CSIR NET MATHEMATICS | Complex Analysis | Zeros & Poles Concepts & Solution June 2019

Hello friends, in this video we are going to discuss the problem of complex analysis from CSIR NET June 2019. Example number 34 from booklet C, marks 3. Let f defined from complex plane C to C be an entire function such that limit z tends to 0, f 1 upon z is equal to infinity. Then which of the following is true? First option, f is constant. Second option, f can have infinitely many zeros. Third option f can have at most finitely many zero and the fourth option f is necessarily nowhere vanishing. Suppose first option is correct f is constant therefore f of z is equal to c so f of 1 upon z is equal to c so limit z tends to 0 modulus 1 upon z has also constant value it cannot be infinity. So first option is wrong without any calculation. For other option we should recall some concept of complex analysis zero and pole. So first zero of an analytic function f defined from c to c is a complex number z such that f of z is equal to zero and pole of f is a zero of one upon f. There is a special definition for pole at infinity. The function f has pole at infinity if f of 1 upon z has pole at z equal to 0 In other words, in the limiting form, limit z tends to 0, f 1 upon z is equal to infinity. But this condition is given in the example. That means the given entire function has pole at infinity. That is the first part. Now, we are going to recall one statement. If f is entire function and f has pole at infinity of order n, then f is a polynomial of degree n. That means f of z is equal to a0 plus a1z plus a2z square plus anz power n. Now in our case the given example is entire as well as pole at infinity. Therefore it should be polynomial and if it is polynomial then 0 of the polynomial means f of z is equal to 0. z says that f of z is equal to 0 will give the zeros of the function. Now polynomial equal to 0. Now, this equation has at most So, the second option which is f can have infinitely many zero is wrong. Third option f can have at most finitely many zero that is correct and the fourth option f is necessarily nowhere vanishing which is wrong. So, third option is correct. Thanks for watching. For more videos like share subscribe press the bell icon of next gen maths.