Question 2. Find the number of zeros of the function \(f ( z ) = 2 z ^ { 5 } + 8 z - 1\) in the annulus \(1 < | z | < 2\)
Solution
There are exactly \(\boxed{4}\) zeros in the annulus, and all are simple.
1. There is exactly one zero in \(|z|<1\).
Let \[ p(z)=2z^5+8z-1. \] On \(|z|=1\), \[ |2z^5-1|\le 3<8=|8z|. \] Rouché’s theorem [@SS03] therefore gives the same number of zeros in the unit disk for \(p\) and \(8z\), namely one, counted with multiplicity. The strict inequality also shows that \(p\) has no zero on \(|z|=1\).
2. There are exactly five zeros in \(|z|<2\).
On \(|z|=2\), \[ |8z-1|\le17<64=|2z^5|. \] A second application of Rouché’s theorem gives the same number of zeros in \(|z|<2\) for \(p\) and \(2z^5\), namely five, counted with multiplicity. Again the strict inequality excludes zeros from the boundary circle.
3. Subtracting the two disk counts gives the annular count.
Because neither boundary circle contains a zero, every zero in \(|z|<2\) lies either in \(|z|<1\) or in \(1<|z|<2\). Hence the annulus contains \[ 5-1=4 \] zeros, counted with multiplicity.
4. Every root is simple, so there are four distinct annular roots.
A multiple root \(a\) would satisfy \(p(a)=p'(a)=0\). The derivative equation \(10a^4+8=0\) gives \(a^4=-4/5\). Substitution in the polynomial equation gives \[ 0=2a\left(-\frac45\right)+8a-1 =\frac{32}{5}a-1, \] so \(a=5/32\) is positive real. This contradicts its negative fourth power. Therefore no multiple root exists, and the preceding count is also a count of distinct zeros.