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)
= z2 sin (1 / z2 )
in
the domain: : 0 < ÷ z ÷ < ¥
We
have (from the hint):
sin(z)
= z - z3 / 3! + z5
/ 5! …..
- (-1) n- 1 z 2n
-1/ (2n -1)! +
Then:
sin
(1/z) = 1/ z - 1/ z3 3! + 1/
z5 5! - 1/ + z7
7! +
……..
And:
sin
(1 / z2 ) = 1/ z2 - 1/ z6 3! + 1/ z 10
5! -
……..
whence (on substituting):
z2
sin (1 / z2 ) = z2
[1/ z2 - 1/ z6 3! + 1/ z 10
5! -
……..
= 1
- 1/ z4 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!
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