Roots of a complex polynomial in the unit disc

Question

I'm trying to solve the following:

Prove that the polynomial $z^3 + 3z^2 + 2$ has 2 roots (counted with multiplicity) in the unit disc.

Am I on the right track? Hope you can help me.

My idea:

• Do something about the $z^3$, since that implies 3 roots
• Using $z= x + i y$
• To expand $z^3: x^3 + 3 i x^2 y - 3 xy^2 - i y^3$

Answer 1

Hints \begin{align} \bullet\hspace{3em}\frac{d}{dx}\big(x^3+3x^2+2\big)&=3x^2+6x\\ &\cases{>0&for $\ x<-2\ $\\ <0&for$\ -2<x<0$\\ >0& for $\ 0<x$} \end{align} Therefore $\ x^3+3x^2+2\ $ increases from $\ {-}\infty\ $ to a local maximum of $6$ over the interval $\ ({-}\infty,-2]\ $, decreases from the local maximum of $6$ to a local minimum of $2$ over the interval $\ [-2,0]\ $, and increases from the local minimum of $2$ to $\ \infty\ $ over the interval $\ [0,\infty)\ $. What does that tell you about the number and location of any real roots?

• The coefficients of the polynomial $\ x^3+3x^2+2\ $ are all real. What can you therefore say about any of its complex roots.

• What is the product of the roots of the polynomial?

Answer 2

Consider the polynomials $$ p(z) = z^3 + 3z^2 + 2 $$ and $$ q(z) = (z+3)(z^2+\frac 23) = z^3 + 3z^2 + \frac 23 z + 2 \, . $$ For $|z| = 1$ is $|z+3| \ge 2$ and $|z^2 + \frac 23| \ge \frac 13$, where equality does not hold simultaneously in both estimates. It follows that $$ |q(z)| > \frac 23 $$ for $|z| = 1$. Now we can apply Rouché's theorem. For $|z| = 1$ is $$ |p(z) - q(z) | = | \frac 23 z| = \frac 23 < |q(z)| $$ and it follows that $p$ and $q$ have the same number of zeros (counted with multiplicity) in the unit disk.

Since $q$ has two zeros in the unit disk (the two square roots of $-2/3$), the same is true for $p$.

Answer 3

(i). for any $\delta \gt 0$ consider $p_\delta(z) :=\frac{1}{1+\delta}z^3 + 3z^2 + 2$
and for $ z\in S^1 = \partial D$
$\big\vert 3z^2\big\vert =3 \gt \frac{1}{1+\delta} + 2 = \big\vert \frac{1}{1+\delta}z^3\big\vert + \big\vert 2\big\vert \geq \big\vert \frac{1}{1+\delta}z^3 + 2\big\vert$
$\implies 3z^2 \text{ and }p_\delta(z)$ have the same number of roots in $D$ per Rouche. That is: $p_\delta(z)$ has exactly $2$ roots in the open unit disc, $D$.

(ii.) Now for $p$ itself, write:
$p(z)=z^3 + 3z^2 + 2=(z-\lambda_1)(z-\lambda_2)(z-\lambda_3)$
$\implies -2=\lambda_1\cdot \lambda_2\cdot \lambda_3$
so $p$ has at least one root (call it $\lambda_3$) with modulus $\gt 1$, i.e. at most two roots $\in \overline D$

Now combining with (i.): the topological continuity of a polynomial's roots [equivalently: Hurwitz Theorem] tells us $p$ has at least two roots in $\overline D$, so $p$ has exactly two roots in $\overline D$.

Finally, $p(1)\neq 0$ and $p(-1)\neq 0$ so any roots on the unit circle come in conjugate pairs (since $p$ has real coefficients) and $\lambda_1 \in \partial D\implies -2=\lambda_1\cdot \lambda_2\cdot \lambda_3 = \lambda_1 \cdot \overline \lambda_1 \cdot \lambda_3 = \lambda_3$ but $p(-2)=6\neq 0$ so this is impossible. Conclude $\lambda_1,\lambda_2 \in D$.

Answer 4

Let $a,b,c$ be the roots of the polynomial $P(z)=z^3 + 3z^2 + 2.$

With the use of Descartes rule of signs we know that $P(z)$ has no positive root, and has one negative root. WLOG this root is $a.$ The remaining two roots are complex conjugate, $b=\overline{c}.$

Since

• the polynomial function is increasing (see the answer by lonza leggiera),
• $\lim\limits_{z\to-\infty}P(z)=-\infty\;$ and $\;P(-3)=-27+27+2>0,$

by continuity of the polynomial function, the negative root satisfies $$a<-3.$$

Further, by Vieta's formula $$abc=-2.$$
Therefore $$|c|^2=\overline{c}c=bc=\frac{-2}{a}=\frac{2}{|a|}<\frac{2}{3},$$ which proves the claim.