Originally Posted by

**abrak** In my textbook, as part of an example, I am given that "$\displaystyle \frac{1}{{{{\left( {z - 1} \right)}^2}}}$ is its own Laurent expansion about z = 1, where it has a double pole."

My reasoning is this:

If we define a punctured disc centred at 1 with radius r > 0, then we can express it as a Laurent series $\displaystyle \sum\limits_{n = - \infty }^\infty {{c_n}{{\left( {z - 1} \right)}^n}} $, where $\displaystyle {c_n} = \frac{1}{{2\pi i}}\int_\gamma {\frac{{f\left( w \right)}}{{{{\left( {w - 1} \right)}^{n + 1}}}}{\rm{ }}} {\rm{d}}w$. Having a double pole means that $\displaystyle {c_{ - 2}} \ne 0$, while $\displaystyle {c_i} = 0$, for all i < -2.

How am I supposed to compute the coefficients of the Laurent series? Am I supposed to use the formula for $\displaystyle {c_n}$ described above? All I seem to be able to do is the following:

$\displaystyle {c_n} = \frac{1}{{2\pi i}}\int_\gamma {\frac{{f\left( w \right)}}{{{{\left( {w - 1} \right)}^{n + 1}}}}{\rm{ }}} {\rm{d}}w = \frac{1}{{2\pi i}}\int_\gamma {\frac{{\left( {\frac{1}{{{{\left( {w - 1} \right)}^2}}}} \right)}}{{{{\left( {w - 1} \right)}^{n + 1}}}}{\rm{ }}} {\rm{d}}w = \frac{1}{{2\pi i}}\int_\gamma {\frac{1}{{{{\left( {w - 1} \right)}^{n + 3}}}}{\rm{ }}} {\rm{d}}w$, so

$\displaystyle {c_{ - 2}} = \frac{1}{{2\pi i}}\int_\gamma {\frac{1}{{{{\left( {w - 1} \right)}^{ - 1}}}}{\rm{ }}} {\rm{d}}w = \frac{1}{{2\pi i}}\int_\gamma {\left( {w - 1} \right){\rm{ }}} {\rm{d}}w$, which does not equal zero (what I am trying to show), and, for example, $\displaystyle {c_{ - 3}} = \frac{1}{{2\pi i}}\int_\gamma {\frac{1}{{{{\left( {w - 1} \right)}^{ - 2}}}}{\rm{ }}} {\rm{d}}w = \frac{1}{{2\pi i}}\int_\gamma {{{\left( {w - 1} \right)}^2}{\rm{ }}} {\rm{d}}w$, which does equal zero (what I am trying to show).

Can anyone shed some light on this?