# Math Help - [SOLVED] Residue of f'/f with zero of order m

1. ## [SOLVED] Residue of f'/f with zero of order m

Happy spring break.

I feel like I get it. But I'm not quite there... Thanks so much!

2. Originally Posted by davismj

Happy spring break.

I feel like I get it. But I'm not quite there... Thanks so much!
Well lets use your definition of $f(z)=g(z)(z-z_0)^m$ so then
$f'(z)=g'(z)(z-z_0)^m+mg(z)(z-z_0)^{m-1}$

Now $\frac{f'(z)}{f(z)}=\frac{g'(z)(z-z_0)^m+mg(z)(z-z_0)^{m-1}}{g(z)(z-z_0)^m}=\frac{g'(z)}{g(z)}+\frac{m}{z-z_0}$

3. Originally Posted by TheEmptySet
Well lets use your definition of $f(z)=g(z)(z-z_0)^m$ so then
$f'(z)=g'(z)(z-z_0)^m+mg(z)(z-z_0)^{m-1}$

Now $\frac{f'(z)}{f(z)}=\frac{g'(z)(z-z_0)^m+mg(z)(z-z_0)^{m-1}}{g(z)(z-z_0)^m}=\frac{g'(z)}{g(z)}+\frac{m}{z-z_0}$
I tried that but erased it because I don't see any conclusion of the g/g' issue.

$g(z) = \frac{f(z)}{(z-z_0)^m}$, taking that derivative and dividing and what not ends up a mess...

4. Originally Posted by davismj
I tried that but erased it because I don't see any conclusion of the g/g' issue.

$g(z) = \frac{f(z)}{(z-z_0)^m}$, taking that derivative and dividing and what not ends up a mess...
$g(z) \text{ and } g'(z)$ are both analytic and $g(z_0) \ne0$ therefore their residue is zero.

i.e $\frac{g'(z)}{g(z)}=\sum_{n=0}^{\infty}a_n(z-z_0)$

5. Originally Posted by TheEmptySet
$g(z) \text{ and } g'(z)$ are both analytic and $g(z_0) \ne0$ therefore their residue is zero.

i.e $\frac{g'(z)}{g(z)}=\sum_{n=0}^{\infty}a_n(z-z_0)$
I really appreciate your help, but please forgive me; my book gave very little motivation or background on residues. I have a vague idea of what they are, I'm not really sure where we're supposed to find them, etc.

Also, I'm not sure why $\frac{g'(z)}{g(z)}=\sum_{n=0}^{\infty}a_n(z-z_0)$. If anything, wouldn't it be equal to $\frac{1}{(z-z_0)}\sum_{n=0}^{\infty}a_n$. How does this relate to residues?

6. Here's an exercise that might help to motivate "residue".

Integrate $\oint (z- z_0)^n dz$ around a circle with center at $z_0$ and radius R where n can be any integer. Do that by writing $z= Re^{i\theta}+ z_0$ and integrating with respect to $\theta$.

You should see that the integral is 0 for all n except n= -1, and, in that case, is $2\pi i$.

Now, suppose f(z) has a "pole of order n" at $z_0$. That means that f(z) is not analytic, so cannot be written as a Taylor's series about $z_0$ but can be written as a "Laurent" series (with negative powers up to -n).

Basically, saying "f(z) has a pole of order n at $z_0$" means that $(z- z_0)^n f(z)$ is analytic at $z_0$ but not for any lower power of $z- z_0$. That, in turn, means that we can write $(z- z_0)^nf(z)$ as a Taylor's series about $z_0$: $(z- z_0)^n f(z)= a_0+ a_1(z-z_0)+ a_2(z- z_0)^2+ a_3(z- z_0)^3+ \cdot\cdot\cdot+ a_{n-1}(z- z_0)^{-1}+ a_n(z- z_0)^n+ \cdot\cdot\cdot$.

Dividing both sides by $(z-z_0)^n$, $f(z)= a_0(z- z_0)^{-n}+ a_1(z- z_0)^{-n+1}+ a_2(z- z_0)^{-n+2}+ a_3(z- z_0)^{-n+3}+ \cdot\cdot\cdot+ a_{n-1}(z- z_0)^{-1}+ a_n+ \cdot\cdot\cdot$.

Now integrate the right side term by term around a path with $z_0$ in its interior. As before, the integral of all powers of x, except one, will be 0. The exception is for the "-1" power- we will have:
$\oint f(z)dz= a_{n-1}(2\pi i)$. That number is the "residue of f(z) at $z= z_0$".

7. Originally Posted by HallsofIvy
Here's an exercise that might help to motivate "residue".
Basically, saying "f(z) has a pole of order n at $z_0$" means that $(z- z_0)^n f(z)$ is analytic at $z_0$ but not for any lower power of $z- z_0$. That, in turn, means that we can write $(z- z_0)^nf(z)$ as a Taylor's series about $z_0$: $(z- z_0)^n f(z)= a_0+ a_1(z-z_0)+ \cdot\cdot\cdot+ a_{n-1}(z- z_0)^{n-1}+ a_n(z- z_0)^n+ \cdot\cdot\cdot$.

Dividing both sides by $(z-z_0)^n$, $f(z)= a_0(z- z_0)^{-n}+ a_1(z- z_0)^{-n+1}+ \cdot\cdot\cdot+ a_{n-1}(z- z_0)^{-1}+ a_n+ \cdot\cdot\cdot$.
Trying to fix the latex. ^^