题目

Problem

\begin{align*} &\,f(z)\\ =&\,z^4+\lambda z^3+14z^2\\ &\,\hspace{2pt}+9\lambda z+45 \end{align*}

where $\lambda\in\mathbb{R}$ .

(a) Show that $f(3\mathrm{i}) = 0$ .

(2)

(b) Hence determine the exact range of values of $\lambda$ for which the equation $f(z) = 0$ has no real solutions.

(5)

题目中文翻译

\begin{align*} &\,f(z)\\ =&\,z^4+\lambda z^3+14z^2\\ &\,\hspace{2pt}+9\lambda z+45 \end{align*}

其中 $\lambda\in\mathbb{R}$ 。

(a) 证明 $f(3\mathrm{i}) = 0$ 。

(2)

(b) 由此确定 $\lambda$ 的精确取值范围，使得方程 $f(z) = 0$ 没有实数解。

(5)

解答

展开

直接把 $z=3\mathrm{i}$ 代入。利用 $\mathrm{i}^2=-1$ 、 $\mathrm{i}^3=-\mathrm{i}$ 、 $\mathrm{i}^4=1$ 化简，可以看到实数项与虚数项分别抵消。

展开

Substituting $z=3\mathrm{i}$ ,

\begin{align*} &\,f(3\mathrm{i})\\ =&\,(3\mathrm{i})^4+\lambda(3\mathrm{i})^3\\ &\,\hspace{2pt}+14(3\mathrm{i})^2\\ &\,\hspace{4pt}+9\lambda(3\mathrm{i})+45\\ =&\,81-27\lambda\mathrm{i}-126\\ &\,\hspace{2pt}+27\lambda\mathrm{i}+45\\ =&\,0, \end{align*}

as required.

展开

由于多项式系数为实数，(a) 的根 $3\mathrm{i}$ 会带来共轭根 $-3\mathrm{i}$ ，所以 $z^2+9$ 是因式。分解后只需令剩余二次因式没有实根，即判别式严格小于零；等号端点会产生重实根，因此不能包含。

展开

Since $f$ has real coefficients and $3\mathrm{i}$ is a root, $-3\mathrm{i}$ is also a root. Hence,

\begin{align*} &\,(z-3\mathrm{i})(z+3\mathrm{i})\\ =&\,z^2+9 \end{align*}

is a factor of $f(z)$ . Since $f$ is monic, write the other quadratic factor as $z^2+pz+q$ . Expanding gives

\begin{align*} &\,(z^2+9)(z^2+pz+q)\\ =&\,z^4+pz^3+qz^2\\ &\,\hspace{2pt}+9z^2+9pz+9q\\ =&\,z^4+pz^3+(q+9)z^2\\ &\,\hspace{2pt}+9pz+9q. \end{align*}

Comparing coefficients with $f(z)$ ,

\begin{align*} p=&\,\lambda,\\ q+9=&\,14,\\ 9p=&\,9\lambda,\\ 9q=&\,45. \end{align*}

Thus $p=\lambda$ and $q=5$ , giving

\begin{align*} &\,f(z)\\ =&\,(z^2+9)(z^2+\lambda z+5)\\ =&\,z^4+\lambda z^3+5z^2\\ &\,\hspace{2pt}+9z^2+9\lambda z+45\\ =&\,z^4+\lambda z^3+14z^2\\ &\,\hspace{2pt}+9\lambda z+45. \end{align*}

The factor $z^2+9$ has no real roots, so $f(z)=0$ has no real solutions precisely when $z^2+\lambda z+5=0$ has no real roots. Thus,

\begin{align*} \lambda^2-4(1)(5)<&\,0\\ \lambda^2<&\,20. \end{align*}

Therefore,

\boxed{-2\sqrt5<\lambda<2\sqrt5}.