1. ## Laurent expansion trouble!

I need to find the Laurent expansion about 0 for the function:

$f(z)=exp(-1/z^4)$

I've done Laurent expansions before for functions involving $exp(z)$, and used the taylor series to expand $exp(z)$, but I'm stumped on what to do for this question. I havn't come across any other examples like this..

2. Originally Posted by gizmo
I need to find the Laurent expansion about 0 for the function:

f(z)=exp(-1/z^4)

I've done Laurent expansions before for functions involving exp(z), and used the taylor series to expand exp(z), but I'm stumped on what to do for this question. I havn't come across any other examples like this..

Replace z with -1/z^4 in the usual Maclaurin series for e^z.

3. that seems blindingly obvious now... sorry :/

4. ok so far my working is as follows:

$f(z)=exp(-1/z^{-4})$

[replace $z$ with $-1/z^{-4}$ in power series for exponential function, to obtain Laurent series]

$f(z)=\sum (-1^n)(z^{-4n})/n!$

[as N $\rightarrow$infinity , the approximation becomes exact for all complex numbers z, except at z=0]

$f(z)=1-(z^{-4})+(z^{-8})/2!-(z^{-12})/3!+...$

I have now written out 4 non-zero terms for this question and the function, is what I've done correct? Would you add anything in?

Thanks. (I couldn't get the z^-4 to display correctly, sorry for the poor coding!)

5. Originally Posted by gizmo
ok so far my working is as follows:

$f(z)=exp(-1/z^{-4})$

[replace $z$ with $-1/z^{-4}$ in power series for exponential function, to obtain Laurent series]

$f(z)=\sum (-1^n)(z^{-4n})/n!$

[as N $\rightarrow$infinity , the approximation becomes exact for all complex numbers z, except at z=0]

$f(z)=1-(z^{-4})+(z^{-8})/2!-(z^{-12})/3!+...$

I have now written out 4 non-zero terms for this question and the function, is what I've done correct? Would you add anything in?

Thanks. (I couldn't get the z^-4 to display correctly, sorry for the poor coding!)
Yes.