1) The
function: f(z) = -1/ (z – 1) (z – 2)

b) Singular points are z = 1 and z = 2 since by inserting the respective values in the partial fraction form one will obtain infinities.

c)
If the function is analytic in the
domains:

**÷**z

**÷**

**< 1; 1 <**

**÷**z

**÷**

**< 2 and: 2 <**

**÷**z

**÷**

**< ¥**

Draw
a sketch showing the different domains.

**d1**is for

**÷**z

**÷**

**< 1;**

**d2**is for : 1 <

**÷**z

**÷**

**< 2**

**d3**is for: 2 <

**÷**z

**÷**

**< ¥**

2)Find the Laurent series that represents the
function:

f(z)
= z

^{2 }sin (1 / z^{2})
in
the domain: : 0 <

**÷**z**÷****< ¥**
We
have (from the hint):

sin(z)
= z - z

^{3}/ 3! + z^{5}/ 5! ….. - (-1)^{n- 1}z^{2n -1}/ (2n -1)! +Then:

sin
(1/z) = 1/ z - 1/ z

^{3}3! + 1/ z^{5}5! - 1/ + z^{7}7! + ……..
And:

sin
(1 / z

^{2}) = 1/ z^{2 }- 1/ z^{6}3! + 1/ z^{10 }5! - ……..whence (on substituting):

z

^{2 }sin (1 / z^{2}) = z^{2 }[1/ z^{2 }- 1/ z^{6}3! + 1/ z^{10 }5! - ……..
= 1
- 1/ z

^{4}3! + 1/ z^{8 }5! - 1/ z^{12 }7! + …..
Which
can be written as 1 followed by a representative series, i.e.:

1 + å

^{¥}_{n = 1 }(-1)^{n}/ (2n + 1)! z^{4n}^{}
Check values
for the term n = 2:

(-1)

^{2 }= 1 and (2n + 1)! = (2(2) + 1)! = 5!
z

^{4n }= z^{4(2) }= 8
So term 2

*in the series*is: 1/ z^{8 }5!
The
reader can check additional terms as he desires to see the series works!

## No comments:

Post a Comment