## Poles or Removable Singularities

hi Maths expert,

Able to help me verify the answer highlighted in red ? thank you

Determine the location and nature of singularities in the finite z plane of the follow function and apply Cauchy Integral Formula.

f(z) =

[z^2 - 1][cos z]
----------------
z (z-1)(z-2)(z-3)

simple pole at z = o

Removable Singularities at z = 1

simple pole at z = 2

simple pole at z = 3

Apply Cauchy Integral Formula for f(z) when C : |z + j | = 5

Because z = 1 is a RS, hence

{ [z - 1][z + 1 ][cos z] / (z-1) }
--------------------------------
z (z-2)(z-3)

[z + 1][cos z]
= ----------------
z (z-2)(z-3)

Using partial fraction to find A, B & C

A = 1 /6
B = 0.624
C = -1.316

= j 2 pi { [0 + 1][cos 0]0.1666 + [2 +1][cos 2] * 0.624 + 4[cos 3] * -1.316 }

----------------------------------------------------------------------

g(z) =

sin 2z
-------
z^15

pole order 14 at z = 0

Apply Cauchy Integral Formula for f(z) when C : |z + 1 | = 2

{sin 2z / z } / z^14

= j 2pi * d^(14-1) f(a) / dz

= j 2pi * 0
----
13!

= 0

Thank you for any help