Laurent Series of 1/[(z-i)(z-2)] at z0=i

[CTID17]

Homework Statement

Find the Laurent series at z₀=i, which is convergent in the annulus A ={z:0<|z-i|<5^1/2 } of

1/[(z-i)(z-2)]

Homework Equations

The Attempt at a Solution

|z-i|/5^1/2 <1

i make
1/[(z-i)(z-2)] = 1/[5^1/2 (z-i)((i-2)/5^1/2 + (z-i)/5^1/2 )
now how do i make it so that i have 1-(z-i)/5^1/2 to use the binomial series?
Is this a right approach to this type of questions? I could use partial factions to get

1/(2-i)*[1/(z-2)-1/(z-i)] but again i don't know how i can take advantage of |z-i|/5^1/2 <1 to use in the binomial series.

Thanks in advance.

 

[HallsofIvy]

Homework Statement

Find the Laurent series at z₀=i, which is convergent in the annulus A ={z:0<|z-i|<5^1/2 } of

1/[(z-i)(z-2)]

Homework Equations

The Attempt at a Solution

|z-i|/5^1/2 <1

i make
1/[(z-i)(z-2)] = 1/[5^1/2 (z-i)((i-2)/5^1/2 + (z-i)/5^1/2 )
now how do i make it so that i have 1-(z-i)/5^1/2 to use the binomial series?
Is this a right approach to this type of questions? I could use partial factions to get

1/(2-i)*[1/(z-2)-1/(z-i)] but again i don't know how i can take advantage of |z-i|/5^1/2 <1 to use in the binomial series.

Thanks in advance.

$$\frac{1}{(z-i)(z-2)}= \frac{1}{z-i}\left(\frac{1}{z-2}\right)$$

$$\frac{1}{z-2}$$
is analytic in the annulus given so it has a Taylor's series about z= i. Find that Taylor's series and divide each term by z-i.

(Hint: write 1/(z-2) as
$$\frac{1}{z-2}= \frac{1}{z- i+ i- 2}= -\frac{1}{(i-2)+ (z-i)}= -\frac{\frac{1}{i-2}}{1- \frac{z-i}{i-2}}$$
and think of geometric series.)

 

[CTID17]

Thanks a lot . I was doing the same thing, except I've forgotten about geometric series :) . Got it now :) .