What is less than 1 here?

So, it has a little bit the look of the 1 over 1- q, but not quite.

We need to get our hands on something that is less than 1 here.

Well, we know that z in absolute value is bigger than 1 and less than 2.

So, how can I rewrite that so I get terms that are less than 1?

Because z is greater than 1, I get that 1 over the absolute value of z, is less than

1 so, if we can get 1 over the absolute value of z in there then I'm good.

I also know that z is less than 2 in absolute value,

which means that z over 2 is less than 1.

So these two facts are going to be crucial for us.

So let's get back to 1 over z- 2.

If I factor out a 2, like I did right here, if I factor out a 2

from the denominator, then my factor becomes 1 over z over 2- 1.

And that's exact in the form 1 over 1- q, except the terms are reversed, so

I'm going to factor up a -1 in addition, and I'm left with 1 over 1- z over 2.

Since z over 2 is less than 1, as we showed over here,

I can now re-write the 1 over 1- z over 2

as a geometric series, where my q is equal to z over 2.

So this first part here becomes minus one half times the sum z over 2 to the k.

You still need to look at the second part, mainly the 1 over z times 1- 1 over z.

That came from 1 over z- 1 and that factored out a z.

The factoring out of the z gave me this 1- 1 over z term of

the denominator and we know that 1 over z is less than 1.

So we're going to again use the geometric series idea.

This z right here stays on the outside and

the 1 over 1- 1 over z becomes the geometric series,

sum k from 0 to infinity 1 over z to the k.

Now that I have all that, my last goal is to write this so

I really recognize it as a Laurent series.

The first thing I want to do is,

I want to bring this factor in front of the series into the series.

So that means this 2 adds another 2 to the denominator,

so the denominator becomes 2 to the k + 1.

The -1 comes inside right here and the z to the k is still there.

And for the second series, I want to bring -1 over z into the series,

so I get the -1 right here and the denominator becomes z to the (k + 1).

Well I only wrote z to the k here and the reason for

that is that I've simply started counting that k = 1.

Since the series goes to infinity,

the powers of z that I will be looking at in this series -1

over z to the k + 1, when k runs from 0 to infinity.

The exponents k + 1 run from 1 through infinity.

And so instead of saying k goes from 0 to infinity and using the exponent k + 1,

I can simply say k goes from 1 to infinity and I use the exponent k.

And my last step is to write the second series in terms of the negative powers

of k because that's really how Laurent series are written.

So, instead of calling this 1 over z to the k,

I recognize that that is equal to z to the -k.

And so instead of k going from 1 to infinity, I'm going to have k go

from -1 to negative infinity, because all these powers here are negative.

And instead, I can just write z to the k, but I use negative indices for my series.

This is what we just showed.

I can now re-off the coefficients of this Laurent series.

The first sum captures all of the positive powers of z.

So for example, a0, which is the coefficient of z to the 0, so

that's what I get k is equal to 0 is simply -1 over 2 to the 0 + 1, so 2.

a1, which is the coefficient I get when k is equal to 1,

it's -1 over 2 squared which is 4, and so forth.

But I also have negative powers of z, and those I find in the second sum here.