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