# Định lý Cauchy-Riemann và áp dụng

Đây là bài học cho Sinh viên ĐH TĐT mà tôi tham gia giảng dạy. Nội dung viết lên đây chỉ là những điều cơ bản thiên về thực hành cho các sinh viên theo học môn này mà không phải theo chuyên ngành toán học. Lưu ý có rất nhiều định nghĩa về một khái niệm, chúng tôi liệt kê để tham khảo. Về phía mình, tôi sẽ chọn một định nghĩa phù hợp để dạy. Đương nhiên các định nghĩa khác đều sẽ tương đương.

Let $\Omega$ be an open set in $\mathbb{C}$ and $f$ a complex-valued function on $\Omega$. The function $f$ is holomorphic at the point $z_0 \in \Omega$ if the quotient

$\displaystyle \frac{f(z_0+h)-f(z_0)}{h}$

converges to a limit when $h \to 0$. Here $h \in \mathbb{C}$ and $h \ne 0$ with $z_0 + h \in \Omega$,so that the quotient is well defined. The limit of the quotient, when it
exists, is denoted by $f'(z_0)$, and is called the derivative of $f$ at $z_0$:

$f'(z_0) = \displaystyle\lim_{z\to z_0}\frac{f(z_0+h)-f(z_0)}{h}$

It should be emphasized that in the above limit, $h$ is a complex number
that may approach $0$ from any direction.
The function $f$ is said to be holomorphic on $\Omega$ if $f$ is holomorphic at every point of $\Omega$. If $\mathbf{C}$ is a closed subset of $\mathbb{C}$, we say that $f$ is holomorphic on $\mathbf{C}$ if $f$ is holomorphic in some open set containing $\mathbf{C}$.
Finally, if f is holomorphic in all of $\mathbb{C}$ we say that $f$ is entire.

Example 1. The function $f(z) = z$ is holomorphic on any open set in
$\mathbb{C}$, and $f'(z) = 1$. In fact, any polynomial
$p(z)=a_0+a_1z+\dots +a_nz^n$ is holomorphic in the entire complex plane and $p'(z)=a_1+2a_2z+\dots +na_nz^{n-1}$.

This follows from Proposition 2.2 below.
Example 2. The function $f(z)=\displaystyle\frac{1}{z}$ is holomorphic on any open set in $\mathbb{C}$ that does not contain the origin, and $f'(z) = -\displaystyle\frac{1}{z^2}$.
Example 3. The function $f(z) = \overline{z}$ is not holomorphic. Indeed, we have
$\displaystyle\frac{f(z_0 + h) - f(z_0)}{h}= \frac{\overline{h}}{ h}$
which has no limit as $h \to 0$ as one can see by first taking $h$ real and then $h$ purely imaginary.

#### The Cauchy-Riemann Equations

Let $f : D \to C$ be a holomorphic function defined over an open set $D$ in the complex plane, and let $\widetilde{D}$ denote the open set in $\mathbb{R}^2$ defined by $\widetilde{D} = \{(x, y) \in \mathbb{R}^2 : x + iy \in D\}$.
Then the holomorphic function $f$ on $D$ determines differentiable real-valued functions $u$ and $v$ on $D$ such that $f (x + iy) = u(x, y) + iv(x, y)$ for all  $(x, y) \in D$. Now if $g$ is a function of a complex variable, defined in the neighbourhood of zero, then $\displaystyle\lim_{h\to 0} g(h)$, if it exists, has the same value
whether $h$ tends to $0$ along the real axis or along the imaginary axis. It follows that if $\displaystyle\lim_{h \to 0} g(h)$ exists then

$\displaystyle\lim_{h \to 0} g(h) = \lim_{t \to 0} g(t) = \lim_{t \to 0} g(it)$,

where $t$ tends to zero through real values only.

On applying this principle to the holomorphic function $f$ , we find that the limit

$f'(z) = \displaystyle\lim_{h\to 0}\frac{f(z+h)-f(z)}{h}$

$= \displaystyle\lim_{h\to 0}\frac{u(x+h,y)+iv(x+h,y)-u(x,y)-iv(x,y)}{h}$

$= \displaystyle\lim_{h\to 0}\frac{u(x+h,y)-u(x,y)+i[v(x+h,y)-v(x,y)]}{h}$

$= \displaystyle\frac{\partial u(x,y)}{\partial x }+ i\frac{\partial v(x,y)}{\partial x }$

and also

$f'(z) = \displaystyle\lim_{h\to 0}\frac{f(z+h)-f(z)}{ih}$

$= \displaystyle\lim_{h\to 0}\frac{u(x,y+h)+iv(x,y+h)-u(x,y)-iv(x,y)}{ih}$

$= \displaystyle\lim_{h\to 0}\frac{v(x,y+h)-v(x,y)-i[u(x,y+h)-u(x,y)]}{h}$

$= \displaystyle\frac{\partial v(x,y)}{\partial y }- i\frac{\partial u(x,y)}{\partial y }$

It follows that the functions $u$ and $v$ must satisfy the partial differential equations

$\displaystyle\frac{\partial u}{\partial x}=\frac{\partial v}{\partial y}\quad , \quad \frac{\partial u}{\partial y}=-\frac{\partial v}{\partial x}$

These equations are referred to as the Cauchy-Riemann equations. Thus to
each holomorphic function f there corresponds a pair of differentiable real-
valued functions u and v, defined over an open subset of $\mathbb{R}^2$ and satisfying the above system of partial differential equations.

The converse is true (provided that the partial derivatives of the functions $u$ and $v$ are continuous).

Vấn đề 1: Tìm hàm giải tích khi biết trước hàm phần thực hoặc hàm phần ảo.

Cho hàm $w = f (z) = u ( x, y) + iv( x, y)$ là giải tích trong miền đơn liên $G$. Phần thực $u(x, y)$ và phần ảo $v(x, y)$ là những hàm điều hoà trong $G$, nghĩa là chúng thoả mãn phương trình:

$\Delta u = \displaystyle\dfrac{\partial^2u}{\partial x^2}+\dfrac{\partial^2u}{\partial y ^2} = 0\quad ; \quad \Delta v =\dfrac{\partial^2v}{\partial x^2}+\dfrac{\partial^2v}{\partial y ^2} = 0$

Ngược lại nếu $u(x, y)$$v(x, y)$ là những hàm điều hoà  thoả mãn điều kiện Cauchy-Riemann thì $w = f (z) = u ( x, y) + iv( x, y)$ là giải tích.

Exercise 1: The function u(x, y) = x2y 2x is the real part of an analytic function. What is the imaginary part (the conjugate harmonic function of
u(x, y)) of this analytic function?

Solution:

u(x, y) = x2y 2xf = u + iv is analytic, v = ?

$\displaystyle \frac{\partial v}{\partial y} = \frac{\partial u}{\partial x} = 2x-1 \Longrightarrow v(x,y)=2xy - y +C(x)$

$\displaystyle \frac{\partial v}{\partial x} = -\frac{\partial u}{\partial y} \Longleftrightarrow 2y+ C'(x) = 2y \Longrightarrow C'(x) = 0 \Longrightarrow C(x)=C$

v = 2xyy + C

Exercise 2:  At which points z the function

$f(z)=\displaystyle\frac{x}{x^2+y^2} -i \frac{y}{x^2+y^2}$

has a derivative? Calculate f ‘(z ).

Solution:

$f(z)=\displaystyle\frac{x}{x^2+y^2} -i \frac{y}{x^2+y^2}$

$\displaystyle\frac{\partial u}{\partial x} = \frac{y^2-x^2}{(x^2+y^2)^2} = \frac{\partial v}{\partial y}$

$\displaystyle\frac{\partial u}{\partial y} = \frac{2xy}{(x^2+y^2)^2} = -\frac{\partial v}{\partial x}\quad , x^2+y^2\ne 0$

$f'(z)=\displaystyle\frac{\partial u}{\partial x} + i \frac{\partial v}{\partial x} = \frac{y^2-x^2}{(x^2+y^2)^2} + i\frac{2xy}{(x^2+y^2)^2} \quad , x^2+y^2\ne 0$

Ví dụ 3: Cho hàm $u(x,y) = \ln(x^2 + y^2 )$ . Tìm hàm giải tích $f(z)$ nhận $u(x,y)$ làm phần thực.

Giải:

$\displaystyle \frac{\partial v}{\partial y} = \frac{\partial u}{\partial x} = \frac{2x}{x^2+y^2} \Longrightarrow v(x,y)=2\arctan\frac{y}{x} +C(x)$

$\displaystyle \frac{\partial v}{\partial x} = -\frac{\partial u}{\partial y} \Longleftrightarrow \frac{-2y}{x^2+y^2}+C'(x)=\frac{-2y}{x^2+y^2}$

$C'(x)=0 \Longrightarrow C(x)=C$.

Vậy: $f(z)= \ln(x^2 + y^2 ) + i(2\arctan\displaystyle\frac{y}{x} +C)$

### Bài tập

Bài 1: Chứng tỏ các hàm sau là hàm điều hòa và tìm hàm giải tích $f (z ) = u + iv$ theo biến $z$ , biết

1. $u(x,y)=x^2y-\displaystyle\frac{1}{3}y^3\quad$$\quad f(-i)=\displaystyle\frac{i}{2}$
2. $v(x,y)=x^3- 3xy^2\quad$$\quad f(1+i)=-i$
3. $v(x,y)=e^y\sin x + 2x$
4. $u(x,y)=e^x\cos y - y$
Exercise 2: Show that the following functions are harmonic and find harmonic conjugates:
1. $x^2-y^2$
2. $xy +3x^2y-y^3$
3. $\arctan\displaystyle\left(\frac{y}{x}\right) , x > 0$
4. $\displaystyle\frac{x}{x^2+y^2}$
5. $e^{x^2-y^2}\cos (2xy)$

Exercise 3: Show that u is  harmonic and find a harmonic conjugate v when:

$u(x,y)=2x-2xy$ Ans: $v(x,y)=2y-y^2+x^2+C$

Vấn đề 2: Xét tính khả vi của hàm phức bằng định lý Cauchy – Riemann.

Exercise 1: Which of the following functions are analytic in the complex plane?

a) $f (z) = \overline{z}^2$,
b) $f (z ) = x^3- 3xy^2+ i(3x^2y -y^3 )$,
c) $f ( z) = |z|$.

Solution: f (z) = u(x, y) + iv (x, y) is analytic in the complex plane, if the 1st order partial derivatives of u and v are continuous and satisfy the Cauchy-Riemann equations

$\displaystyle\frac{\partial u}{\partial x}=\frac{\partial v}{\partial y}\quad , \quad \frac{\partial u}{\partial y}=-\frac{\partial v}{\partial x}$

a) $f (z) = \overline{z}^2 = (x-iy)^2=x^2-y^2-2xyi$,

$u(x,y)=x^2-y^2 \quad , v(x,y)=-2xy$

$\displaystyle\frac{\partial u}{\partial x}=2x \ne \frac{\partial v}{\partial y}= -2y$, $f$ is not analytic.

b) $u(x,y)=x^3-3xy^2$ , $v(x,y)=3x^2y -y^3$

$\displaystyle\frac{\partial u}{\partial x}=3x^2-3y^2 = \frac{\partial v}{\partial y}\quad , \quad \frac{\partial u}{\partial y}=-6xy=-\frac{\partial v}{\partial x}$, f is analytic.

c) $f ( z) = |z| = \sqrt{x^2+y^2}$.

$\displaystyle\frac{\partial u}{\partial x}=\frac{x}{\sqrt{x^2+y^2}} \ne \frac{\partial v}{\partial y}= 0$, $f$ is not analytic.

 Definition. A function $f$ is said to be analytic at a point $z_0 \in \mathbb{C}$ if it is diﬀerentiable at every $z$ in some $\epsilon$-neighbourhood of the point $z_0$. The function $f$ is said to be analytic in a region if it is analyticat every point in the region. The function $f$ is said to be entire if it is analytic in $\mathbb{C}$.
 A function $f$ of a complex variable is said to be analytic (or holomorphic, or regular ) in an open set $S$ if it has a derivative at every point of $S$. If $S$ is not an open set, then we say $f$ is analytic in $S$ if $f$ is analytic in an open set containing $S$. We call $f$ analytic at the point $z_0$ if $f$ is analytic in some neighborhood of $z_0$. It is important to note that while diﬀerentiability is deﬁned at a point, analyticity is deﬁned on an open set. If a function $f$ is analytic on the whole complex plane, then it is said to be entire
Copyright $\copyright$ David R. Wilkins 1989–2008