1. ## Laurent expansions

In my textbook, as part of an example, I am given that " $\frac{1}{{{{\left( {z - 1} \right)}^2}}}$ 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 $\sum\limits_{n = - \infty }^\infty {{c_n}{{\left( {z - 1} \right)}^n}}$, where ${c_n} = \frac{1}{{2\pi i}}\int_\gamma {\frac{{f\left( w \right)}}{{{{\left( {w - 1} \right)}^{n + 1}}}}{\rm{ }}} {\rm{d}}w$. Having a double pole means that ${c_{ - 2}} \ne 0$, while ${c_i} = 0$, for all i < -2.

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

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

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

Can anyone shed some light on this?

2. Originally Posted by abrak
In my textbook, as part of an example, I am given that " $\frac{1}{{{{\left( {z - 1} \right)}^2}}}$ 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 $\sum\limits_{n = - \infty }^\infty {{c_n}{{\left( {z - 1} \right)}^n}}$, where ${c_n} = \frac{1}{{2\pi i}}\int_\gamma {\frac{{f\left( w \right)}}{{{{\left( {w - 1} \right)}^{n + 1}}}}{\rm{ }}} {\rm{d}}w$. Having a double pole means that ${c_{ - 2}} \ne 0$, while ${c_i} = 0$, for all i < -2.

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

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

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

Can anyone shed some light on this?

Of course $c_{-2}\neq 0$ : you're given that it is 1...!

$\frac{1}{(z-1)^2} =...0\cdot (x-1)^{-3}+1\cdot (z-1)^{-2} + 0\cdot (z-1)^{-1} + ....$

Tonio

3. Originally Posted by tonio
Of course $c_{-2}\neq 0$ : you're given that it is 1...!

$\frac{1}{(z-1)^2} =...0\cdot (x-1)^{-3}+1\cdot (z-1)^{-2} + 0\cdot (z-1)^{-1} + ....$

Tonio
Oh, of course!

So it's much easier than I thought: my function's Laurent expansion has only one non-zero term - that being the (-2)th term of the power series. That makes sense, thank you!

4. Notice that $\int_{\gamma } \frac{1}{(z-1)^{n+3}}dz = 0$ if $n\neq -2$ and $2\pi i$ if $n=-2$.

1) $-2+3\neq -1$
2) $\int_{\gamma } (z-1)dz =0$ (Cauchy's theorem)