CSIR JUNE 2011 PART C QUESTION 79 SOLUTION (Analytic functions and Identity theorem application)

Let $\Bbb D = \{z \in \Bbb C : |z| < 1\}$ be the unit disc. Let $f : \Bbb D \to \Bbb C$ be an analytic function satisfying $$f(\frac{1}{n}) = \frac{2n}{3n+1}.$$ Then
1. $f(0) = \frac{2}{3}$,
2. $f$ has simple pole at $z = -3$,
3. $f(3) = \frac{1}{3}$,
4. no such $f$ exists.

I am spending my crucial Ph.D. time in writing this blog to help others to achieve in Mathematics. So please encourage me by following the blog by email. Also, visit my blog every day and share the solutions with friends. Thank you.
Solution: We will explicitly calculate the function $f$ then the options are easy to verify.
Identity Theorem: Let $\Bbb D = \{z \in \Bbb C : |z| < 1\}$ be the unit disc. Let $f,g : \Bbb D \to \Bbb C$ be two analytic function. If  $f(\frac{1}{n}) = g(\frac{1}{n})$ for all $n \in \Bbb N$ then $f \equiv g$ (i.e., $f(z) = g(z)$ for all $z \in \Bbb D$). 

Replace $n$ by $\frac{1}{n}$ in $\frac{2n}{3n+1}$ we get $\frac{2}{3 + n}$. Let $g : \Bbb D \to \Bbb C$ be an analytic function defined by $g(z) = \frac{2}{3+z}$ (Note that $g$ has no singularities in $\Bbb D$). Then it is given in the problem that  $f(\frac{1}{n}) = g(\frac{1}{n})$ for all $n \in \Bbb N$. Therefore by the above said identity theorem we have $f(z) = g(z)$ for all $z \in \Bbb D$. Hence $$f(z) = \frac{2}{3+z}.$$
option 1: (True) Clearly $f(0) = \frac{2}{3}$.
option 2: (True) Clearly $-3$ is a simple pole of $f(z)$.
option 3: (True) Clearly $f(3) = \frac{1}{3}$
option 4: (False)  The function $f(z) = \frac{2}{3+z}$ is one such function. 
Click here for more problems. Share to your groups:
FOLLOW BY EMAIL TO GET NOTIFICATION OF NEW PROBLEMS. SHARE YOUR DOUBTS IN THE COMMENTS BELOW. ALSO, YOU CAN SUGGEST PROBLEMS TO SOLVE WHICH WILL BE SOLVED IMMEDIATELY.

No comments:

Post a Comment

Featured Post

NBHM 2020 PART A Question 4 Solution $$\int_{-\infty}^{\infty}(1+2x^4)e^{-x^2} dx$$

Evaluate : $$\int_{-\infty}^{\infty}(1+2x^4)e^{-x^2} dx$$ Solution : $$\int_{-\infty}^{\infty}(1+2x^4)e^{-x^2} dx = \int_{-\infty}^{\inft...

Popular Posts