Visualization of complex functions and basic theorems of algebra

Complex function:

The world we live in is three-dimensional, and three coordinate information is needed to clearly describe the position of an object. Therefore, humans can easily understand the shape of objects below three dimensions. The two-dimensional Cartesian coordinate system and the three-dimensional coordinate system are used for analysis. Math tools for such problems.

But for complex functions, the situation is different.

Take the following complex function as an example

$f(z)=z^3 \quad z\in C$

It is impossible to draw its image in a two-dimensional or three-dimensional Cartesian coordinate system like a real variable function. Because you want to draw a pair of complex numbers $(z,f(z))$ , you need 4 dimensions, two for drawing $z=x+iy$ and two for drawing $f (z) = u+iv$ .

Expand the function so that

$\\ f (z) = X + Yi \\ z = x + yi$ ,

You can get:

$\\X+Yi=(x+yi)^3=(x^2-y^2+2xyi)(x+yi)\\=x^3-xy^2+2x^2yi+x^2yi-y^3i-2xy^2=(x^3-3xy^2)+(3x^2y-y^3)i$

and so:

$\left\{\begin{matrix} X(x,y)=x^3-3xy^2\\ Y(x,y)=3x^2y-y^3 \end{matrix}\right.$

Therefore, it is obvious that a complex variable function of one variable can be decomposed into two functions of two variables, so no matter in two-dimensional or three-dimensional Cartesian coordinate system, it cannot represent four-dimensional

$x, y, X (x, y), Y (x, y)$

The location of the coordinates.

Two-dimensional complex space can be interpreted as real, but four-dimensional space. The space dimension is not enough. The Cartesian coordinate system is represented by two real axes, which is a bit wasteful. In fact, one axis is enough for one-dimensional real number space. Up.

In matlab, it is achieved by decomposing the real and imaginary parts of the function. The real part is represented by the z-axis of the three-dimensional Cartesian coordinate system, the imaginary part is represented by color, and xOy represents the entire complex plane, which is drawn like this:

Color analysis is not intuitive. Is there a way to express the value of the imaginary axis through the coordinate intercept?

The problem is transformed into, if you use the three-dimensional Cartesian coordinate system to identify the four-dimensional function, there are two ways:

1. Use two Cartesian coordinate systems to draw separately $X (x, y), Y (x, y)$

2. In a Cartesian coordinate system, draw separately $X (x, y), Y (x, y)$

This is a bit like becoming a domain, the issue of space and time interchange, the two cannot be achieved at the same time, this is also the universal law of nature.

The second method is used below, using geogebra to

$\left\{\begin{matrix} X(x,y)=x^3-3xy^2\\ Y(x,y)=3x^2y-y^3 \end{matrix}\right.$

Respectively drawn in a Cartesian coordinate system:

The graph is a bit messy, but generally it can be seen that green and magenta are graphs of two different functions. Green is X(x,y) and magenta is Y(x,y). On the left is the entire three-dimensional graph on the XOY plane. In the three-dimensional projection, the two graphics are exactly the same, but the angle is rotated.

The lines on the XOY plane are the contour lines of the graph, which are also the projections of the points with the same real and imaginary parts on the XOY plane. Each projection has three branches, which are distributed symmetrically on a circle around the central origin. The two figures are completely similar, so there are two groups of six projection lines in the XOY plane,

The coordinates of the intersection of each two contour lines from different functions in the complex plane represent the equation

$f(z)=z_0, \quad z_0\in C$

One solution of is, that is, the three vertices of the equilateral triangle in the figure, so you can see that for any $z_0$ equation, there are three solutions, evenly distributed around the circumference.

When $z_0 = 0$ , that is

$\left\{\begin{matrix} x^3-3xy^2=0\\ 3x^2y-y^3=0\end{matrix}\right.=>\left\{\begin{matrix} x(x-\sqrt{3}y)(x+\sqrt{3}y)=0\\ y(\sqrt{3}x-y)(\sqrt{3}x+y)=0\end{matrix}\right.=>\left\{\begin{matrix} x=0,y=\frac{\sqrt{3}}{3}x, y=-\frac{\sqrt{3}}{3}x\\ x=0,y=\sqrt{3}x, y=-\sqrt{3}x \end{matrix}\right.$

The resulting equation is three asymptotes for each image.

Mold surface:

From a geometric point of view, a $with$ vertical coordinate vertical line is placed on each complex point , and its length is equal to the modulus of the polynomial f(z) at this point. The endpoints of these vertical coordinates form a certain surface M. We call it a polynomial The modular surface of f(z), so that the real part and imaginary part of the complex variable function can be considered together instead of separately

Plot the real and imaginary parts.

Modular surfaces have some properties:

1. In all complex planes, the modulus surface will not drop below the complex plane, and the modulus cannot be negative.

2. For any point z on the complex plane, there is a point on this surface, and only one point is located above this point or the point itself, indicating that it is in contact with the complex plane.

3. When z continuously moves on the complex plane, the vertical coordinates of the points on this curved surface $\left \| f(z)\right \|$ change continuously.

4. Since the roots of a polynomial are either real roots or conjugate complex roots, because if the complex number $a$ is a root of the polynomial, then:

$f(a)=0=>f(\bar{a})=\bar{f}(a)=\bar{0}=0$

$\bar{a}$ It is also a root, so the M surface must be symmetric about the real axis.

Assume

$z = x + yi$

For one-off functions

$f (z) = z + a = x + yi + a = (x + a) + yi$

The modular surface has a contact point with the complex plane, and it is fixed on the x-axis, the coordinates are

$(-a,0)$

For quadratic complex functions:

$f(z)=z^2+az+b=x^2-y^2+ax+b +(2xy+ay)i$

There are two contacts with the complex plane, either conjugate complex roots or two real roots, which also conforms to the distribution of quadratic function roots.

Draw below

$f (z) = z ^ 3 + az ^ 2 + bz + c$

The model surface:

then:

$\\f(z)=(x+yi)^3+a(x+yi)^2+b(x+yi)+c=\\(x^3-3xy^2)+(3x^2y-y^3)i+a(x^2-y^2)+a(2xy)i + bx+byi +c =\\ (x^3-3xy^2+ax^2-ay^2+bx+c)+(3x^2y-y^3+2axy+by)i$

and so

$\\ \left \| f(z) \right \|=\left \| (x^3-3xy^2+ax^2-ay^2+bx+c)+(3x^2y-y^3+2axy+by)i \right \|=\sqrt{(x^3-3xy^2+ax^2-ay^2+bx+c)^2+(3x^2y-y^3+2axy+by)^2}$

The graphics are:

Three real root situations:

octave finds the arithmetic solution:

For the case of four times:

$\\f(z)=z^4+az^3+bz^2+cz+d=\\x^4+4x^3yi+ax^3-6x^2y^2+3ax^2yi+bx^2-4xy^3i-3axy^2+2bxyi+cx+y^4-ay^3i-by^2+cyi+d=\\ (x^4+y^4-6x^2y^2+ax^3-3axy^2+bx^2-by^2+cx+d)+(4x^3y-4xy^3+3ax^2y-ay^3+2bxy+cy)i$

The corresponding mold surface shape is shown below. There are four points in contact with the complex plane, which are the zero points.

It can be seen from the figure that the order of the equation and the number of roots seem to have a certain corresponding relationship. In fact, the order is equal to the root tree. This is the basic theorem of algebra. The n contact points between the modular surface equation and the complex plane are zero, as if there is The n pillars support the entire surface, constantly changing the values of a, b, c, and d. The relative positions of these three pillars will change in the complex plane, but they are always in contact with the complex plane. According to D'Alembert’s lemma ( Understanding is good, it has little to do with understanding), any minimum point on the surface is 0, and no minimum is a positive non-zero value.

This means that there are always three points on the complex plane, so that $\left \| f(z)\right \|$ = 0. Since the complex number is equal to zero, it is synonymous with its modulus equal $\left \| f(z)\right \|$ to zero. The necessary and sufficient condition for being 0 is

$\dpi{120} \left \| f(z)\right \|=0<=>f(z)=0$

Equivalent to, polynomial

$f(z)$ ＝０

There are three roots.

This is determined by the basic theorem of algebra: the basic theorem of algebra says that any unary n-degree polynomial equation with complex coefficients has at least one (n≥1) in the complex number domain. It follows that the n-th degree polynomial equation with complex coefficients has And there are only n roots (multiple roots are calculated by multiplicity). The fundamental theorem of algebra plays a fundamental role in algebra and even mathematics as a whole. It is said that there are more than 200 proof methods for the proof of the basic theorems of algebra.

The fundamental theorem of algebra emphasizes that there is one root instead of n roots. The reason for this is that multiple roots can be derived from one root. Let:

$f(z)=z^n+a_1z^{n-1}+\cdots++a_n$

Let it $z_0$ be a root of f(z), use

$z-z_0$

$f(z)$ Divide by , because the division is of $R$ degree 1, the remainder is a constant (according to the polynomial divisor rule, if the polynomial with z is left as the remainder, the divisible rule can continue until only the constant term remains), so:

$f(z)=(z-z_0)f_1(z)+R$

Where $f_1(z)$ is a polynomial of degree n-1, and R is a constant, $z_0$ substitute it with:

$f(z_0)=(z_0-z_0)f_1(z)+R=R$

due to

$f(z_0)=0$

and so

R=0

which is

$f(z)=(z-z_0)f_1(z)$

In the same way, if the fundamental theorem of algebra is correct, it must

$f_1(z)=(z-z_1)f_2(z)$

..............

Therefore, if there is 1 root, there must be n roots, and the basic theorem of algebra only needs to explain the case of 1 root.

The n roots are not symmetrically distributed along the real axis in the entire complex plane. In addition, there are no other shape requirements (such as whether they must be in a circle? The five roots in the following figure are complex with a real part of -2 On a straight line, it is obviously impossible to be a circle).

Try to understand in a geometric way:

For the form:

$f(z)=z^n+c_1z^{n-1}+\cdots+c_{n-1}z+c_n \qquad c_1\quad c_2 \quad \cdots \quad c_n$ 　Plural

Rewrite the form as:

$f(z)=z^n(1+\frac{c_1}{z}+\frac{c_2}{z^2}+\cdots+\frac{c_{n-1}}{z^{n-1}}+\frac{c_{n}}{z^{n}})$

Make a few points:

1. $f(z)$ is a mapping from the complex number domain to the complex number domain.

$\\\left \| f(z) \right \|=\left \| z^n(1+\frac{c_1}{z}+\frac{c_2}{z^2}+\cdots+\frac{c_{n-1}}{z^{n-1}}+\frac{c_{n}}{z^{n}}) \right \|<=\\ \left \| z^n \right \|(1+\left \| \frac{c_1}{z} \right \|+\left \| \frac{c_2}{z^2} \right \|+\cdots+\left \| \frac{c_{n-1}}{z^{n-1}} \right \|+\left \| \frac{c_{n}}{z^{n}} \right \|)\approx \left \| z^n \right \|$

Next, suppose a circle on the complex plane $C_r$ , the center of the circle is at the origin, and the radius r is large enough.

Since r is large enough, when z $C_r$ is moving up, it is $f(z)$ mainly determined by its higher-order terms and will be very close to

$g(z)=z^n$

When z revolves $C_r$ around once, according to the complex multiplication rule

$\\z=r(cos(\theta)+isin(\theta))\\ g(z)=r^n(cos(n\theta)+isin(n\theta))$

At that $\theta=2\pi$ time, $g(z)$ the argument of motion $2n \ pi$ , that is, rotate n times.

Since r is large enough, $f(z)$ and $g(z)$ close enough, it also rotates n times.

Then, imagine

$r\rightarrow 0$

When the radial diameter of z gradually shrinks to the origin of the complex plane,

$f(z)\rightarrow c_n$

When r is large enough, the origin and $c_n$ both are contained $C_r$ inside (always controllable, as long as r is large enough, this condition must be met), then $f(z)$ during the shrinking process, the origin must be swept, that is to say

Exist at least once
$f(z)=0$

Proven.

Combined with the M surface graph,

$\left \| f(0) \right \|=\left \| c_n \right \|$

and

$\left \| f(r(cos\theta+isin\theta)) \right \|=$ It is large enough
and it has been proven above that it will pass through the origin.

Therefore, during the shrinking process, the M surface must be in "friction" contact with the complex plane!

Think of the following trajectory problem introduced in the principle of automatic control, you can also use the M surface theory to understand.

E.g:

For the open loop transfer function is

　　 $H(s)=\frac{K}{s(0.5s+1)}$

The control system, as shown in the figure below:

The closed-loop transfer function is:

$G(s)=\frac{H(s)}{1+H(s)}=\frac{2K}{s^2+2s+2K}$

So the characteristic equation

$s^2+2s+2K=0$

The roots are:

$\\s_1=-1+\sqrt{1-2K} \\s_2=-1-\sqrt{1-2K}$ 　　

If you want to know how the stability of the control system changes as the open-loop gain K changes, you can draw the M surface of the special equation:

another

$s=x+yi$

then

$f(s)=(x+yi)^2+2(x+yi)+2K=(x^2-y^2+2x+2K)+(2xy+2y)i$

Therefore, the M surface equation is:

$M(x,y)=\sqrt{(x^2-y^2+2x+2K)^2+(2xy+2y)^2}$

K=0, the image is two different real roots:

K=0.5, two identical real roots

When K>0.5, it is two separate complex roots.

Therefore, the follow trajectory on the complex plane is actually:

kaihuanfenzi=[0,2];
kaihuanfenmu=[1,2,0];
sys=tf(kaihuanfenzi,kaihuanfenmu);
rlocus(sys);
axis([-8 2 -2 2]);

Therefore, the stability of the control system is actually the root distribution of the polynomial in the complex plane.

The combination of number and shape can reflect the deep connection between algebra and geometry. Geogebra is a powerful tool for the combination of number and shape to analyze problems.

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