Scroll Top
19th Ave New York, NY 95822, USA

数学竞赛代写|Stanford大学数学夏令营SUMaC代写

Multiplication and addition are defined in the most obvious way subject to the convention that $i^{2}=-1$. Thus,
$$
(a+i b)+(c+i d)=(a+c)+i(b+d)
$$
and
$$
(a+i b)(c+i d)=a c+i a d+i b c+i^{2} b d=(a c-b d)+i(b c+a d)
$$
Every non zero complex number, $a+i b$, with $a^{2}+b^{2} \neq 0$, has a unique multiplicative inverse.
$$
\frac{1}{a+i b}=\frac{a-i b}{a^{2}+b^{2}}=\frac{a}{a^{2}+b^{2}}-i \frac{b}{a^{2}+b^{2}}
$$
You should prove the following theorem.
Theorem 2.13.1 The complex numbers with multiplication and addition defined as above form a field satisfying all the field axioms listed on Page 9 .
The field of complex numbers is denoted as $\mathrm{C}$. An important construction regarding complex numbers is the complex conjugate denoted by a horizontal line above the number. It is defined as follows.
$$
\overline{a+i b} \equiv a-i b
$$
What it does is reflect a given complex number across the $x$ axis. Algebraically, the following formula is easy to obtain.
$$
(\overline{a+i b})(a+i b)=a^{2}+b^{2} .
$$
Definition 2.13.2 Define the absolute value of a complex number as follows.
$$
|a+i b| \equiv \sqrt{a^{2}+b^{2}}
$$
Thus, denoting by $z$ the complex number $z=a+i b$,
$$
|z|=(z \bar{z})^{1 / 2}
$$
Be sure to verify the last claim in this definition. With this definition, it is important to note the following. Be sure to verify this. It is not too hard but you need to do it.
Remark $2.13 .3:$ Let $z=a+i b$ and $w=c+i d .$ Then $|z-w|=\sqrt{(a-c)^{2}+(b-d)^{2}}$. Thus the distance between the point in the plane determined by the ordered pair, $(a, b)$ and the ordered pair $(c, d)$ equals $|z-w|$ where $z$ and $w$ are as just described.
For example, consider the distance between $(2,5)$ and $(1,8)$. From the distance formula which you should have seen in either algebra or calculus, this distance is defined ass
$$
\sqrt{(2-1)^{2}+(5-8)^{2}}=\sqrt{10}
$$
On the other hand, letting $z=2+i 5$ and $w=1+i 8, z-w=1-i 3$ and so
$$
(z-w)(\overline{z-w})=(1-i 3)(1+i 3)=10
$$
so $|z-w|=\sqrt{10}$, the same thing obtained with the distance formula.
Notation 2.13.4 From now on I will sometimes use the symbol $\mathrm{F}$ to denote either $\mathbb{C}$ or $\mathbb{R}$, ruther than fussing over which one is meant becuuse it often does not make any difference.
乘法和加法以最明显的方式定义,遵循 $i^{2}=-1$ 的约定。因此, $$ (a+i b)+(c+i d)=(a+c)+i(b+d) $$ 和 $$ (a+i b)(c+i d)=a c+i a d+i b c+i^{2} b d=(a c-b d)+i(b c+a d) $$ 每个具有 $a^{2}+b^{2} \neq 0$ 的非零复数 $a+i b$ 都有一个唯一的乘法逆元。 $$ \frac{1}{a+ib}=\frac{ai b}{a^{2}+b^{2}}=\frac{a}{a^{2}+b^{2}}-我 \frac{b}{a^{2}+b^{2}} $$ 你应该证明以下定理。 定理 2.13.1 上面定义的具有乘法和加法的复数形成一个满足第 9 页列出的所有域公理的域。 复数域记为$\mathrm{C}$。关于复数的一个重要结构是复共轭,由数字上方的水平线表示。定义如下。 $$ \overline{a+i b} \equiv a-i b $$ 它的作用是在 $x$ 轴上反映给定的复数。代数上,下面的公式很容易得到。 $$ (\overline{a+i b})(a+i b)=a^{2}+b^{2} 。 $$ 定义 2.13.2 定义复数的绝对值如下。 $$ |a+ib| \equiv \sqrt{a^{2}+b^{2}} $$ 因此,用 $z$ 表示复数 $z=a+i b$, $$ |z|=(z \bar{z})^{1 / 2} $$ 请务必验证此定义中的最后一个声明。有了这个定义,重要的是要注意以下几点。请务必验证这一点。这不是太难,但你需要这样做。 备注 $2.13 .3:$ 令 $z=a+i b$ 且 $w=c+i d .$ 则 $|z-w|=\sqrt{(a-c)^{2}+(b-d)^{2}}$。因此,平面中由有序对 $(a, b)$ 和有序对 $(c, d)$ 确定的点之间的距离等于 $|zw|$ 其中 $z$ 和 $w$ 是一样的描述。 例如,考虑 $(2,5)$ 和 $(1,8)$ 之间的距离。从你应该在代数或微积分中看到的距离公式中,这个距离被定义为 $$ \sqrt{(2-1)^{2}+(5-8)^{2}}=\sqrt{10} $$ 另一方面,让 $z=2+i 5$ 和 $w=1+i 8, z-w=1-i 3$ 等等 $$ (z-w)(\overline{z-w})=(1-i 3)(1+i 3)=10 $$ 所以 $|z-w|=\sqrt{10}$,用距离公式得到的结果是一样的。 符号 2.13.4 从现在开始,我有时会使用符号 $\mathrm{F}$ 来表示 $\mathbb{C}$ 或 $\mathbb{R}$,而不是大惊小怪,因为它经常没有任何区别。

real analysis代写analysis 2, analysis 3请认准UprivateTA™. UprivateTA™为您的留学生涯保驾护航。

抽象代数Galois理论代写

偏微分方程代写成功案例

代数数论代考

组合数学代考

统计作业代写

集合论数理逻辑代写案例

凸优化代写

统计exam代考

Related Posts

Leave a comment