- How many zeros does the polynomial \(z ^ { 6 } - 2 z ^ { 5 } + 7 z ^ { 4 } + z ^ { 3 } - z + 1\) have in the open unit disc \(D = \{ z : | z | < 1 \} ?\)
Solution
Let \(p(z)=z^6-2z^5+7z^4+z^3-z+1\).
1. On \(\abs z=1\), \(\abs{p(z)-7z^4}<\abs{7z^4}\).
\(\abs{z^6-2z^5+z^3-z+1}\le1+2+1+1+1=6<7\).
2. Q.E.D.
By step 1 and Rouché’s theorem, \(p\) has as many zeros in \(\abs z<1\) as \(7z^4\), namely \(\boxed{4}\), counted with multiplicity.