Question

the poles of F(z) = 1 / (z - 2)^3 (z - 3)^2 is z=2 and z=3 with orders__ and __ respectively.

A. 2,3
B. 3,2
C. 2,2
D. 3,3​

Answer 1

Answer:

no

no

no

it is a good question

Answer 2

The poles of F(z) = 1/(z - 2)³ (z - 3)² is z = 2 and z = 3 with orders 3 and 2 respectively.

Given :

The function

$\displaystyle \sf{ F(z) = \frac{1}{ {(z - 2)}^{3} {(z - 3)}^{2} } }$

To find :

The poles of F(z) = 1/(z - 2)³ (z - 3)² is z = 2 and z = 3 with orders __ and __ respectively.

A. 2 , 3

B. 3 , 2

C. 2 , 2

D. 3 , 3

Solution :

Step 1 of 2 :

Write down the given function

Here the given function is

$\displaystyle \sf{ F(z) = \frac{1}{ {(z - 2)}^{3} {(z - 3)}^{2} } }$

Step 2 of 2 :

Find poles the function

The poles the function are given by

$\displaystyle \sf{ {(z - 2)}^{3} {(z - 3)}^{2} = 0}$

$\displaystyle \sf{ \implies z = 2,2,2,3,3}$

So z = 2 and z = 3 with orders 3 and 2 respectively

Hence the correct option is B. 3 , 2