CSIR NET MATHEMATICS | Complex Analysis | Shortest Solution Open Mapping Theorem June 2019

Hello friends, in this video we are going to discuss the problem of complex analysis from CSIR NET June 2019. Example number 40 from booklet C marks 3. Consider the function f defined from C to C complex plane to complex plane given by f of z is equal to e power z which of the following is false. f of z belongs to C such that modulus z less than 1 is not open set. Second option. f of modulus z less than equal to 1 is not open, f of mod z is equal to 1 is closed and f of modulus z greater than 1 is an unbounded open set. So here we want to find the false statement and the statements are about the images of set which is open or closed. So for that we have one result associated with the complex function and that is open mapping theorem. Suppose f is analytic and non-constant in open set G then f is open. Then the image of open set is open for non-constant analytic function. Now here given f is equal to e power z it is analytic. So, it satisfies the first condition as well as the non-constant. Now, image of open set is open. We want to find the wrong statement. Now, look at the first option mod z less than 1 which is an open disc and it is an open set in complex plane. So, its image should be open by the open mapping theorem. But here, the statement is not open. So, first statement is wrong and it is a problem of part b. Therefore, it is only one correct answer and which is first one and other answers are wrong. So this example has one correct answer first one where image of open set or open disk is not open. Thanks for watching. For more videos, like, share, subscribe, press the bell icon of Next Gen Maths