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Find the Laurent series of the form $\sum\limits_{n = -\infty}^{\infty}c_nz^n$ for $f(z) = \dfrac{z^2}{(z - 1)(z - 3)}$ that converges in an annulus containing the point $z = 2$, and state precisely where this Laurent series converges.}

By the method of partial fractions, (how does the 1 effect the solution?)

$$

f(z) = \frac{z^2}{(z - 1)(z - 3)} = 1 + \frac{9}{2z - 6} - \frac{1}{2z - 2}.

$$

So $f(z)$ has simple poles at $z = 3$ and $z = 1$.

$f(z)$ will be analytic when

$$

|z| < 1,\quad 1 < |z| < 3,\quad\text{and}\quad 3 < |z|.

$$

However, $z = 2$ is only in the annulus $1 < |z| < 3$.

The series will converge when

$$

\left|\frac{1}{z}\right| < 1\quad\text{and}\quad \left|\frac{z}{3}\right| < 1.

$$

Continuing like so what do I do to account for the constant 1? Would I just proceed by expression each fraction of the expansion as the appropriate geometric series?

So I would have:

$$\frac{1}{2z - 2} = \frac{1}{2z}\frac{1}{1 - \frac{1}{z}} = \frac{1}{2}\sum_{n = 0}^{\infty}\left(\frac{1}{z}\right)^{n + 1}

$$

Then doing the other fraction so the solution would be

$$

f(z) = 1 - \frac{1}{2}\left[9\sum_{n = 1}^{\infty}\left(\frac{z}{3}\right)^n + \sum_{n = 1}^{\infty}\left(\frac{1}{z}\right)^{n}\right]

$$

I know how to solve the other series that isn't the question so don't present the solution. The question is in regards to how 1 would or wouldn't affect writing f as a Laurent series. Would this be correct?

By the method of partial fractions, (how does the 1 effect the solution?)

$$

f(z) = \frac{z^2}{(z - 1)(z - 3)} = 1 + \frac{9}{2z - 6} - \frac{1}{2z - 2}.

$$

So $f(z)$ has simple poles at $z = 3$ and $z = 1$.

$f(z)$ will be analytic when

$$

|z| < 1,\quad 1 < |z| < 3,\quad\text{and}\quad 3 < |z|.

$$

However, $z = 2$ is only in the annulus $1 < |z| < 3$.

The series will converge when

$$

\left|\frac{1}{z}\right| < 1\quad\text{and}\quad \left|\frac{z}{3}\right| < 1.

$$

Continuing like so what do I do to account for the constant 1? Would I just proceed by expression each fraction of the expansion as the appropriate geometric series?

So I would have:

$$\frac{1}{2z - 2} = \frac{1}{2z}\frac{1}{1 - \frac{1}{z}} = \frac{1}{2}\sum_{n = 0}^{\infty}\left(\frac{1}{z}\right)^{n + 1}

$$

Then doing the other fraction so the solution would be

$$

f(z) = 1 - \frac{1}{2}\left[9\sum_{n = 1}^{\infty}\left(\frac{z}{3}\right)^n + \sum_{n = 1}^{\infty}\left(\frac{1}{z}\right)^{n}\right]

$$

I know how to solve the other series that isn't the question so don't present the solution. The question is in regards to how 1 would or wouldn't affect writing f as a Laurent series. Would this be correct?

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