Can someone help me find how many roots does this polynom : $p(z)=z^4-5z+1$ have on set $|z|>2$ using Rouché's theorem?
I'm thinking like this: This polynomial has 4 roots(fundamental theorem of algebra) , and by applying Rouché's theorem I can find that $p$ has one root on unit disc $|z|<1$ , (since the largest coefficient is at $z$).
In similar way , I got that $p$ has 4 roots on $|z|=2$ , since by using $g(z)=z^{4}$ ,I have $|p-g|=|-5z+1|<10+1=11<2^4=|g(z)|$ , so by Rouché's theorem $p$ has 4 roots in $|z|<2$.
Meanwhile, I have to find number of roots in $|z|>2$ , and by the proof above, I would say that $p$ has NO roots on $|z|>2$ (since it has 4 in total , and I got that all of them are in the interior of this set).
But,the result has to be 3-polynom $p$ has 3 roots on $|z|>2$. Can someone please help me with this?I don't know what I'm doing wrong...
Thank you in advance!
What you have done is correct.If $|z| >2$ then $|p(z)| \geq |z^{4}-5z| -1=|z|(|z|^{3} -5)-1>2(8-5)-1 >0$ so there is no root for $|z| >2$.