Suppose that $f$ has a pole of order $m$ at $a$ and $g$ is holomorphic in $D(a;r)$ (open disc centered at a with radius $r>0$). Then at $a$ the function $fg$

(i) has pole of order $n-m$ if $g$ has zero of order $n$ ($n < m$) at $a$.

(ii) has removable singularity if $g$ has zero of order $n$ ($n > m$) at $a$.


Since $f$ has a pole of order $m$, we can write it as $$ f(z) = \lim_{z \to a} \frac{D}{(z-a)^m}, $$ where $D$ is a non-zero constant.

Similarly, $$ g(z) = \lim_{z \to a} {(z-a)^n}h(z), $$ where $h(z) \neq 0$.

Hence, \begin{align*} fg(z) &= \lim_{z \to a} \frac{D}{(z-a)^m} \times \lim_{z \to a} (z-a)^n h(z) \\ &= \lim_{z \to a} \frac{D}{(z-a)^m}{(z-a)^n}h(z). \end{align*}

For (i), since $m > n$, we have $(z-a)^{m-n}$ in the denominator, hence $fg$ has a pole of order $m-n$.

For (ii), since $m < n$, we have $(z-a)^{n-m}$ in the numerator, hence $fg$ should have zero of order $n-m$.

But why is there a removable singularity? Please explain.

  • 1
    $\begingroup$ For what it's worth, your limit criteria are probably not what you mean. Normally, one says $f$ has a pole of order $n > 0$ at $a$ if there exists a holomorphic function $h$ such that $$f(z) = \frac{h(z)}{(z - a)^{n}},\qquad h(a) \neq 0,$$for all $z$ in some deleted open disk about $a$, and similarly for a zero of order $m$. Still, the idea of your calculation is correct. :) $\endgroup$ – Andrew D. Hwang Sep 20 '16 at 12:40

In case (ii), the product $fg$ isn't defined at $z=a$, since $f$ isn't.

Hence, at least formally, $fg$ has a singularity at $z=a$, but your computation shows that it is removable. (Putting $fg(a) = 0$ gives you a holomorphic function with a zero of order $n-m$ at $z=a$.)

  • $\begingroup$ I appreciate for trying to help..but sorry..I cannot understand fg is not defined at a...will u be please little more specific.. $\endgroup$ – Sam Christopher Sep 20 '16 at 11:17
  • $\begingroup$ @SamChristopher in complex analysis, we consider the analytic continuation of functions, so (when it exists) $fg(a) = \lim_{z \to a} fg(z)$ $\endgroup$ – reuns Sep 20 '16 at 11:20
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    $\begingroup$ @SamChristopher $f$ has a pole at $z=a$, so in particular $f$ is not defined at $a$. $\endgroup$ – mrf Sep 20 '16 at 13:09
  • $\begingroup$ Great man....i got it...but i also thought otherwise since f has a pole of order m, the expansion laurentz series has (z-a) upto (z-a)^m in denominator, so when multiplying (z-a)^n, we get (z-a)^{n-m} in the numerator, so the principal terms of laurentz series is 0 not even C_0 term so by adding f(0)=0..it has removable singularity...I dont know whether I am right or not... $\endgroup$ – Sam Christopher Sep 20 '16 at 13:36

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