Suppose f(z)=sin(1/cos(1/z))be a complex function.
Then what type of singularity f has at z=0 and what will be the residue of f at z=0?
thanks. regards
Where did you get this question from? It seems like it isn't even "well-posed": has a bunch of essential singularities at the points for all , and these singularities accumulate at , so for starters is not even an isolated singularity, and every closed curve that contains the origin in its interior will contain an infinite number of other singularities.
The function has in z=0 an essential singularity. Setting the function becomes...
(1)
... and the residue in z=0 is the term in s of its Taylor expansion, that is...
(2)
The computation of the derivative is left to You and the result is that the limit (2) is zero...
Kind regards
Originally Posted by sorv1986...
... I am a little bit confused. Why r u calculating the limit (2). Besides this, if s-> 0, then z->∞. You are calculating the residue of f(z) at z=∞ if i am right.How did you get z=0 is an essential singularity?... sorry if i misunderstood...
I do hope that the following example will clarify the procedure to obtain an easy solution of Your question. Let's consider the function...
(1)
The (1) is the Laurent expansion of the function around z=0 and, because the infinite number of term in , it has in z=0 an essential singularity. By definition the residue in z=0 is the coefficient of the tem in of the Laurent expansion and in this case in (1) You see that is r=1. Setting the (1) becomes...
(2)
Now the (2) is the Taylor expansion of the analytic function around s=0 and, of course, the coefficient of the term in in (1) and the coefficient of the term in s in (2) are the same. But the coefficient of the Taylor expansion of an analytic function f(s) around s=0 is f'(0) and that justifies the formula...
(3)
Kind regards