Power Series for log z: Finding Singularity at z=0

[Harrisonized]

Homework Statement

Does there exist a power series expansion of log z around z=0? If so, what is it? If not, demonstrate that it is impossible.

Homework Equations

[PLAIN]http://img839.imageshack.us/img839/4839/eq1.gif

Here is the expansion of log z around z=1.

The Attempt at a Solution

I'm actually trying to find the value of the singularity at z=0 (the coefficient of 1/(z-a) around z=a), if it's possible. I have no idea what to do.

 

[I like Serena]

Welcome to PF, Harrisonized!

It is impossible to find the coefficient of 1/(z-a) if a=0, with z=0.
This means it is not possible to make an expansion at 0.

 

[Harrisonized]

Thank you for the welcome.

Why is it impossible? I tried putting the function into Wolfram|Alpha and shrinking the results to 0. Here's what happened:

[PLAIN]http://img6.imageshack.us/img6/7266/msp180819gf50a73e94c05h.gif

[PLAIN]http://img814.imageshack.us/img814/6152/msp355719gf4d225i9ahfee.gif

[PLAIN]http://img37.imageshack.us/img37/200/msp218319gf52gfh2fadihb.gif

Here's from the imaginary axis:

[PLAIN]http://img833.imageshack.us/img833/9853/msp347219gf4f1dgd6a2e17.gif

[PLAIN]http://img607.imageshack.us/img607/7232/msp210519gf4ifefdd61bgd.gif

[PLAIN]http://img232.imageshack.us/img232/3484/msp145319gf5304af95gbb8.gif

[PLAIN]http://img811.imageshack.us/img811/3433/msp103419gf53i0b5cd6h31.gif

This isn't a homework problem by the way. I just felt like the question wasn't worthy of real discussion.

 

[I like Serena]

The expansion around 0 is:
ln(z) = ln(0) + z * 1/0 + ...

Neither ln(0) nor 1/0 are defined.
As you can see in your results, the coefficients approach infinity.

 

[Harrisonized]

But e^(1/z) isn't defined at 0, yet a power series can still be found for e^(1/z) about z=0. In fact, e^(-∞)=0, so that's the equivalent of e^(1/z) approached to 0 from the negative real axis. Maybe there's some way of inverting e^(1/z) = 1+z^(-1)+z^(-2)/2!+z^(-3)/3!+... ?

 

[I like Serena]

But e^(1/z) isn't defined at 0, yet a power series can still be found for e^(1/z) about z=0. In fact, e^(-∞)=0, so that's the equivalent of e^(1/z) approached to 0 from the negative real axis. Maybe there's some way of inverting e^(1/z) = 1+z^(-1)+z^(-2)/2!+z^(-3)/3!+... ?

I am not aware of an expansion of e^(1/z) at 0.
I believe the one you mention is at z=infinity.
The expansion of e^(1/z) at 0 has a coefficient for the first derivative that is undefined (approaches infinity).

 

[Ray Vickson]

The function f(z) = ln(z) has a *branch point* at z = 0. For any r > 0 (no matter how small) we get different results for ln(-r), depending on whether we approach x=-r from positive imaginary values or negative imaginary values. This could not happen if f(z) had a convergent power series around zero.

RGV