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I have never done a Laurent series nor have we went over it in class but I guess I am supposed to know it perfectly already. The explanation in the book isn't that great.

Find the Laurent series of the form $\sum\limits_{n = -\infty}^{\infty}c_nz^n$ for $f(z) = \dfrac{33}{(2z - 1)(z + 5)}$ that converges in an annulus containing the point $z = -3i$

So I have used partial fractions

$$

f(z) = \frac{33}{(2z - 1)(z + 5)} = \frac{6}{2z - 1} - \frac{3}{z + 5}

$$

What am I supposed to do next?

Is this it

$$

\frac{-6}{1-2z}=-6(1+2z+(2z)^2+\cdots )\quad\text{and}\quad

\frac{3/5}{1-\left(-z/5\right)} = \frac{3}{5}\left(1+\frac{-z}{5}+\left(\frac{-z}{5}\right)^2+\cdots \right)

$$

Find the Laurent series of the form $\sum\limits_{n = -\infty}^{\infty}c_nz^n$ for $f(z) = \dfrac{33}{(2z - 1)(z + 5)}$ that converges in an annulus containing the point $z = -3i$

So I have used partial fractions

$$

f(z) = \frac{33}{(2z - 1)(z + 5)} = \frac{6}{2z - 1} - \frac{3}{z + 5}

$$

What am I supposed to do next?

Is this it

$$

\frac{-6}{1-2z}=-6(1+2z+(2z)^2+\cdots )\quad\text{and}\quad

\frac{3/5}{1-\left(-z/5\right)} = \frac{3}{5}\left(1+\frac{-z}{5}+\left(\frac{-z}{5}\right)^2+\cdots \right)

$$

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