review

Three forms of plural numbers

1. Algebraic form: $z=a+bi(a,b\in R),a=\Re(z),b=\Im(z)$
2. Trigonometric form: $z=r(\cos\theta+i\sin\theta),(r\ge0)$-$\arg(z)\in[0,2\pi)$represents the main value of the argument, $Arg(z)\in\R$represents the argument
   • $z_1z_2=r_1r_2[\cos(\theta_1+\theta_2)+i\sin(\theta_1+\theta_2)$
   • $\frac{z_1}{z_2}=\frac{r_1}{r_2}[\cos(\theta_1-\theta_2)+i\sin(\theta_1-\theta_2)$
   • $z^n=r^n(\cos n\theta+i\sin n\theta)$
   • $\sqrt[n]{z}=\sqrt[n]{r}(\cos\frac{\theta+2k\pi}{n}+i\sin\frac{\theta+2k\pi}{n}),k=0,1,2,...,n-1$
3. Exponential form: $e^{i\theta}=\cos\theta+i\sin\theta$

Application of plural numbers

1. $S_\triangle=\frac{1}{2}|Im(\overline{z_1}z_2+\overline{z_2}z_3+\overline{z_3}z_1)|$
2. Necessary and sufficient conditions for four points to be a circle: $\frac{z_3-z_1}{z_4-z_1}:\frac{z_3-z_2}{z_4-z_2}\in R$
3. $\triangle z_1z_2z_3,\triangle w_1w_2w_3$Similar necessary and sufficient conditions: $\frac{z_3-z_2}{z_2-z_1}=\frac{w_3-w_2}{w_2-w_1}$

brush questions

Example 1

If $z\in C,|z-2|\le1$, find the maximum and minimum values of $|z|$, and the range of $\arg(z)$

Solution: The point represented by $z$in the complex plane is on a circle with center $(2,0)$and radius $1$.

Since there is $||z|-2|\le|z-2|\le|z|+|2|$:

So $-1\le|z|-2\le1,1\le|z|\le3$

Combining numbers and shapes, $\arg(z)\in[0,\frac{\pi}{6}]\cup[\frac{11}{6}\pi,2\pi)$

Example 2

The complex number $z$satisfies $\arg(z+3)=\frac{6}{5}\pi$, find $\min|z+6|+|z-3i|$.

Solution: It is easy to know that the trajectory represented by $\arg(z+3)=\frac{6}{5}\pi$is $y=-\frac{1}{2}(x+3)(x\lt-3)$

The obtained $|z+6|+|z-3i|$is equivalent to the sum of the distances from the corresponding points of $z$to $(-6,0),(0,3)$in the complex plane.

Obviously $\min |z+6|+|z-3i|=\sqrt{6^2+3^2}=3\sqrt{5}$(the shortest line segment between two points).

Example 3

Given $|z-2i|\le1$, find $\max\arg(z-4i)$Apparently $\max\arg(z-4i)=\frac{5}{3}\pi$

Example 4

A line $l$passes through the origin. The parabola $C$has its vertex at the origin and its focus on the positive $x$-axis. The reflections of $A(-1,0)$and $B(0,8)$across $l$both lie on $C$. Find the equations of $l$and $C$.

Let parabola $C:y^2=2px(p\gt0)$and straight line $l:y=kx$.

If you consider calculating the symmetry point coordinates and then bringing them into the parabola, the calculation will be extremely complicated.

We establish a complex plane and assume that $\vec{OA'},\vec{OB'}$corresponds to the complex number $x_1+y_1i,x_2+y_2i$.

Apparently $\vec{OA'}\perp\vec{OB'},|OA'|=1,|OB'|=8$

Therefore $x_2+y_2i=8i(x_1+y_1)$, then $\begin{cases} x_2=-8y_1,\\ y_2=8x_1 \end{cases}$

Bringing $A',B'$into the parabola:

$$\begin{gathered} y_1^2=2px_1 (1)\\ y_2^2=2px_2 (2)\\ x_2=-8y_1 (3)\\ y_2=8x_1 (4)\\ \frac{(1)(2)}{(3)(4)}:-y_1y_2=4p^2\\ (1)(4):y_1^2=\frac{p}{4}y_2\\ y_1=-p,y_2=4p\\ x_1=\frac{p}{2}\\ x_1^2+y_1^2=(\frac{p}{2})^2+p^2=1,p=\frac{2\sqrt{5}}{5}\\ C:y^2=\frac{4\sqrt{5}}{5}x\\ \end{gathered}$$

Find the equation of the straight line below:

$$\begin{gathered} x_1=\frac{\sqrt{5}}{5},y_1=-\frac{2\sqrt{5}}{5}\\ -\frac{1}{k}=\frac{y_1-0}{x_1-(-1)}\\ k=\frac{x_1+1}{-y_1}=\frac{\frac{p}{2}+1}{p}=\frac{1+\sqrt{5}}{2}\\ l:y=\frac{1+\sqrt{5}}{2}x \end{gathered}$$

Example 5

If point $A(3,0),B$is on ellipse $\frac{x^2}{4}+y^2=1$, point $A,B,C$and three points are arranged in a clockwise direction, and $\triangle ABC$is an equilateral triangle, find the trajectory of point $C$.

Assume $C(x,y)$, then the geometric relationship $[(x-3)+yi](\cos60\degree+i\sin60\degree)=(x_B-3)+y_Bi$

Then $\begin{cases} x_B=3+\frac{x-3}{2}-\frac{\sqrt{3}}{2}y=\frac{x-\sqrt{3}y+3}{2}\\ y_B=\frac{\sqrt{3}(x-3)}{2}+\frac{y}{2}=\frac{\sqrt{3}x+y-3\sqrt{3}}{2} \end{cases}$

Enter the elliptic equation: $$(\frac{x-\sqrt{3}y+3}{2})^2+4(\frac{\sqrt{3}x+y-3\sqrt{3}}{2})^2=4$$

Further simplify the coefficients:

$$(x-\sqrt{3}y+3)^2+4(\sqrt{3}x+y-3\sqrt{3})^2=16$$

Example 6

Assume $x, y \in R$, $z_1 = 2 - \sqrt{3}x + xi$, $z_2 = \sqrt{3}y - 1 + (\sqrt{3} - y)i$, known $|z_1| = |z_2|$, $\arg \frac{z_1}{z_2} = \frac{\pi}{2}$,

(1) Find $\left( \frac{z_1 + z_2}{2} \right)^{100}$

(2) Assume $z = \frac{z_1 + z_2}{2}$, find the number of elements in the set $A = \left\{ x \mid x = z^{2k} + z^{-2k}, k \in \Z \right\}$.

(1)

Because $\arg(\frac{z_1}{z_2})=\frac{\pi}{2},|z_1|=|z_2|$

So $z_1=iz_2$

$$(2-\sqrt{3}x)+xi=i[(\sqrt{3}y-1)+(\sqrt{3}-y)i]$$

This gives: $$\begin{cases} 2-\sqrt{3}x=y-\sqrt{3},\\ x=\sqrt{3}y-1 \end{cases}$$

Solving gives: $$\begin{cases} x=\frac{1+\sqrt{3}}{2}\\ y=\frac{1+\sqrt{3}}{2} \end{cases}$$

Therefore: $$\begin{cases} z_1=\frac{1-\sqrt{3}}{2}+\frac{1+\sqrt{3}}{2}i\\ z_2=\frac{1+\sqrt{3}}{2}+\frac{\sqrt{3}-1}{2}i \end{cases}$$

Hence: $$\begin{gathered} \left( \frac{z_1 + z_2}{2} \right)^{100}\\ =(\frac{1}{2}+\frac{\sqrt{3}}{2}i)^{100}\\ =(\cos60\degree+i\sin60\degree)^{100}\\ =\cos6000\degree+i\sin6000\degree\\ =\cos240\degree+i\sin240\degree\\ =-\frac{1}{2}-\frac{\sqrt{3}}{2}i \end{gathered}$$

Consider $(\frac{1}{2}+\frac{\sqrt{3}}{2}i)^{3}=-1$for easier calculation.

(2)

$$\begin{gathered} x=z^{2k}+z^{-2k}\\ =(\cos60\degree+i\sin60\degree)^{2k}+(\cos60\degree+i\sin60\degree)^{-2k}\\ =2\cos(120k)\degree \end{gathered}$$Considering $k\in\Z$, there are 2 possible $x=2\cos(120k)$. $|A|=2$

Let $\omega=-\frac{1}{2}+\frac{\sqrt{3}}{2}i$

There is $z=-\overline{w}$

$$\begin{gathered} x=z^{2k}+z^{-2k}\\ =(-\overline{\omega})^{2k}+(-\overline{\omega})^{-2k}\\ =(\overline{\omega})^{2k}+(\overline{\omega})^{-2k}\\ =\omega^k+\omega^{-k} \end{gathered}$$

Considering $k=3m,3m+1,3m+2$, we get that $x$has two possible values.

Example 7

Assume complex numbers $z = \cos\theta + i\sin\theta(0 < \theta < \pi)$, $w = \frac{1 - (\overline{z})^4}{1 + z^4}$, and $|w| = \frac{\sqrt{3}}{3}$, $\arg w < \frac{\pi}{2}$, and find $\theta$.

$$\begin{gathered} w\\ =\frac{(1-\cos4\theta)+i\sin4\theta}{(1+\cos4\theta)+i\sin4\theta}\\ =\frac{2\sin^{2}2\theta+2\sin2\theta\cos2\theta i}{2\cos^{2}2\theta+2\sin2\theta\cos2\theta i}\\ =\tan2\theta\frac{\sin2\theta+i\cos2\theta}{\cos2\theta+i\sin2\theta}\\ =\tan2\theta\frac{\cos(\frac{\pi}{2}-2\theta)+i\sin(\frac{\pi}{2}-2\theta)}{\cos2\theta+i\sin2\theta}\\ =\tan2\theta[\cos(\frac{\pi}{2}-4\theta)+i\sin(\frac{\pi}{2}-4\theta)] \end{gathered}$$

Requirements: $|\tan2\theta|=|w|=\frac{\sqrt{3}}{3},2\theta=\pm\frac{1}{6}\pi+k\pi$

Case 1

$$\tan2\theta=\frac{\sqrt{3}}{3}$$$$\theta=\frac{1}{12}\pi\text{ or }\frac{7}{12}\pi$$$$w=\frac{\sqrt{3}}{3}(\cos\frac{\pi}{6}+i\sin\frac{\pi}{6}),\arg(w)=\frac{\pi}{3}\lt\frac{\pi}{2}$$

Situation 2

$$\tan2\theta=-\frac{\sqrt{3}}{3}$$$$\theta=\frac{5}{12}\pi\text{ or }\frac{11}{12}\pi$$$$w=-\frac{\sqrt{3}}{3}(\cos\frac{5}{6}\pi+i\sin\frac{5}{6}\pi)=\frac{\sqrt{3}}{3}(\cos\frac{11\pi}{6}+i\sin\frac{11\pi}{6})\\ \arg(w)\gt\frac{\pi}{2}$$

To sum up, $\theta=\frac{1}{12}\pi\text{ or }\frac{7}{12}\pi$

Example 8

Proof: $\cos\frac{\pi}{7} - \cos\frac{2\pi}{7} + \cos\frac{3\pi}{7} = \frac{1}{2}$

Construct an equation: $z^7-1=0$

By de Mauve's formula: $z_k=\cos\frac{2k\pi}{7}+i\sin\frac{2k\pi}{7},k=0,1,2,...,6$

According to Vedic theorem: $z_0+z_1+z_2+...+z_6=0$

The real part is equal to 0:

$$\begin{gathered} 1+\cos\frac{2}{7}\pi+\cos\frac{4}{7}\pi+\cos\frac{6}{7}\pi+\cos\frac{8}{7}\pi+\cos\frac{10}{7}\pi+\cos\frac{12}{7}\pi\\ =1-2(\cos\frac{\pi}{7} - \cos\frac{2\pi}{7} + \cos\frac{3\pi}{7})=0 \end{gathered}$$

Q.E.D

If you consider using trigonometric identity transformation:

$$\begin{gathered} \cos\frac{\pi}{7} - \cos\frac{2\pi}{7} + \cos\frac{3\pi}{7}\\ =\cos\frac{\pi}{7}+\cos\frac{3\pi}{7}+\cos\frac{5\pi}{7} \end{gathered}$$

For the sum of trigonometric functions of equal angles, the solution is: multiply by ** twice the sine of half the difference angle **

$$\begin{gathered} 2\sin\frac{\pi}{7}(\cos\frac{\pi}{7}+\cos\frac{3\pi}{7}+\cos\frac{5\pi}{7})\\ =\sin\frac{2\pi}{7}+\sin0+\sin\frac{4\pi}{7}-\sin\frac{-2\pi}{7}+\sin\frac{6\pi}{7}-\sin\frac{4\pi}{7}\\ =\sin\frac{\pi}{7} \end{gathered}$$

Q.E.D

Example 9

Simplify $\frac{\sin x + \sin 3x + \sin 5x + \cdots + \sin(2n-1)x}{\cos x + \cos 3x + \cos 5x + \cdots + \cos(2n-1)x}$.

$$\begin{gathered} \frac{\sin x + \sin 3x + \sin 5x + \cdots + \sin(2n-1)x}{\cos x + \cos 3x + \cos 5x + \cdots + \cos(2n-1)x}\\ =\frac{2\sin x[\sin x + \sin 3x + \sin 5x + \cdots + \sin(2n-1)x]}{2\sin x[\cos x + \cos 3x + \cos 5x + \cdots + \cos(2n-1)x]}\\ =\frac{-[(\cos2x-\cos0)+(\cos4x-\cos2x)+(\cos6x-\cos4x)+...+(\cos2nx-\cos(2n-2)x]}{(\sin2x-\sin0)+(\sin4x-\sin2x)+(\sin6x-\sin4x)+...+(\sin2nx-\sin(2n-2)x)}\\ =\frac{1-\cos2nx}{\sin2nx}\\ =\frac{2\sin^2nx}{2\sin nx\cos nx}\\ =\tan{nx} \end{gathered}$$

How to consider complex numbers: let $z=\cos x+i\sin x$

So there are: $\begin{cases} \cos nx=\frac{z^n+\overline{z}^n}{2},\\ \sin nx=\frac{z^n-\overline{z}^n}{2i} \end{cases}$

So the numerator:

$$\begin{gathered} \sin x + \sin 3x + \sin 5x + \cdots + \sin(2n-1)x\\ =\frac{1}{2i}[(z+z^3+...+z^{2n-1})-(\overline{z}+\overline{z}^3+...+\overline{z}^{2n-1})]\\ =\frac{1}{2i}[\frac{z^{2n+1}-z}{z^2-1}-\frac{\overline{z}^{2n+1}-\overline{z}}{\overline{z}^2-1}]\\ =\frac{1}{2i}[\frac{z^{2n+1}-z}{z^2-z\overline{z}}-\frac{\overline{z}^{2n+1}-\overline{z}}{\overline{z}^2-z\overline{z}}]\\ =\frac{1}{2i}[\frac{(z^{2n}-1)+(\overline{z}^{2n}-1)}{z-\overline{z}}]\\ =\frac{1}{2i}\frac{(z^n-\overline{z}^n)^2}{z-\overline{z}}=(\frac{z^n-\overline{z}^n}{2i})^2\frac{2i}{z-\overline{z}}=\frac{\sin^2 nx}{\sin x} \end{gathered}$$

The same denominator:

$$\begin{gathered} \cos x + \cos 3x + \cos 5x + \cdots + \cos(2n-1)x\\ =\frac{1}{2}[(z+z^3+...+z^{2n-1})+(\overline{z}+\overline{z}^3+...+\overline{z}^{2n-1})]\\ =\frac{1}{2}[\frac{(z^{2n}-1)-(\overline{z}^{2n}-1)}{z-\overline{z}}]\\ =\frac{1}{2}[\frac{(z^n+\overline{z}^n)(z^n-\overline{z}^n)}{z-\overline{z}}]\\ =(\frac{z^n+\overline{z}^n}{2})(\frac{z^n-\overline{z}^n}{2})(\frac{2i}{z-\overline{z}})=\frac{\sin nx\cos nx}{\sin x} \end{gathered}$$

Divide to get $\tan nx$

Example 10

Familiar: $\tan\frac{\alpha}{2}=\frac{1-\cos\alpha}{\sin\alpha}=\frac{\sin\alpha}{1+\cos\alpha}$

Use the sum ratio theorem:

$$\frac{1 + \sin\alpha - \cos\alpha}{1 + \sin\alpha + \cos\alpha} = \tan\frac{\alpha}{2}$$

$$\frac{1 + 2\sin\alpha - \cos\alpha}{1 + \sin\alpha + 2\cos\alpha} = \tan\frac{\alpha}{2}$$

$$\frac{\sqrt{3} + \sqrt{2}\sin\alpha - \sqrt{3}\cos\alpha}{\sqrt{2} + \sqrt{3}\sin\alpha + \sqrt{2}\cos\alpha} = \tan\frac{\alpha}{2}$$