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By admin 数学代写 2022年2月22日

Theorem $2.13 .5$ Let $z, w \in \mathbb{C}$. Then
$$
|w+z| \leq|w|+|z|,|| z|-| w|| \leq|z-w| .
$$
Proof: First note $|z w|=|z||w|$. Here is why: If $z=x+i y$ and $w=u+i v$, then
$$
\begin{aligned}
&|z w|^{2}=|(x+i y)(u+i v)|^{2}=|x u-y v+i(x v+y u)|^{2} \
=&(x u-y v)^{2}+(x v+y u)^{2}=x^{2} u^{2}+y^{2} v^{2}+x^{2} v^{2}+y^{2} u^{2}
\end{aligned}
$$
Now look at the right side.
$$
|z|^{2}|w|^{2}=(x+i y)(x-i y)(u+i v)(u-i v)=x^{2} u^{2}+y^{2} v^{2}+x^{2} v^{2}+y^{2} u^{2}
$$
the same thing. Thus the rest of the proof goes just as before with real numbers. Using the results of Problem 6 on Page 36 , the following holds.
$$
\begin{aligned}
|z+w|^{2} &=(z+w)(\bar{z}+\bar{w})=z \bar{z}+z \bar{w}+w \bar{z}+w \bar{w} \
&=|z|^{2}+|w|^{2}+z \bar{w}+\overline{\overline{w z z}} \
&=|z|^{2}+|w|^{2}+2 \operatorname{Re} z \bar{w} \
& \leq|z|^{2}+|w|^{2}+2|z \bar{w}|=|z|^{2}+|w|^{2}+2|z||w| \
&=(|z|+|w|)^{2}
\end{aligned}
$$
and so $|z+w| \leq|z|+|w|$ as claimed. The other inequality follows as before.
$$
|z| \leq|z-w|+|w|
$$
and so $|z|-|w| \leq|z-w|=|w-z|$. Now do the same argument switching the roles of $z$ and $w$ to conclude
$$
|z|-|w| \leq|z-w|,|w|-|z| \leq|z-w|
$$
which implies the desired inequality.
Since $\mathbb{R} \subseteq \mathbb{C}$ and the absolute value is consistently defined, the inequality holds also on $\mathbb{R}$.
2.14 Dividing Polynomials
It will be very important to be able to work with polynomials, especially in subjects like linear algebra and with the technique of partial fractions. It is surprising how useful this junior high material will be. In this section, a polynomial is an expression. Later, the expression will be used to define a function. These two ways of looking at a polynomial are very different.
Definition 2.14.1 A polynomial is an expression of the form $p(\lambda)=$
$$
a_{n} \lambda^{\mathrm{n}}+a_{n-1} \lambda^{n-1}+\cdots+a_{1} \lambda+a_{0},
$$
$a_{n} \neq 0$ where the $a_{1}$ are (real or complex) numbers. Two polynomials are equal means that the coefficients match for each power of $\lambda$. The degree of a polynomial is the largest power of $\lambda$. Thus the degree of the above polynomial is $n$. Addition of polynomials is defined in the usual way as is multiplication of two polynomials. The leading term in the above polynomial is $a_{n} \lambda^{n}$. The coefficient of the leading term is called the leading coefficient. It is called a monic polynomial if $a_{n}=1$. A root of a polynomial $p(\lambda)$ is $\mu$ such that $p(\mu)=0$. This is also called a zero.

定理 $2.13 .5$ 令 $z, w \in \mathbb{C}$.然后 $$ |w+z| \leq|w|+|z|,|| z|-| w|| \leq|z-w| . $$ 证明：首先注意$|z w|=|z||w|$。原因如下：如果 $z=x+i y$ 且 $w=u+i v$，那么 $$ \开始{对齐} &|z w|^{2}=|(x+i y)(u+i v)|^{2}=|x u-y v+i(x v+y u)|^{2} \\ =&(x uy v)^{2}+(x v+yu)^{2}=x^{2} u^{2}+y^{2} v^{2}+x^{2} v^{2}+y^{2} u^{2} \end{对齐} $$ 现在看右边。 $$ |z|^{2}|w|^{2}=(x+iy)(xi y)(u+iv)(ui v)=x^{2} u^{2}+y^{2} v^{2}+x^{2} v^{2}+y^{2} u^{2} $$ 同一件事情。因此，其余的证明就像以前一样使用实数。使用第 36 页问题 6 的结果，以下成立。 $$ \开始{对齐} |z+w|^{2} &=(z+w)(\bar{z}+\bar{w})=z \bar{z}+z \bar{w}+w \bar{z} +w \bar{w} \\ &=|z|^{2}+|w|^{2}+z \bar{w}+\overline{\overline{w z z}} \\ &=|z|^{2}+|w|^{2}+2 \operatorname{Re} z \bar{w} \\ & \leq|z|^{2}+|w|^{2}+2|z \bar{w}|=|z|^{2}+|w|^{2}+2|z|| w| \\ &=(|z|+|w|)^{2} \end{对齐} $$ 所以 $|z+w| \leq|z|+|w|$ 如所声称的。另一个不等式如前所述。 $$ |z| \leq|z-w|+|w| $$ 所以 $|z|-|w| \leq|z-w|=|w-z|$。现在做同样的论据切换 $z$ 和 $w$ 的角色来得出结论 $$ |z|-|w| \leq|z-w|,|w|-|z| \leq|z-w| $$ 这意味着所需的不平等。由于 $\mathbb{R} \subseteq \mathbb{C}$ 和绝对值是一致定义的，所以不等式也适用于 $\mathbb{R}$。 2.14 多项式相除能够使用多项式非常重要，特别是在线性代数和部分分数技术等科目中。令人惊讶的是，这种初中材料会有多么有用。在本节中，多项式是一个表达式。稍后，表达式将用于定义函数。这两种查看多项式的方式非常不同。定义 2.14.1 多项式是 $p(\lambda)=$ 形式的表达式 $$ a_{n} \lambda^{\mathrm{n}}+a_{n-1} \lambda^{n-1}+\cdots+a_{1} \lambda+a_{0}， $$ $a_{n} \neq 0$ 其中 $a_{1}$ 是（实数或复数）数字。两个多项式相等意味着系数匹配 $\lambda$ 的每个幂。多项式的次数是 $\lambda$ 的最大幂。因此上述多项式的次数为$n$。多项式的加法以通常的方式定义，就像两个多项式的乘法一样。上述多项式中的首项是$a_{n} \lambda^{n}$。前导项的系数称为前导系数。如果$a_{n}=1$，则称为一元多项式。多项式 $p(\lambda)$ 的根是 $\mu$ 使得 $p(\mu)=0$。这也称为零。