(3) . By identiy theorem, if we take g(z)= 2 /(3+z) then f(z)= g(z) but g(z) have no singularity in domain, neither pole nor removebale so how -3 is pole of f(z). Kya identity theorem me function exist karne ke bad codomain se singularity check ki jati hai ya domain se , yadi domain se check hoti hai to -3 is not pole , and 3 is also does not belong the domain D fir se only option a is wright aa rha hai . Kya kare?
Sir in first question option d is also correct because you have given open interval.... And at 34:00 time only option a is correct and b and c option is wrong because f(3)= 1/3 is true but 3 doesn't belong in the domain D.. similar for simple pole -3 does not belong in domain D....... Please correct sir.. i am confused..
Sir in the gate question in time stamp 39:55 , how limit point of Zn is 1, it is zero for all. Isn't we find limit point of images in co-domain??!! Kindly please take a look and reply, otherwise it gets confusing
THANK YOU SIR.....VIDEOS ARE VERY VERY HELPFUL
We wait for your video....
Sir your video is really helpful.. ♥️
Thank you sir your video very helpful and amazing
Your videos are very helpful sir
Most helpful video for net and GATE exam ❤
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Very useful lecture
8:23 option d is correct...as the set contains limit points.
Limit point (-1,1) ka [-1,1]. Aayega jo ki 1 usk domain me nhi h isliye wrong hua
Option b ka limit point 0 kaise hua@@Harish-Dhar-Deewan
@@mayachetri2959b me f(z) me ,z=1/2 radius ka circle bnakr limit point nikalte to usk domain me aayeg isliye 0 hua
@@Harish-Dhar-Deewan
but sir if I am not wrong Theorem says " if one limit point is there means it will works,"
Useful 🙏
Sir your explanation is very good. Thankyou sir.
Videos are very helpful
Really awesome❤❤❤
Thank you very well sir
Very clearly explained sir
Your vedio is help full for us
Osm....🙏🙏
Sir please upload video on pyq of several variables calculus of csir specially
Thank you sir
Thank you so much sir🙏🌷🌷🌷🙏☀️☀️☀️
Thank you sir ...very helpful video
Thanks Sir❤❤
Please re discuss options b &d of question 1. 🙏
Thank you so much sir 🙏
@8:16 could u plz check the 4th option ,,, I think its a correct option .
Thank you
1:09:2 sinz kayse polynomial hoga polynomial ka degree finite hota ha to sinz ka degree e define nhi hoga to why sinz polynomial function
Upload more videos for csir net Dec 2023 sir
Sir 55:18 me function f ka range C-{-infinity,0} de rakha ha to (0, infinity) point skip karega na to function constant ho jayega by pickerd theorem
Thank you so much sir
Thankyou sir ❤
48.10 ....tell us the valiid reason... function is boundes and entire they why it's not bounded ...the theorem says that entire and bounded function are constant
Thank you sir ❤❤
There some doubt please sir reply 👍👍
1:09:00 csir net june 2018 ans ,I think all options are not correct 2,4
in question( 22 minutes) , z=2 is not a simple pole cause doimain is unit disc here & domain doesnt contain 2
Sir after csir net please upload video on whole pyqs for IIT jam MA. 🙏🙏🙏🙏it's a request sir plz.
Sir. I have one doubt in 33:56. In option 3, 3 does not belong to the domain, |z| less than 1. Then how can we choose the given function? Can we take any value for z??
32:40 Sir please solve my doubt if function is f( z)=2/(3+z) then how we can say -3 is pole of f while -3 is not in the domain, and second 3 is also does not belong the domain D then how we calculate value of f(3) . By identiy theorem, if we take g(z)= 2 /(3+z) then f(z)= g(z) but g(z) have no singularity in domain, neither pole nor removebale so how -3 is pole of f(z). Kya identity theorem me function exist karne ke bad codomain se singularity check ki jati hai ya domain se , yadi domain se check hoti hai to -3 is not pole , and 3 is also does not belong the domain D fir se only option a is wright aa rha hai . Kya kare?
Same doubt, plz explain
Sir in first question option d is also correct because you have given open interval.... And at 34:00 time only option a is correct and b and c option is wrong because f(3)= 1/3 is true but 3 doesn't belong in the domain D.. similar for simple pole -3 does not belong in domain D....... Please correct sir.. i am confused..
this men has not enough knowlege to correct this
Great
33:50 in option 3
Z not belong to domain
Sir in the gate question in time stamp 39:55 , how limit point of Zn is 1, it is zero for all. Isn't we find limit point of images in co-domain??!!
Kindly please take a look and reply, otherwise it gets confusing
Continuing...
And in December '12 question in the time stamp 56:00, range of f cannot take 0, but f(z)=z^2 will take 0 as 0 belongs to the region D
Sir dec 2017 1:03 par sir n=2 lene par only a correct h baki discard ho rhe sir
Sir numerical ek sath pless 🙏
Sir plz pyq of Algebra ka bhi aradonaa
At 22:13 ,he ask which of the following is not correct!!
Please hindi m question explain kiya kro sir
Maximum likelihood estimate podunga sir
Trigonometric functions are not polynomial!!!!!
Grean bord please. White board is not cleared sir
20:23 Next problem of June 2012
Question is asked only not correct sir
But you say correct options sir
june 2011 ka solution galat hai sir answer bhi gal;at hai apka
f(z) =sinz or cosz are entire function also it's bounded.
Thank you sir