I want to use Rouches Theorem which says Given a jordan curve $\gamma $ in a domain $U$, if two holomorphic functions $f,g$ on $U$ satisfies $|f(z)-g(z)|<|g(z)|$ for all $z\in Im(\gamma)$ then they have the same number of zeros in $int(U)$.

Suppose $g(z)=z^4$ and $f(z)=z^4-5z+1$ and we have $|f(z)-g(z)|\leq |-5z|+1=11<|g(z)|=16$, hence $z^4-5z+1$ has all its roots inside $D_2(0)$.

But I am having trouble figuring out how many of the roots lies within the unit disk, I dont know where go from here. Can someone give me a hint? Thanks!

  • $\begingroup$ what book is this from? Factoring Rouche through Jordan Curves is strange and the statement is not even true when $U$ isn't simply connected (unless there is some special definition underlying 'interior'). $\endgroup$ Dec 7, 2022 at 19:01
  • $\begingroup$ @user8675309 Hi Thanks for the comment this is the book Complex Analysis by Saeed Zakeri and I think the meaning of $int(\gamma)$ is the bounded component of $U-Im(\gamma)$. And the Theorem 3.46 I am referring to. $\endgroup$
    – Remu X
    Dec 7, 2022 at 19:37
  • 1
    $\begingroup$ If $U$ has holes in it, e.g. $U:=\mathbb C-\big\{0\big\}$ then the 'theorem' breaks, e.g. $g(z)=z^2$ and $f(z) = z^2-1$ with $\gamma(t)=2\cdot \exp\big(2\pi i \cdot t\big)$ for $t \in [0,1]$ -- one has 2 zeros and the other none, in the bounded component of $U$. Stating $U$ is simply connected remedies this. $\endgroup$ Dec 7, 2022 at 22:57

1 Answer 1


Let $f(z)=z^{4}-5z$ and $g(z)=z^{4}-5z+1$. On $|z|=1$ we have $ |g(z)| \geq 5|z|-(|z^{4}|+1)=3>1=|f(z)-g(z)|$. So $g(z)$ has the same number of roots as $f(z)=z(z^{3}-5)$ which is $1$.


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