# Find the number of zeroes of a function $z^3+z^2 +z+1$ lying inside $|z|=2$ by Rouche's theorem? [closed]

This function have mamy possible values. Which one we should select?

• Note that $(z-1)(z^3+z^2+z+1) = (z^4-1)$ implies that all (three) roots lie on the unit circle. Commented Nov 21, 2019 at 16:07
• What do you mean it has many possible values? It's a polynomial. Commented Nov 21, 2019 at 16:59

If $$\lvert z\rvert=2$$, then\begin{align}\lvert z^2+z+1\rvert&\leqslant\lvert z\rvert^2+\lvert z\rvert+1\\&\leqslant4+2+1\\&<8\\&=\lvert z\rvert^3.\end{align}So, by Rouché's theorem, your function has as many zeros in the given region as $$z^3$$, which has $$3$$ zeros there.

Hint For $$|z|=2$$ you have.

$$|z^3| > |z^2+z+1|$$

Apply the Rouche theorem.

On $$|z| =2$$, $$|z^3|>|z^2+z+1|$$.

Hence by Rouche's Theorem, the number of zeros of $$z^3+z^2+z+1$$ is equal to the number of zeros of $$z^3$$ inside $$|z|=2$$.

The zeros of $$z^3$$ are just 0, and all three of them lie inside $$|z|=2$$, hence the number of zeros of $$z^3+z^2+z+1$$ is equal to three.