Answer:
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.
<u> cos(z)
</u>
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) __<u>cos(z</u>)___ = _<u>cos(πn)__</u>
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) __<u>cos(z)__</u> = __<u>cos(2πni) </u>= −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
