Find the number of roots of the polynomial $P(z)=z^5+2z^3+3$ in the closed unit disk $\{z\colon |z|\le 1\}$.

Question

Find the number of roots of the polynomial $P(z)=z^5+2z^3+3$ in the closed unit disk $\{z\colon |z|\le 1\}$.

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My solution:

Set $f(z)=3$.

For every $0<\delta<1$ and $|z|=\delta$, we have $|P(z)-f(z)|\le \delta^5+2\delta^3<3\delta<3=|f(z)|$. Hence $P(z)$ and $f(z)=3$ has the same number of zeros, which is $0$, inside the open unit disk.

Now consider the behavior of $P(z)$ when $|z|=1$. If we have $P(z)=0$, then $z^5=-2z^3-3$, and $1=|z^5|=|-2z^3-3|$ when $|z|=1$ which forces $-2z^3=2$ and $z^3=-1$.

Clearly, $z=-1$ is a zero of $P(z)$.

For the other two roots of $z^3=-1$, note that: $$ P(z)=z^2\cdot z^3+2z^3+3=z^2-2+3=z^2+1=0\implies z=\pm 1 $$

Therefore we conclude that there is only one root, $z=-1$, of $P(z)$ in the closed unit disk.

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However, I am asking for other solutions, such as applying the symmetric version of Rouché''s theorem like in this post. Thank you.

Answer 1

Let $z^5+2z^3+3=0$ with $|z|\le 1. $ Then $3=|-3|=|z^5+2z^3|=|z|^3\cdot |z^2+2|\le |z^2+2|\le |z|^2+|2|=|z|^2+2\le 3.$

So $|z|=1$. So let $z=e^{it}$ with $t\in \Bbb R.$ Then $-3=Re (e^{5it}+2e^{3it})=\cos 5t+2\cos 3t.$

Since $\cos 5t\ge -1$ and $2\cos 3t\ge -2,$ therefore $\cos 5t=\cos 3t=-1$. This implies $z^5=z^3=-1,$ hence $z^2=1,$ hence $z=\pm 1.$

So with $P$ standing for $(\,|z|\le 1\land z^5+2z^3+3=0\,)$ we have $$P\implies (z\in \{-1,1\}\land z^5+2z^3+3=0)\implies (z=-1)\implies P.$$

Answer 2

$f(z)=z^5+2z^3+3 \Rightarrow f'(z)=5z^4+6z^2>0 \Rightarrow f(z)$ is an increasing function. So $f(z)=0$ can have at most oone real root and it is $z=-1.$

Take $$-3=z^5+2z^3 \Rightarrow 3=|z^5+2z^3|\le |z|^5+2|z|^3 \le 3.$$ This proves that at least one root of this equation is s.t. |z|=1. Other roots cannot not satisfy $|z|<1$, so they will lie outside the unit disk i.e.,$z>1$

In fact, one can numerically check that apart from -1, this equation has two pairs non-real complex conjugate roots roots such that $|z|>1.$

Answer 3

Note that on $|z| < 1$ there are no roots as $|z^5+2z^3| < 3$ and so the roots of $z^5+2z^3+3$ on $|z| \leq 1$ lie on the boundary $|z| = 1$. Let $z = cos(\theta) + isin(\theta)$ satisfy the polynomial relation $P(z)$ stated above.

We must have $|(e^{i\theta})^5 + 2(e^{i\theta})^3|^2 = 9$ and hence $|e^{i2\theta}+2|^2 = 9$ which is exactly when $\theta \equiv 0$ $ mod(\pi)$ so there are atmost two roots of $P(z)$ in $|z| = 1$ at $z=1$ and $z=-1$; but only $z = - 1$ is a root. Your conclusion should be correct.