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If $f$ has a pole of order $m$ at $z_0$ find the order of the pole of $g(z) = \frac{f'(z)}{f(z)}$ at $z_0$. What is the coefficient of $(z-z_o)^{-1}$ in the Laurent expansion for $g(z)$.

M Since $f$ has a pole of order $m$--does that mean that $f'$ has a pole of order $m-1$? How are they related? Also, when multiplying functions, you multiply the order of the poles. Is that correct? What about dividing functions? Also, how is this related to finding the Laurent coefficient?

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    $\begingroup$ "$f$ has a pole of order $m$ at $z_0$" means that $(z-z_0)f(z)$ is analytic and nonzero in a neighborhood of $z_0$; thus its Laurent series there looks like $f(z) = \frac{c_{-m}}{(z-z_0)^m} + \frac{c_{-(m-1)}}{(z-z_0)^{m-1}} + \cdots$. (So, for example, $f'$ will have a pole of order $m+1$.) Hint: use the fact that $g(z) = \frac d{dz}(\log f(z))$. $\endgroup$ – Greg Martin Apr 26 '16 at 3:13
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    $\begingroup$ @GregMartin You mean $(z-z_0)^m f(z)$ is analytic. $\endgroup$ – Rick Sanchez Apr 26 '16 at 3:17
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If you write down the Laurent series carefully, I suppose these answers will come like A,B,C to you.

Since $f$ has a pole of order $m$, it follows that $f(z)=(z-z_0)^{-m}h(z)$, where $h(z)$ is an analytic function on the domain where $f$ is analytic, and the point $z_0$. Now, we can write down $\frac{f'(z)}{f(z)}$ as: $$ \frac{f'(z)}{f(z)} = \frac{(-m)(z-z_0)^{-m-1}h(z) + h'(z)(z-z_0)^{-m}}{(z-z_0)^{-m}h(z)} $$

There is only one factor at the bottom of $(z-z_0)$, that is to say, $$ \frac{f'(z)}{f(z)} = \frac{(-m)}{(z-z_0)} + \frac{h'(z)}{h(z)} $$

The first function has a pole at $z_0$ of order $1$, and the second one doesn't have a pole at $z_0$ because $h(z_0) \neq 0$. Hence the order of $z_0$ as a pole of $f$ is $1$.

When we multiply fractions, we add the pole factors. To see this, suppose $f$ and $g$ have poles at $z_0$ or orders $m$ and $n$ respectively. Now, note that by definition, $f(z)=(z-z_0)^{-m}f_1(z)$ and $g(z)=(z-z_0)^{-n}g_1(z)$ where $f_1$ and $g_1$ are holomorphic at $z_0$ and don't vanish at $z_0$. Just multiply these to see that $fg(z)=(z-z_0)^{-(m+n)}f_1g_1(z)$, where $f_1g_1$ is holomorphic at $z_0$ and doesn't vanish at $z_0$. This tells us the order of $z_0$ as a pole of $fg$ is $-(m+n)$.

For dividing functions, see that you get $m-n$ when $m>n$, and the pole vanishes if $m \leq n$.

Actually, when you Laurent expand $f$ around $z_0$, the negative coefficient $c_m$ will be non-zero, but all coefficients of terms less than $-m$ will all be zero. When you differentiate this relation and get the term $f'$ then see what the result is for yourself about the order of $z_0$ as a pole of $f'$. You will see that the order actually increases by $1$, it would now be $m+1$.

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  • $\begingroup$ If $f$ has a pole of order $m$, then $f(z)=\frac{h(z)}{(z-z_0)^m}\ne (z-z_0)^mh(z)$, where $h$ is analytic. $\endgroup$ – Mark Viola Apr 26 '16 at 3:21
  • $\begingroup$ I've got the minuses mixed up. I will sort it out soon enough. $\endgroup$ – астон вілла олоф мэллбэрг Apr 26 '16 at 3:22
  • $\begingroup$ Yes, and the ratio $f'/f$ is flawed also. $\endgroup$ – Mark Viola Apr 26 '16 at 3:24
  • $\begingroup$ Corrected that as well. Please check once more as I would like to avoid silly mistakes. $\endgroup$ – астон вілла олоф мэллбэрг Apr 26 '16 at 3:27
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    $\begingroup$ All seems well. +1 $\endgroup$ – Mark Viola Apr 26 '16 at 3:28

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