On |z|<1 define $f(z)=z-25z^3+z^5/5!-z^7/7!+z^9/9!-z^{11}/11!$. P:f has three zeros counting multiplicity in the unit disk. Q:f has one zero in 1/2<|z|<1. Choose truth values.
On |z|=1, the −25z³ term has modulus 25, exceeding the sum of the other term moduli (<1.01). On |z|=1/2, its modulus is 25/8=3.125, while the remainder is <0.501. Rouché gives three zeros in each disk. Neither circle has a zero, so the intervening annulus has none. **A**.
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