Complex Analysis CSIR NET July 2026 Memory-based Question

[00:00:00]Hello students. Welcome to the next lecture on the memory based questions CSIR NET July 2026. Myself Dr. Harish Garg, you can follow and subscribe my YouTube channel. The first eight part of the CSIR NET July 2026, you can find at my YouTube channel. Now today I will explain you the few questions related to the complex analysis.

[00:00:24]The first question is if capital D is my unit disk that means capital of Z Z is less than of the one is my unit disk and phi is my holomorphic function such that phi of half is zero, phi of minus one is four by five and your target is to find the bound of some holomorphic function.

[00:00:48]Which things come in your mind when you look about the unit disk and the bound?

[00:00:53]That is called as a Schwarz lemma.

[00:00:55]Fine? But because the function value is given at some particular point, then we can apply the Schwarz Pick Lemma. On your Schwarz Lemma and the Pick Lemma come in the equality of the value they give, they always provide you the bound, find upper bound. So what is the result for the Schwarz Pick Lemma?

[00:01:13]f of Z minus f of a divided by one minus complex conjugate of f of a into f of Z which is less than or equal to Z minus a divided by one minus where where f of a is my zero. Fine? Now look at that. A is my half.

[00:01:40]Fine? Now in this case instead of the f, it is a phi, so I can substitute the value phi of one over three minus phi of half. Phi of half is my zero. One minus zero is less than or equal to Z A / 1 - complex conjugate of half is half into 1 by 6. So, that will be less than equal to -1/6 / 5/6.

[00:02:12]The left-hand side will be 5 of 1/3. So, what is the answer? That will be 1/5.

[00:02:18]So, the upper bound is 1/5. So, you can see 7/5 is more than 1/5. A C and D option cancel. B is the right answer of the problem by using the Schwarz pick lemma.

[00:02:32]Let me know in the comment box.

[00:02:38]Look at the second question.

[00:02:39]The power series is convergent at some point, divergent at some point, then the radius of the convergent.

[00:02:46]Whenever you have a power series let's say center is my C and once the power series is convergent at point Z not, then the radius of convergent is always given greater than equal to Z not minus C. While it is divergent at Z not, then R must be less than or equal to Z not minus C. In this case, the center is my zero. Fine? Now, look at that. It is convergent at 1 + iota / 2. Therefore, R must be greater than or equal to 1 + iota / root 2. So, that will be 1. Root 2 / root 2 is 1. Similarly, it is divergent at 1 - iota / root 2. Again, it comes to be 1. So, what does it means? 1 is less than equal to R is less than equal to 1. It implies R is equal to 1. So, C is the right answer of the problem.

[00:03:55]Fine. For more detail, you must watch about my lecture PYQ on the radius of convergent at my YouTube channel Dr. Harish Garg.

[00:04:14]If you have some more questions on the complex analysis, real, calculus of variation, and many more, you can send the comment box and subscribe my YouTube channel. I will upload the video for you very soon. Till then, best of luck, students.

[00:04:28]Happy learning.