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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.

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If the exponent has more than one character, you can use curly brackets; compare $z^{10}$ and $z^10$, which was typeset as $z^{10}$ and $z^10$. –  Martin Sleziak Apr 17 '12 at 6:42

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I don't see why it follows from the steps you listed. But here is a proof:

Using Rouché's theorem on $|z|=1$ with $-6z^6$ as the dominant term, you can see that $z^{10}-6z^6+3z^4-1$ has exactly $6$ roots inside the unit circle. Then using the intermediate value theorem on the real axis from $z=1$ to $z=3/2$, you find a real root in that interval; since the polynomial is even, there's a real root between $-3/2$ and $-1$ as well.

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Why does this exhaust all the possible roots inside the circle $|z| = 3/2$? –  user90182312 Apr 17 '12 at 23:59
    
This has accounted for 8 of the 10 roots, and you already found the other 2 roots on the imaginary axis with $3/2 < |iy| < 2$. –  Greg Martin Apr 18 '12 at 6:11

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