f(z) = (z^22)/(z(z+1)(z2)) = 1/z  1/(z+1) + 1/(z2)
a) Compute the Laurent Series of f(z) on 0 < z < 1
b) Compute the Laurent Series of f(z) on 1 < z < 2
I'm very confused as to what the question wants. Please help!
a) Compute the Laurent Series of f(z) on 0 < z < 1
b) Compute the Laurent Series of f(z) on 1 < z < 2
I'm very confused as to what the question wants. Please help!

You can rearrange the three terms in a variety of ways. What you want to do here is to find arrangements that allow you to use a generalization of a geometric series.
For z > 0, 1/z is it's own Laurent series. I hope that is clear. It has precisely the form
∞
Σ a_n zⁿ
n=∞
where a_(1) =1 and all other a_n = 0.
Note that you can rewrite the second term in a couple of similar (yet different) ways.
1/(z + 1) = 1/(1  (z)), or
1/(z + 1) = 1/[z(1 + 1/z)] = (1/z)/(1  (1/z)).
The first has series
∞
Σ (1)ⁿ zⁿ = 1/(1 + z), convergent for z < 1.
n=0
The latter has series
∞
Σ (1)ⁿ (1/z)^(n+1) = 1/(1 + z), convergent for 1/z < 1 i.e. for z > 1.
n=0
The last term can similarly be arranged in a couple of ways.
1/(z  2) = 1/(2  z) = (1/2)/(1  z/2), or
1/(z  2) = 1/[z(1  2/z)] = (1/z)/(1  2/z).
Each of these can be expressed by a ("geometric like") series
∞
Σ  zⁿ/2^(n+1) = 1/(z  2), convergent for z/2 < 1 i.e. z < 2
n=0
or
∞
Σ  2ⁿ/z^(n+1) = 1/(z  2), convergent for 2/z < 1 i.e. z > 2.
n=0
Okay, just pick amongst these to get the desired interval of convergence. It is okay if part of the series converges on a larger domain, you just want to be sure that all parts converge at least on the indicated domain.
a) 0 < z < 1, Choose the first series for each of the last two terms.
∞
Σ  [(1)ⁿ + 1/2^(n+1)] zⁿ + 1/z.
n=0
b) 1 < z < 2, Choose the second series for the term 1/(z+ 1) and the first for the term 1/(z  2)
∞ . . . . . . . . . . . . . . . . . . . ∞
For z > 0, 1/z is it's own Laurent series. I hope that is clear. It has precisely the form
∞
Σ a_n zⁿ
n=∞
where a_(1) =1 and all other a_n = 0.
Note that you can rewrite the second term in a couple of similar (yet different) ways.
1/(z + 1) = 1/(1  (z)), or
1/(z + 1) = 1/[z(1 + 1/z)] = (1/z)/(1  (1/z)).
The first has series
∞
Σ (1)ⁿ zⁿ = 1/(1 + z), convergent for z < 1.
n=0
The latter has series
∞
Σ (1)ⁿ (1/z)^(n+1) = 1/(1 + z), convergent for 1/z < 1 i.e. for z > 1.
n=0
The last term can similarly be arranged in a couple of ways.
1/(z  2) = 1/(2  z) = (1/2)/(1  z/2), or
1/(z  2) = 1/[z(1  2/z)] = (1/z)/(1  2/z).
Each of these can be expressed by a ("geometric like") series
∞
Σ  zⁿ/2^(n+1) = 1/(z  2), convergent for z/2 < 1 i.e. z < 2
n=0
or
∞
Σ  2ⁿ/z^(n+1) = 1/(z  2), convergent for 2/z < 1 i.e. z > 2.
n=0
Okay, just pick amongst these to get the desired interval of convergence. It is okay if part of the series converges on a larger domain, you just want to be sure that all parts converge at least on the indicated domain.
a) 0 < z < 1, Choose the first series for each of the last two terms.
∞
Σ  [(1)ⁿ + 1/2^(n+1)] zⁿ + 1/z.
n=0
b) 1 < z < 2, Choose the second series for the term 1/(z+ 1) and the first for the term 1/(z  2)
∞ . . . . . . . . . . . . . . . . . . . ∞
12
keywords: series,Laurent,question,Laurent series question!