Mathemagic: How to find what kind of singularities the function have!

SINGULARITIES

There are following main terms about singularity.

1.    Zeros:  An analytic function f(z) is said to have a zero of order m if f(z) is expressible as,

f(z) = (z-a)^mφ(z)

where, φ(z) is analytic and φ(a) ≠ 0.

(a) f(z) is said to have a simple zero at z=a, if z=a is a zero of order 1.

Example:  find the zeros of (z+1)²/(z²+1).

Solution: let f(z) = (z+1)² φ(z) , notice that here a=-1 and m=2.

      Where φ(z) = 1/(z²+1). And also φ(z) is analytic and φ(-1)=1/2 ≠ 0.

So zeros of f(z) is given by (z+1)² = 0

OR,    z=-1,-1

Hence we can say z=-1 is a zero of order 2.

2.    Limit point of zeros:  Limit points of zeros is an isolated essential singularity.

Example: what kind of singularity of the function sin[1/(1-z)] at z=1 ?

Solution: let f(z)= sin[1/(1-z)]

            Zeros of f(z) are given by sin[1/(1-z)] = 0

So, 1/(1-z) = nπ or 1-z = 1/nπ

z = 1-(1/n π)  where, n=0, ±1, ±2, …

taking limit n→∞ , we get z=1.

So  we can say, z=1 is a limit points of zeros.

Hence z=1 is an isolated essential singularity.

3.    Singular points: A singularity (or singular points) of a function is the point at which

the function ceases be analytic.

Example: if f(z) = 1/(z-2),  then z=2 is a singularity of f(z). because at z=2, function is not analytic.

Note:: here z=2 is also an “isolated singularity”.

Example: function f(z)=1/z is analytic everywhere except at z=0.

       So z=0 is an isolated singularity.

4.    Removable singularity: A singularity z=a is called removable of f(z) if  

      Lim(z→a)  f(z) exists finitely.

Example: f(z) = sinz/z. then lim(z→0) sinz/z = 1

       Because   sinz = z-z³/3!+z⁵/5!- …  = z(1-z²/3!+z⁴/5!- …)

Hence we can say, z=0 is a removable singularity of f(z).

5.         Poles: A function f(z) is said to have pole of order m if it is expressible as,

f(z) = φ(z) / (z-a)^m

where, φ(z) is analytic and φ(a) ≠ 0.

Example: discuss the nature of singularity of f(z)=1/[z(1-z²)].

Solution: singularities(or poles) of f(z) is given by,

                                    z(1-z²) = 0

                                                or,  z = 0, -1, 1

hence z = 0, -1, 1 are simple poles (or poles of order 1).

6.  Limit points of poles:  limit point of poles is an non-isolated essential singularity.

Example: what kind of singularity has the function?

Solution: f(z)=1/[cos(1/z)] at z=0.

                 Poles of f(z) are given by, cos(1/z) = 0 = cos(π/2)

                                                                       Or, 1/z = 2nπ ± π/2

                                                                            z = 1/(2nπ ± π/2)

                                                                       taking limit n→∞

                                                                           z=0

so z=0 is the limit point of poles.

Hence z=0 is an non-isolated essential singularity.