Find the isolated singularities of the following functions, and determine whether they are removable, essential, or poles. Determine the order of any pole, and find the principal part at each pole

Determine the order of any pole, and find the principal part at each pole

Step-by-step explanation:

z cos(z ⁻¹ ) : The only singularity is at 0.

Using the power series  expansion of cos(z), you get the Laurent series of cos(z −1 ) about 0. It is an  essential singularty. So z cos(z ⁻¹ ) has an essential singularity at 0.

z ⁻²  log(z + 1) : The only singularity in the plane with (−∞, −1] removed

is at 0. We have

                              log(z + 1) = z −  z ²/ 2  +  z ³/ 3

So

z ⁻²  log (z + 1)  =  z ⁻¹ −  1 /2  +  z/ 3

So at 0 there is a simple pole with principal part 1/z.

z ⁻¹  (cos(z) − 1)  The only singularity is at 0. The power series expansion

of cos(z) − 1    about   0 is    z ² /2 − z ⁴ /4,    and so the singularity is removable.

   cos(z)    

sin(z)(e z−1)     The singularities are at the zeroes of sin(z) and of e z − 1,

i.e.,  at   πn and i2πn   for integral n.    These zeroes are all simple, so for

n ≠ 0    we  get simple poles and at   z = 0    we get a pole of order 2.     For n ≠ 0, the residue  of the simple pole at  πn is

  lim (z − πn)      __cos(z)___ =    _cos(πn)__

    z→πn              sin(z)(e z − 1)       cos(πn)(e nπ − 1) =  1 e nπ  −  1

For n ≠ 0, the residue of the simple pole at 2πni is

lim (z − 2πni)   __cos(z)__  =  __cos(2πni)  = −i coth(2πn)

 z→2πni                     sin(z)(e z − 1)         sin(2πni)

For the pole of order 2 at z = 0   you can get the principal part by plugging

in power series for the various functions and doing enough of the division to  get the    z ⁻² and z⁻¹    terms. The principal part is z⁻² −  1/ 2  z ⁻¹

in simplified version   it would be x ≤-4

step-by-step explanation:

multiply each term in x ≤-4 by -1