How many roots does $z^{10} - 6z^6 + 3z^4 - 1$ have inside the circle of radius $3/2$?
A solution uses Rouche's theorem on $|z| = 2$ with $z^{10}$ and $-6z^6 + 3z^4 - 1$ to conclude that there are 10 solutions inside the circle of radius 2. Then by evaluating along the imaginary axis 3/2 < |iy| < 2 they use the intermediate value theorem to find at least two more roots. They then conclude that there are exactly two roots in the annulus 3/2 < |z| < 2 (hence 8 roots inside the circle $|z| = 3/2$). I don't see why there must be exactly two roots.