Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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
up vote 1 down vote accepted

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.

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.