### 数学代写|复变函数作业代写Complex function代考|Fundamental Concepts

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• Foundations of Data Science 数据科学基础

## 数学代写|复变函数作业代写Complex function代考|Elementary Properties of the Complex Numbers

We take for granted the real numbers, which will be denoted by the symbol $\mathbb{R}$. Then we set $\mathbb{R}^{2}={(x, y): x \in \mathbb{R}, y \in \mathbb{R}}$. The complex numbers $\mathbb{C}$ consist of $\mathbb{R}^{2}$ equipped with some special algebraic operations. Namely, one defines
\begin{aligned} (x, y)+\left(x^{\prime}, y^{\prime}\right) &=\left(x+x^{\prime}, y+y^{\prime}\right) \ (x, y) \cdot\left(x^{\prime}, y^{\prime}\right) &=\left(x x^{\prime}-y y^{\prime}, x y^{\prime}+y x^{\prime}\right) . \end{aligned}
You can check for yourself that these operations of $+$ and $\cdot$ are commutative and associative.

It is both conventional and convenient to denote $(1,0)$ by 1 and $(0,1)$ by $i$. We also adopt the convention that, if $\alpha \in \mathbb{R}$, then
$$\alpha \cdot(x, y)=(\alpha, 0) \cdot(x, y)=(\alpha x, \alpha y) .$$
Then every complex number $(x, y)$ can be written in one and only one way in the form $x \cdot 1+y \cdot i$ with $x, y \in \mathbb{R}$. We usually write the number even more succinctly as $x+i y$. Then our laws of addition and multiplication become
\begin{aligned} (x+i y)+\left(x^{\prime}+i y^{\prime}\right) &=\left(x+x^{\prime}\right)+i\left(y+y^{\prime}\right), \ (x+i y) \cdot\left(x^{\prime}+i y^{\prime}\right) &=\left(x x^{\prime}-y y^{\prime}\right)+i\left(x y^{\prime}+y x^{\prime}\right) . \end{aligned}
Observe that $i \cdot i=-1$. Moreover, our multiplication law is consistent with the real multiplication introduced in line $(*)$.

The symbols $z, w, \zeta$ are frequently used to denote complex numbers. Unless it is explicitly stated otherwise, we always take $z=x+i y, w=$ $u+i v, \zeta=\xi+i \eta$. The real number $x$ is called the real part of $z$ and is

written $x=\operatorname{Re} z$. Likewise $y$ is called the imaginary part of $z$ and is written $y=\operatorname{Im} z$.

The complex number $x-i y$ is by definition the conjugate of the complex number $x+i y$. We denote the conjugate of a complex number $z$ by the symbol $\bar{z}$. So if $z=x+i y$, then $\bar{z}=x-i y$.

Notice that $z+\bar{z}=2 x, z-\bar{z}=2 i y$. You should verify for yourself that
\begin{aligned} \overline{z+w} &=\bar{z}+\bar{w} \ \overline{z \cdot w} &=\bar{z} \cdot \bar{w} \end{aligned}
A complex number is real (has no imaginary part) if and only if $z=\bar{z}$. It is imaginary (has zero real part) if $z=-\bar{z}$.

The ordinary Euclidean distance of $(x, y)$ to $(0,0)$ is $\sqrt{x^{2}+y^{2}}$. We also call this number the modulus (or absolute value) of the complex number $z=x+i y$, and we write $|z|=\sqrt{x^{2}+y^{2}}$. Notice that
$$z \cdot \bar{z}=x^{2}+y^{2}=|z|^{2} .$$
You should check for yourself that the distance from $z$ to $w$ is $|z-w|$. Verify also that $|z \cdot w|=|z| \cdot|w|$ (square both sides). Also $|\operatorname{Re} z| \leq|z|$ and $|\operatorname{Im} z| \leq|z|$.

Let $0=0+i 0$. If $z \in \mathbb{C}$, then $z+0=z$. Also, letting $-z=-x-i y$, we notice that $z+(-z)=0$. So every complex number has anditive inverse.
Since $1=1+i 0$, it follows that $1 \cdot z=z \cdot 1=z$ for every complex number $z$. If $|z| \neq 0$, then $|z|^{2} \neq 0$ and
$$z \cdot \frac{\bar{z}}{|z|^{2}}=\frac{|z|^{2}}{|z|^{2}}=1 .$$
So every nonzero complex number has a multiplicative inverse. It is natural to define $1 / z$ to be the multiplicative inverse $\bar{z} /|z|^{2}$ of $z$ and, more generally, to define
$$\frac{z}{w}=z \cdot \frac{1}{w}=\frac{z \bar{w}}{|w|^{2}} \quad \text { for } w \neq 0 .$$
You can also see that $\overline{z / w}=\bar{z} / \bar{w}$.
Observe now that multiplication and addition satisfy the usual distributive, associative, and commutative (as previously noted) laws. So $\mathbb{C}$ is a field. It follows from general properties of fields, or you can just check directly, that every complex number has a unique additive inverse and every nonzero complex number has a unique multiplicative inverse. Also $\mathbb{C}$ contains a copy of the real numbers in an obvious way:
$$\mathbb{R} \ni x \mapsto x+i 0 \in \mathbb{C} .$$

## 数学代写|复变函数作业代写Complex function代考|Further Properties of the Complex Numbers

We first consider the complex exponential, which we define as follows:
(1) If $z=x$ is real, then
$$e^{z}=e^{x} \equiv \sum_{j=0}^{\infty} \frac{x^{j}}{j !}$$
as in calculus.
(2) If $z=i y$ is pure imaginary, then
$$e^{z}=e^{i y} \equiv \cos y+i \sin y$$
(3) If $z=x+i y$, then
$$e^{z}=e^{x+i y} \equiv e^{x} \cdot(\cos y+i \sin y)$$
Parts (2) and (3) of the definition, due to Euler, may seem somewhat arbitrary. We shall now show, using power series, that these definitions are

perfectly natural. We shall wait until Section $3.2$ to give a careful presentation of the theory of complex power series. So the power series arguments that we are about to present should be considered purely formal and given primarily for motivation.
Since, as was noted in (1), we have
$$e^{x}=\sum_{j=0}^{\infty} \frac{x^{j}}{j !}$$
then it is natural to attempt to define
$$e^{z}=\sum_{j=0}^{\infty} \frac{z^{j}}{j !} .$$
If we assume that this series converges in some reasonable sense and that it can be manipulated like the real power series with which we are familiar, then we can proceed as follows:
If $z=i y, y \in \mathbb{R}$, then
\begin{aligned} e^{i y}=& \sum_{j=0}^{\infty} \frac{(i y)^{j}}{j !} \ =& 1+i y-\frac{y^{2}}{2 !}-\frac{i y^{3}}{3 !}+\frac{y^{4}}{4 !} \ \quad & \quad+\frac{i y^{5}}{5 !}-\frac{y^{6}}{6 !}-\frac{i y^{7}}{7 !}+\frac{y^{8}}{8 !}+\cdots \ =&\left(1-\frac{y^{2}}{2 !}+\frac{y^{4}}{4 !}-\frac{y^{6}}{6 !}+\frac{y^{8}}{8 !}-\cdots\right) \ =& \quad \cos y+i \sin y \end{aligned}
By formal manipulation of series, it is now easily checked, using the definition (*), that
$$e^{a+b}=e^{a} e^{b}, \quad \text { any } a, b \in \mathbb{C} .$$
Then for $z=x+i y$ we have
\begin{aligned} e^{z} &=e^{x+i y}=e^{x} e^{i y} \ &=e^{x}(\cos y+i \sin y) \end{aligned}
giving thus a formal “demonstration” of our definition of exponential.
To stress that there is no circular reasoning involved here, we reiterate that the definition of the complex exponential is that, for $z=x+i y$,
$$e^{z}=e^{x}(\cos y+i \sin y) .$$

## 数学代写|复变函数作业代写Complex function代考|Complex Polynomials

In the calculus of real variables, polynomials are the simplest nontrivial functions. The purpose of this section is to consider complex-valued polynomials of a complex variable, with the idea of seeing what new features appear. Later we shall use the discussion as motivation for considering more general functions.

There are several slightly different ways of looking at polynomials from the complex viewpoint. One way is to consider polynomials in $x$ and $y$, $(x, y) \in \mathbb{R}^{2}$, with complex coefficients: for example, $(2+i) x y+3 i y^{2}+5 x^{2}$. Such polynomials give functions from $\mathbb{R}^{2}$ to $\mathbb{C}$, which we could equally well think of as functions from $\mathbb{C}$ to $\mathbb{C}$, with $(x, y)$ determined by $z=x+$ iy. Another kind of polynomial that we can consider is complex-coefficient polynomials in the complex variable $z$, for example, $i+(3+i) z+5 z^{2}$. These also give functions from $\mathbb{C}$ to $\mathbb{C}$. A polynomial in $z$ gives rise naturally to a polynomial in $x$ and $y$ by substituting $z=x+i y$ and expanding. For instance
\begin{aligned} i+(3+i) z+5 z^{2} &=i+(3+i)(x+i y)+5(x+i y)^{2} \ &=i+3 x-y+i x+3 i y+5 x^{2}+10 i x y-5 y^{2} \ &=i+(3+i) x+(3 i-1) y+5 x^{2}+(10 i) x y-5 y^{2} \end{aligned}
It is an important and somewhat surprising fact that the converse of this expansion process does not always work: there are many polynomials in $x$ and $y$ that cannot be written as polynomials in $z$. Let us consider a specific simple example: the polynomial $x$ itself. If it were true that
$$x=P(z)=P(x+i y)$$
for some polynomial $P(z)$ in $z$, then $P$ would have to be of first degree. But a first degree polynomial $a z+b=a x+i a y+b$ cannot be identically equal to $x$, no matter how we choose $a$ and $b$ in $\mathbb{C}$ (see Exercise 35 ). What is really going on here?

One way to write a polynomial in $x$ and $y$ in complex notation is to use the substitutions
$$x=\frac{z+\bar{z}}{2}, y=\frac{z-\bar{z}}{2 i},$$
where $\bar{z}=x-i y$ as in Section 1.1. The point of the previous paragraph is that, when a polynomial in $x, y$ is converted to the $z, \bar{z}$ notation, then there will usually be some $\bar{z}$ ‘s in the resulting expression, and these $\bar{z}$ ‘s may not cancel out.
For example,
$$x^{2}+y^{2}=\left(\frac{z+\bar{z}}{2}\right)^{2}+\left(\frac{z-\bar{z}}{2 i}\right)^{2}$$

\begin{aligned} &=\frac{z^{2}}{4}+\frac{z \bar{z}}{2}+\frac{\bar{z}^{2}}{4}-\frac{z^{2}}{4}+\frac{z \bar{z}}{2}-\frac{\bar{z}^{2}}{4} \ &=z \cdot \bar{z} \end{aligned}
You can check for yourself that there is no polynomial expression in $z$, without any $\bar{z}$ ‘s, that equals $x^{2}+y^{2}$ : the occurrence of $\bar{z}$ is required.
Of course, sometimes one can be lucky and there will not be any $\bar{z}$ ‘s:
\begin{aligned} x^{2}-y^{2}+2 i x y &=\left(\frac{z+\bar{z}}{2}\right)^{2}-\left(\frac{z-\bar{z}}{2 i}\right)^{2}+2 i\left(\frac{z+\bar{z}}{2}\right)\left(\frac{z-\bar{z}}{2 i}\right) \ &=\frac{z^{2}}{4}+\frac{z \bar{z}}{2}+\frac{\bar{z}^{2}}{4}+\frac{z^{2}}{4}-\frac{z \bar{z}}{2}+\frac{\bar{z}^{2}}{4}+\frac{2 i\left(z^{2}-\bar{z}^{2}\right)}{2 \cdot 2 i} \ &=z^{2} . \end{aligned}
In this example, all the $\bar{z}$ terms cancel out.

## 数学代写|复变函数作业代写Complex function代考|Elementary Properties of the Complex Numbers

(X,是)+(X′,是′)=(X+X′,是+是′) (X,是)⋅(X′,是′)=(XX′−是是′,X是′+是X′).

(X+一世是)+(X′+一世是′)=(X+X′)+一世(是+是′), (X+一世是)⋅(X′+一世是′)=(XX′−是是′)+一世(X是′+是X′).

R∋X↦X+一世0∈C.

## 数学代写|复变函数作业代写Complex function代考|Further Properties of the Complex Numbers

(1) 如果和=X是真实的，那么

(2) 如果和=一世是是纯虚数，那么

(3) 如果和=X+一世是， 然后

## 数学代写|复变函数作业代写Complex function代考|Complex Polynomials

X=磷(和)=磷(X+一世是)

X=和+和¯2,是=和−和¯2一世,

X2+是2=(和+和¯2)2+(和−和¯2一世)2=和24+和和¯2+和¯24−和24+和和¯2−和¯24 =和⋅和¯

X2−是2+2一世X是=(和+和¯2)2−(和−和¯2一世)2+2一世(和+和¯2)(和−和¯2一世) =和24+和和¯2+和¯24+和24−和和¯2+和¯24+2一世(和2−和¯2)2⋅2一世 =和2.

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