# Roots of a complex polynomial in the unit disc

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 $$z^3: x^3 + 3 i x^2 y - 3 xy^2 - i y^3$$
• This feels like it would involve Rouché's theorem. May 24 at 10:48
• Expanding by $z = x + iy$ could work, since the real and imaginary parts have to separately equal 0. Also, for unit disc, magnitude of complex number should be $\leq 1$. So if you use $z = r e^{i\theta}$, maybe you can show that $r \leq 1$ for the two roots, thereby proving they lie within the disc. The question wording feels vague whether the third root is multiplicity, or would lie outside the unit disc. May 24 at 11:18

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\ -20& for \ 0 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?

• My first thought was “Rouché!” but this is also an elegant argument (+1). May 24 at 11:35
• Thank you. Why is it that we just can take the derivative of the function? May 24 at 11:55
• @user1183561 I'm not sure I understand what you're asking. All polynomials are everywhere differentiable, so you can always take derivatives of them if you need to. The reason why I took the derivative of this polynomial is because it provides you with information that enables you to answer your question. May 24 at 13:08
• @user1183561 Maybe you're asking what the derivative $f'$ means when $f:\mathbb C\to\mathbb C$, and you're asking the very reasonable question whether it's still true that $f(x+\epsilon)\approx f(x)+\epsilon f'(x)$ when things are complex, and whether the rules of differentiation are the same. They are. May 24 at 20:47

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 $$p$$.

• Typos? Should the middle term in $p(z)$ be $3z^2$ like in the question, and should the first factor in the definition of $q(z)$ be $(z+3)$ so that all terms become "plus" when it is expanded? Is a square missing in $|z + \frac 23|$? May 24 at 19:26
• @JeppeStigNielsen: Yes indeed, thanks. May 24 at 20:04

(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 $$\lambda_1,\lambda_2 \in D$$.

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.