Let \[ f(z)=z^4+\frac{z^3}{4}-\frac14. \] How many zeros does \(f\) have in \[ \left\{z\in\mathbb C:\frac12<|z|<1\right\}? \]
Solution
1. The polynomial \(f\) has exactly four zeros, counted withmultiplicity, in the disk \(\abs z<1\).
On \(\abs z=1\), \[ \begin{aligned} \abs{f(z)-z^4} &= \left|\frac{z^3}{4}-\frac14\right| \\ &\le \frac14+\frac14 \\ &= \frac12 < 1 = \abs{z^4}. \end{aligned} \] By Rouché’s theorem, \(f\) and \(z^4\) have the same number of zeros in \(\abs z<1\). The polynomial \(z^4\) has four zeros there, counted with multiplicity.
2. The polynomial \(f\) has no zero in the closed disk\[ \abs z\le\frac12. \]
If \(\abs z\le1/2\), then the reverse triangle inequality gives \[ \begin{aligned} \abs{f(z)} &= \left|z^4+\frac{z^3}{4}-\frac14\right| \\ &\ge \frac14-\abs z^4-\frac{\abs z^3}{4} \\ &\ge \frac14-\frac1{16}-\frac1{32} \\ &= \frac5{32} > 0. \end{aligned} \] Thus \(f(z)\ne0\) throughout the closed disk.
3. Every zero of \(f\) in the unit disk lies in the annulus\[ \frac12<\abs z<1. \]
Step 2 excludes all zeros with \(\abs z\le1/2\), while step 1 counts the zeros with \(\abs z<1\). Hence the four zeros counted in step 1 all lie in the stated annulus.
4. Therefore the number of zeros in the annulus, counted withmultiplicity, is\[ \boxed{4}. \]
This is exactly the count from step 3.
5. Q.E.D.
Step 4 answers the problem.