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.