2
$\begingroup$

Using Rouche's Theorem one can show that $f(z) = z^4-6z+3$ has 3 zeros on the annulus $1 <|z|<2$. How can I show that the multiplicity of each zero is equal to $1$.

I tried to look to the derivative of $f(z)$, because to show that the zeros are simple it is enough to show that if $f(z_0) = 0$ then $f'(z_0) \not = 0$, however I am not sure how to finish this...

$\endgroup$

1 Answer 1

Reset to default
3
$\begingroup$

The multiple zeros of any polynomial $f(z)$ are zeros of both $f$ and $f'$, and therefore of $\gcd(f,f')$. Use the Euclidean algorithm to show that is $1$.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .