PYQs on Entire Function| Complex Analysis | CSIR NET 2011 to 2023

[@aaqibwani7406]

I recommend and suggest all aspirants, first go through theory part of all lectures....then i am sure you will enjoy the most @dr. Harish Garg's video's.....Most helpful .....unparalleled to other contents on you tube. Thanks Dr. Harish.

[@Justyourkaran]

very lucid explanation.. it is proving to be really helpuful in our revision for June 24 attempt. Thankyou Harish sir for all your efforts.

[@sheelalakra1251]

Very helpful lecture

[@user-lo2qx8ow9e]

thankyou sir for this well explained lecture.

[@Radharanilove-y5d]

Outstanding 🎉🎉

[@amrutavibhandik1346]

Nice explained sir👍... from these video confidence level boost🙂

[@Diksha-eu4tg]

These videos are very very helpful for us sir thankyou so much for your valuable efforts sir 🙏

[@charitylyngdoh8912]

Thank you so much Sir, your explanation is really excellent, thanks alot Sir 🙏 🙏

[@priyankasaini3612]

Sir your teaching method is best

[@Ekshivbhakat]

Sir you are the best and you are helping us thank you so much for your efforts 🙏

[@Studyforcsirnetmath]

Thanks sir 🙏 ☺️

[@udaymusale9104]

Thank You so much sir...

[@Zeelkumar31]

Sir your vidio is to much helpful for us ❤.!!

[@sheemanitinsan7235]

Nice explanation 🙌🏻🙏🏻♥️

[@Hasnainmaths]

apka jawab nhii king of mathematics dr harish sir💗💗

[@PriyaSinghJRAMathematicalScien]

Thank 🎉🎉🎉uhh

[@ChandanKumar-dn9lz]

excellent ....Thank you sir .

[@ashokisaini8427]

Very well explained sir👍👍

[@ronneysaini7763]

Thank you very Much Sir❤❤

[@simonrai7300]

Thank you Sir ❤❤❤ Let's do it.

[@meghasharma8761]

ThAnku so much sir 😊😊😊 Keep it up ❤❤

[@kapilkumarsingh8236]

Thanks sir

[@SiyaRam-ih2kd]

Nice sir gi

[@Easymathscsir]

Sir ur videos are very helpful for us ..can u upload descriptive paper questions in higher mathematics sir

[@knowledgefactory33]

Sir kindly improve the quality of sound... Thanks.. Your lectures are highly qualitative..❤

[@shankarkubal5259]

Sir in dec 2013 question (52.14) , e^z is bounded in the unit circle no, although it is unbounded on C. And continues functions maps bounded sets to bounded sets, so b should be correct option no?

[@celinjoseph-x4s]

Thank you sir

[@shubhamkumartiwari2379]

Sir, In a question CSIR net june 2015 at 1:19:29 , you had selected all options as a correct option mistakenly. Option (c) is not correct

[@DrHarishGarg]

Yes, C is incorrect... I select mistakenly, as i had given example of f(z)=e^z which is non-constant.

[@NabhiSam]

I request you sir in so many question you provide wrong information please correct them althougt your videos are help full

[@madavsharma9253]

Thanku very much sir

[@RiteshYadavDelta]

Excellent 🎉 8:19

[@aaqibwani7406]

Sir at 1:49.23, how you use maximum modulus principle as it is only about real part of f, please look

[@PIMMATHS]

Sir limit continuity and u.c in real analysis aplod sir

[@PIMMATHS]

Thnks sir ji

[@Infinity-t1d]

Sir, at 27:26 in the question 1 option is true , as all the coefficients of image of D is real

[@krishnakantadutta5374]

sir please upload singularities in complex analysis

[@meghasharma8761]

Sir please make vdo on numerical analysis pyq

[@Rahul_Prajapati-n9n]

Good afternoon sir, are all the videos in the probability and statistics playlist arranged by sequence of content? If not then can you please arrange them whenever you get time, it would be really helpful for us🙏🙏🙏

[@DrHarishGarg]

All are already arranged in the sequence wise....

[@Rahul_Prajapati-n9n]

@@DrHarishGarg ok sir, thank you🙏🙏

[@madavsharma9253]

Thanku sir

[@sanjitha_dhivin]

Sir in 54:12 minutes e^z doesn't follow option A?

[@Radharanilove-y5d]

Sir ... If f is entire function and img f >=0 Then it means v>=0 So how we can say that v is bounded ... If it is less than zero ...than how it is bounded bcz negative tak to jayega?

[@DrHarishGarg]

Dear... See the tips given in the starting of the video... Whenever f is entire and either Real f or Imag f is bounded then f is constant..

[@naveenpondara4903]

@50:50 How f/g is entire ... Sin z and cos z are entire but tan z is not entire

[@shankarkubal5259]

g(z) is not equal to zero is given in the question, where as cosz becomes zeroes at certain points.

[@SabaKhan-gh1vx]

Non constant entire function is not bounded

[@aweirdhimalayanbiker]

Non constant entire function bounded kese ho skta h? 🤔🤔🤔🤔🤔 analytic or entire function same h kya??? Kux v tym khrb toh mt kro yrr plz

[@NabhiSam]

How csir net dec2016 question is bounded

[@barmanIItpkd]

Has to unbounded

[@aweirdhimalayanbiker]

Ye sir short cut ke chkkr me baketi krte h bhyi ek subject pdao pr ache se maine apna bahut tym waste kiya h enke videos pe phle mujhe lgta tha ki mai glt hu pr nhi ye mhaaanubhau glt pdate h

[@Zeelkumar31]

Why bcz p(z) is polynomial & it is continuous and image of a bdd set under continuous map is bdd .​@@NabhiSam

[@PujaJangra-jx6eg]

Just set up , itni hi problem h sir se to q dkhtye ho vdo

[@deepakrao1941]

At 1:35 sinz is not bound function

[@RanjangamingYT7477]

Sir 1:43:28 me function differantiable nhi hoga please check kar Lena sir

[@shankarkubal5259]

Yes, b is correct option. Because function cannot be analytic due to presence of conjugate

[@RanjangamingYT7477]

​@@shankarkubal5259z=0 pe function differantiable nhi ha

[@smaranikabiswal3439]

​@@shankarkubal5259question say which is false So option d is correct

[@shankarkubal5259]

@@smaranikabiswal3439 yes u are right, thanks😊

[@prernamishra9079]

Kindly improve your voice

[@SahibaanClasses]

Sir please correct padhayiye

[@icms2410]

38:30 you prove option (C) as analytic but it only satisfy CR-eq and it is not differentiable at z = 0.

[@professorragavan9783]

Thank you sir