Finding Laurent Series and Residues for Complex Functions

[Incand]

Homework Statement

Find four terns of the Laurent series for the given function about $z_0=0$. Also, give the residue of the function at the point.
a) $\frac{1}{e^z-1}$
b) $\frac{1}{1-\cos z}$

Homework Equations

The residue of the function at $z_0$ is coefficient before the $1/(z-z_0)$ term in the Laurent expansion.

The definition of the coefficients for $f(z)$ (possibly of use)
$a_k = \frac{1}{2\pi i} \int_{|w-z_0|=s} \frac{f(w)}{(w-z_0)^{k+1}}dw, \; \; \; k=0,\pm 1, \dots$

Theorem:
If $F$ and $G$ are analytic functions on the disc $\{z:|z-z_0|<r_0\}$ with $g(z_0)=0$ but $G'(z_0) \ne 0$. Then
$\text{Res}\left( \frac{F}{G}; z_0\right) = \frac{F(z_0)}{G'(z_0)}$.

The Attempt at a Solution

Starting with a) using the definition seems very impractical so there's probably an easier way to find the coefficient but I'm not sure how. It seems computing the residue is easier using the theorem above. $\text{Res}\left(\frac{1}{e^x-1};0\right) = \frac{1}{e^0} = 1$ and hence one term of the Laurent expansion is $\frac{1}{z}$.

As for how to actually get the Laurent series I have no idea. I tried to match the coefficients by putting $g(z) = \frac{a_{-1}}{z} + a_0 + a_1z + a_2z^2+\dots$ since we have a pole of order $1$. We can then write
$g(z)(e^x-1) = 1$ and we know the expansion of $e^x$ so we have
$\left( a_{-1} + \sum_0^\infty a_kz^k \right) \left( \sum_1^\infty \frac{z^k}{k!} \right) = 1$. But that doesn't seem to help me at all.

 

[Samy_A]

Homework Statement

Find four terns of the Laurent series for the given function about $z_0=0$. Also, give the residue of the function at the point.
a) $\frac{1}{e^z-1}$
b) $\frac{1}{1-\cos z}$

Homework Equations

The residue of the function at $z_0$ is coefficient before the $1/(z-z_0)$ term in the Laurent expansion.

The definition of the coefficients for $f(z)$ (possibly of use)
$a_k = \frac{1}{2\pi i} \int_{|w-z_0|=s} \frac{f(w)}{(w-z_0)^{k+1}}dw, \; \; \; k=0,\pm 1, \dots$

Theorem:
If $F$ and $G$ are analytic functions on the disc $\{z:|z-z_0|<r_0\}$ with $g(z_0)=0$ but $G'(z_0) \ne 0$. Then
$\text{Res}\left( \frac{F}{G}; z_0\right) = \frac{F(z_0)}{G'(z_0)}$.

The Attempt at a Solution

Starting with a) using the definition seems very impractical so there's probably an easier way to find the coefficient but I'm not sure how. It seems computing the residue is easier using the theorem above. $\text{Res}\left(\frac{1}{e^x-1};0\right) = \frac{1}{e^0} = 1$ and hence one term of the Laurent expansion is $\frac{1}{z}$.

As for how to actually get the Laurent series I have no idea. I tried to match the coefficients by putting $g(z) = \frac{a_{-1}}{z} + a_0 + a_1z + a_2z^2+\dots$ since we have a pole of order $1$. We can then write
$g(z)(e^x-1) = 1$ and we know the expansion of $e^x$ so we have
$\left( a_{-1} + \sum_0^\infty a_kz^k \right) \left( \sum_1^\infty \frac{z^k}{k!} \right) = 1$. But that doesn't seem to help me at all.

Looks good, at first sight.
You have $g(z) = \frac{1}{z} + a_0 + a_1z + a_2z^2+\dots$
Then $g(z)(e^z-1)=1$ gives $(\frac{1}{z} + a_0 + a_1z + a_2z^2+\dots)(z+\frac{z²}{2}+\frac{z³}{6}+ \frac{z^4}{24}+\dots)=1$.
That should be enough to find 4 terms of the Laurent series (the terms up to $z²$).

 

[Incand]

Thanks! That was easier than I expected, seems I gave up on the finish line.
Posting the rest of the solution in case anyone is curious:
Matching coefficients:
z: $1/2 +a_0 = 0 \Longrightarrow a_0 = -1/2$
z^2: $1/6+a_0/2+a_1 = 0 \Longrightarrow a_1 = 1/12$
z^3: $1/24+a_0/6+a_1/2+a_2 = 0 \Longrightarrow a_2 = 0$.

I believe I'm able to do b) myself now!