Analyticity of Complex Function f(z)=1/(z^2-1)

[tthivanka]

f(z)=1/(z^2-1)
Is this function analytical or not

 

[snipez90]

It is analytic on the set {z | z =/= 1 and z =/= -1}, that is, it is analytic away from its poles.

 

[tthivanka]

could u please show me how It is analytic on the set {z | z =/= 1 and z =/= -1}

 

[snipez90]

Sure, the simplest way is to realize that nothing has really changed in this problem from real analysis. Although complex differentiability is much more powerful than real differentiability, the algebra of taking derivatives is still the same. Thus we can differentiate polynomials and rational functions just as we did in calculus, and this would immediately give you the answer.

 

[slider142]

could u please show me how It is analytic on the set {z | z =/= 1 and z =/= -1}

A function that is complex differentiable on an open set is analytic on that set and its power series converges on the largest open disc that does not contain a singularity/pole. The singuarities of this function are at z = 1 and z = -1.

 

[mathwonk]

They are using a big theorem that differentiable implies analytic. It is also elementary to expand this function about any point a other than 1,-1, by setting z = (z-a+a), expanding z^2 in terms of (z-a), and then using the geometric series. I.e. when c is not zero, 1/[c + f(z-a)] equals (1/c)[1/{1 - (-1/c)(f(z-a))}], and you know how to expand 1/(1-anything) as 1 + anything + (anything)^2+...