Addition and subtraction

Law of addition?

Complex numbers are added according to the following rules: let z 1 = a+bi and z2 = c+di be any two complex numbers, then their sum is (a+bi)+(c+di) = (a+c)+(b+d) i. The sum of the two complex numbers is still a complex number, but its real part is the original two complex numbers. ?

The addition of complex numbers satisfies the commutative law and associative law.

That is, for any complex number Z 1, Z2 and Z3, there are: z1+z2 = z2+z1; (z 1+z2)+z3 = z 1+(z2+z3)。 Subtraction rule?

The subtraction of complex numbers is carried out according to the following rules: let z 1 = a+bi and z2 = c+di be any two complex numbers, then their difference is (a+bi)-(c+di) = (a-c)+(b-d) i. The difference between the two complex numbers is still a complex number, and its real part is the original two complex numbers. ?

2 multiplication and division

Multiplication rule?

Multiplication of complex numbers is specified according to the following rules:?

Let z 1=a+bi, z2=c+di(a, b, c, d∈R) be any two complex numbers, then their product (a+bi) (c+di) = (AC-BD)+(BC+ad) i?

In fact, it is to multiply two complex numbers, which is similar to the multiplication of two polynomials. Expand: ac+adi+bci+bdi? Because of me? =- 1, so the result is (AC-BD)+(BC+AD) i. The product of two complex numbers is still a complex number. Law of division?

Definition of complex number division: the complex number x+yi(x, y∈R) satisfying (c+di)(x+yi)=(a+bi) is called the quotient operation method of complex number a+bi divided by complex number c+di: division can be converted into multiplication, and the denominator can be multiplied by the * * yoke of the denominator at the same time.

① let the complex number a+bi(a, b∈R) be divided by c+di(c, d∈R), and its quotient is x+yi(x, y∈R), that is, (a+bi)÷(c+di)=x+yi?

*( x+yi)(c+di)=(CX-dy)+(dx+cy)I . ∴(cx-dy)+(dx+cy)i=a+bi.？

According to the definition of complex equation, cx-dy=a, dx+cy=b?

Solve this equation group and get x=(ac+bd)/(c? +d？ )y=(bc-ad)/(c？ +d？ )?

So there are: (a+bi)/(c+di)=(ac+bd)/(c? +d？ ) +i (BC-AD) /(c? +d？ )

② Use * * * yoke complex number to realize denominator (see the right figure):?

Comments: ① It is a conventional method; ② When using the simplified irrational number fraction we learned in junior high school, the denominator is physico-chemistry, while the complex numbers c+di and c-di are equivalent to the dual formulas we learned in junior high school. Their product is 1, which is a rational number, and (c+di) (c-di) = C2+D2 is a positive real number. Therefore, we can make the denominator real. Put this method into practice.

How to solve the problem that the complex plane is too big? As far as high school mathematics is concerned, it is similar to solving the problem of plane analytic geometry.

Complex number solution of plane geometry problem

Complex number is one of the important contents of high school mathematics. In high school mathematics, there are many math problems. If we can turn the problems into complex problems according to their specific characteristics, then this kind of mathematical problems can often be proved by complex solutions.

The basic idea of solving plane geometry by complex number method is to represent points on the complex plane with complex numbers, and then use the related properties of the modulus and amplitude angle of complex numbers, the geometric meaning of complex number operation and the condition of complex number equality to turn geometric problems into complex number problems to deal with.