The Argument Principle

Suppose a function $f$ is meromorphic on an open set that contains a circle $C$ and its interior. Further assume that $f$ has no zeroes on $C$ (but may have zeroes in the interior of $C$). Then,

$$\frac{1}{2\pi i} \oint_C \frac{f'(z)}{f(z)} dz \ =$$

the number of zeroes of $f$ in the interior of $C$ (counted with multiplicity) minus the number of poles of $f$ in the interior of $C$ (counted with multiplicity).

Rouche's Theorem

Suppose that $f$ and $g$ are holomorphic functions on a domain and let $C$ be a circle whose interior also lies in the domain. If $|f(z)-g(z)| \lt |f(z)|$ for all $z$ on $C$, then $f$ and $g$ have the same number of zeroes in the interior of $C$, counting multiplicities.

My Question

Why does the Argument Principle require $f$ to have no zeroes on $C$, whereas this requirement is relaxed for Rouche's Theorem?

Thank you.

  • $\begingroup$ It should be $|f-g|<|g|$ or $|f-g|<|f|$ on $C$, not $|f-g|<|f|+|g|$ on $C$. $\endgroup$ Jan 3, 2013 at 22:26
  • $\begingroup$ If $f$ has a zero on $C$, the (correct) condition in Rouche's theorem can't be satisfied. $\endgroup$
    – mrf
    Jan 3, 2013 at 22:35
  • $\begingroup$ @Mercy, thanks, corrected. $\endgroup$
    – Conan Wong
    Jan 3, 2013 at 22:54
  • $\begingroup$ @mrf, not sure about your comment. Various sources I've seen do not have this assumption. $\endgroup$
    – Conan Wong
    Jan 3, 2013 at 22:55
  • 1
    $\begingroup$ @Mercy: Rouché's theorem also holds with the estimate $|f-g| < |f|+|g|$ on $C$. This is sometimes useful, apart from being nicely symmetric. See Remmert, page 390 $\endgroup$
    – Martin
    Jan 4, 2013 at 11:45

1 Answer 1


In the Argument Principle: If $f$ has a zero $z_0 \in C$, then $z_0$ is a pole for the quotient $f'/f$. Therefore you cannot define $\int_C f'(z)/f(z)\,dz$.

In Rouche's Theorem: The inequality $|f-g|<|f|$ on $C$ implies that $f$ and $g$ have no zeroes on $C$. In fact, if $g$ (resp. $f$) had a zero $z_0 \in C$, then $|f(z_0)|<|f(z_0)|$ (resp. $|g(z_0)|<0$), which is a contradiction.

  • $\begingroup$ Yes, and if there's some roots of $f$ on $C$, as $f$ is not zero everywhere and hence its roots doesn't have accumulation point, you can find a greater $C'$ whose interior contains $int(C)$ and then form the integral $\int_{C'}f'/f$, Am I right ? $\endgroup$ Nov 23, 2014 at 19:10

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