# Laurent series

1. Sep 7, 2009

### squaremeplz

1. The problem statement, all variables and given/known data

describe the laurent series for the function

$$f(z) = z^3 cos(\frac {1}{z^2})$$

b) use your answer to part a to compute the contour integral

$$\int z^3 cos(\frac {1}{z^2}) dz$$

where C is the unit counter-clockwise circle around the origin.

2. Relevant equations

3. The attempt at a solution

a)

$$f(z) = z^3 * \sum_{n=0}^\infty \frac {(-1)^n}{(2n)!} * ( \frac {1}{z^2} )^2^n$$

$$f(z) = \sum_{n=0}^\infty \frac {(-1)^n}{(2n)!} * \frac {1}{z^n}$$

b) so would I just evaluate

$$\sum_{n=0}^1 \frac {(-1)^n}{(2n)!} * \frac {1}{z^n}$$

Last edited: Sep 8, 2009
2. Sep 8, 2009

### gabbagabbahey

Ermmm...

$$z^3\left( \frac {1}{z^2} \right)^{2n}=\frac{1}{z^{4n-3}}\neq\frac{1}{z^n}$$

Not quite, the Residue at $z=0$ will be given by the coefficient of the $\frac{1}{z}$ term in the Laurent series...what is that coefficient?...What does that make the integral?

3. Sep 8, 2009

### squaremeplz

the coefficient is

$$\frac {(-1)^n}{(2n)!}$$

so it would be 1?

$$1 + \sum_{n=0}^\infty \frac {(-1)^n}{(2n)!} * \frac {1}{z^4^n^-^3}$$

?

sorry, a bit confused about te residue part.

Last edited: Sep 8, 2009
4. Sep 8, 2009

### gabbagabbahey

$\frac {(-1)^n}{(2n)!}$ is the coefficient of the $\frac{1}{z^{4n-3}}$ term....you want to find the coefficient of the $\frac{1}{z^{1}}$ term....so, for what value of $n$ does

$$\frac{1}{z^{4n-3}}=\frac{1}{z^{1}}$$ ?

Plug that value of $n$ into $\frac {(-1)^n}{(2n)!}$ to get the coefficient of that term. That coefficient will be equal to the residue of $f(z)$ at $z=0$.

5. Sep 8, 2009

### squaremeplz

hmm

so for n = 1 $$\frac {1}{z^4^n^-^1} = \frac {1}{z^1}$$

then

$$\frac {(-1)^1}{(2(1))!} = \frac {-1}{2}$$

then by the residue thrm.

$$\int z^3 * cos(\frac {1}{z^2}) = 2*\pi*i * \frac {-1}{2}$$

$$= -\pi*i$$

thanks

6. Sep 8, 2009

### gabbagabbahey

Looks good to me.

7. Sep 8, 2009