Laurent Series for f(z): Computing Contour Integral

[squaremeplz]

Homework Statement

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.

Homework Equations

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

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

 

[gabbagabbahey]

a)

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

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

Ermmm...

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

b) so would I just evaluate

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

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?

 

[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.

 

[gabbagabbahey]

the coefficient is

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

so it would be 1?

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

 

[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.

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

$$= -\pi*i$$

is this the right answer?

thanks

 

[gabbagabbahey]

Looks good to me.

 

[squaremeplz]

thanks! that was very helpful.