#CSIRNET MATHEMATICS NOV 2020 | 30th Nov | Section C | Complex Analysis | Schwarz-Pick Theorem

MindRiddlex
6 followers
#CSIRNET MATHEMATICS NOV 2020 | 30th Nov | Section C | Complex Analysis | Schwarz-Pick Theorem
CSIR NET, Answer Key Nov 2020, Mathematical Science, Helpful those who are preparing for NET, GATE, SET, SLET exams
Complex Analysis | Schwarz-Pick Theorem
Visit our website
AI Generated·May contain errors
Video Summary
AI GeneratedThis video provides a step-by-step mathematical solution to a complex analysis problem from the CSIR NET November 2020 exam. The tutorial focuses on finding the possible values of the modulus of the derivative of a holomorphic map at a specific point using Schwarz's Pick Theorem.
Hello Friends! This is Mahesh Ullekar and you are watching Next Gen Math.
If you are not subscribed my channel then please subscribe the channel and press the bell icon for
the notification of new video. In this video, we are going to discuss the problem CSINATE
November 2020 Problem of Complex Analysis. Question. Let D is a subset of C be an open
unit disc. Consider the family F of all homomorphic maps, F defined from D to D such that F of
0 equal to 1 by 2 for F belongs to family of F. The possible values of modulus F dash
0 are, we have again 4 options, 1st option 7 by 8, 2nd option 5 by 6, 3rd option 3 by
4 and the 4th option is 1. If you want notes of this solution then please visit our website
www.nextgenmath.com. The description link is available. Here we are going to use Schwarz's
fifth theorem which is a variant of Schwarz's lemma. Statement is a holomorphic function
F defined from D to D must satisfy inequality modulus F dash Z divided by 1 minus modulus
F of Z square less than equal to 1 upon 1 minus square of modulus Z for every Z belongs
to D. Here D is an open unit disc so 0 must be an element of D and this inequality is
also satisfied 0. Then we have modulus F dash 0 divided by 1 minus square of modulus
F of 0 less than equal to 1 upon 1 minus 0. Then we have modulus of F dash 0 divided by
1 minus square of modulus of F of 0 less than equal to 1. Now we have given in the data
F of 0 equal to 1 by 2. Put that value in the inequality. Then we have modulus F dash
0 divided by 1 minus square of 1 by 2 less than equal to 1. So 1 minus 1 by 4 equal to
3 by 4. Therefore we have modulus F dash 0 divided by 3 by 4 less than equal to 1. That
means modulus F dash 0 less than equal to 3 by 4. So for this particular family of function
the modulus F dash 0 has value less than equal to 3 by 4. From this 4 option only option
number 3 satisfy this condition and hence option number 3 is only the correct answer.
Thanks for watching. For more videos like, share, subscribe, press the bell icon of NextGenMS.
