# Show that the equation $z^4-5z+1=0$ has all its roots in the disk $D_2(0)$ and how many of these roots are in the unit disk

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!

• 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'). Dec 7, 2022 at 19:01
• @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. Dec 7, 2022 at 19:37
• 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. Dec 7, 2022 at 22:57

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 $$1$$.