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