61637 1059@qq.com, can you send me the answers to your math elective course 2-3?

Elective course 2-2 3. 1.2

First, multiple choice questions

1. The false proposition in the following proposition is ()

A. the modulus of a complex number is a non-negative real number.

The necessary and sufficient condition for a complex number to be equal to zero is that its modulus is equal to zero.

C. the modulus equality of two complex numbers is a necessary condition for the equality of these two complex numbers.

D. complex number z1>; The necessary and sufficient condition of z2 is | z 1 | > | z2 |

[answer] d

【 Analysis 】 ① The module | z | = A2+B2 ≥ 0 of any complex number Z = A+Bi (A, b∈R) will always hold. ∴ A is correct;

(2) What is the condition that complex number equals z = 0? a=0b=0。 ? | z |z|=0, so b is correct;

③ if z 1 = a 1+b 1i, z2 = a2+b2i (a 1, b 1, a2, b2∈R).

If z 1 = z2, then a 1 = a2, b 1 = B2, ∴| z 1 | = | z2 |

On the other hand, from | z 1 | = | z2|, we cannot deduce z 1 = z2.

For example, when z 1 = 1+3i, z2 = 1-3i | z 1 | = | z2 |, then c is correct;

④ Two complex numbers that are not all zero cannot compare sizes, but the modules of any two complex numbers can always compare sizes. D is wrong, so I chose D.

2. Given a and b∈R, the positional relationship between two points corresponding to complex numbers A-Bi and -A-Bi on the complex plane is ().

A. About X-axis symmetry

B. About Y-axis symmetry

C. symmetry about the origin

D. symmetry of straight line y = x

[answer] b

[Analysis] The two corresponding points of complex numbers A-bi and -a-bi on the complex plane are (a, -b) and (-a, -b) which are symmetric about Y. 。

3. The following conclusion is correct ()

On the complex plane, the x axis is called the real axis and the y axis is called the imaginary axis.

B. No two complex numbers can compare sizes.

C. if the real number a corresponds to the pure imaginary number ai, then the real number set and the pure imaginary number set are in one-to-one correspondence.

The square root of D.- 1 is i.

[Answer] A.

[Analysis] Two imaginary numbers cannot be compared to exclude B. When a = 0, ai is a real number, excluding C, and the square root of-1 is I, excluding D, so choose A.

4. The point corresponding to the complex number z = (a2-2a)+(a2-a-2) i is on the imaginary axis, then ().

A.a ≠ 2 or a≠ 1

B.a ≠ 2 or a ≦- 1

C.a = 2 or a = 0.

Digital and analog =0

[answer] d

[Resolution] According to the meaning of the question, A2-2A = 0, A2-A-2 ≠ 0.

The solution is a = 0.

5. There is _ _ _ _ _ in the following formula. ()

①3i>2i ②|2+3i| >|-2-3i |③I2 & gt； (-i)2

④ | z | = | z| (where z is the complex number of the * * * yoke of the complex number z)

a . 0 b . 1

C.2 D.3

[answer] b

[Resolution] The imaginary numbers 3i and 2i are not comparable in size; |2+3i|= 13,|-2-3i|= 13,∴|2+3i|=|-2-3i|； I2 =- 1，(-I) 2 =- 1，∴ I2 = (-I) 2。 Let z = A+Bi (A, b∈R), then | z | = A2+B2, | z | = A2.

6. if the complex number z 1 = a+2i (a ∈ r), z2 = 2+i and | z 1 | < |z2|, then the range of a is ().

A.( 1，+∞)b .(-∞，- 1)

C.(- 1， 1)d .(-∞，- 1)∞( 1，+∞)

[answer] c

[resolution] ∵| z1| <| z2 |, ∴ a2+4 < 5,∴a2+4<； 5,

∴- 1 < a < 1。 So, C.

7. In the complex plane, the complex number represented by the vector OA→ is 1+I, and the vector O ′ A→ is obtained by shifting OA→ to the right by one unit, so the complex numbers corresponding to the point A→ are () respectively.

A. 1+i， 1+I b . 2+i，2+I

C. 1+i，2+I ` d . 2+I， 1+i

[answer] c

[Resolution] After the vector OA→ the starting point O'( 1, 0) is shifted to the right by one unit,

∫OA′→= OO ′→+ O′A′→= OO ′→+ OA→=( 1，0)+( 1， 1)=(2， 1)，

∴ Point A' corresponds to the complex number 2+I, o' a' → = OA →

∴O'A'→ the corresponding complex number is1+i. Therefore, C.

At the age of 8.23

A. The first quadrant B. The second quadrant

C. The third quadrant D. The fourth quadrant

[answer] d

[resolution ]∫23 < m < 1, ∴ 3m-2 > 0, m- 1 0 and m2-3m-3 >; 0,∴m= 15.

14. If the complex number z satisfies | z+3-3i | = 3, then the maximum and minimum values of |z| are _ _ _ _ _ _ _.

[Answer] 33, 3

[Analysis] | z+3-3i | = 3 represents a circle with c (-3,3) as the center and 3 as the radius, so |z| represents the distance from the point on the circle to the origin. Obviously, the maximum value of |z| is | oc |+3 = 23+3 = 33, and the minimum value is | oc |-3 =.

Third, answer questions.

15. If the inequality m2-(m2-3m) i.

[Resolution] According to the meaning of the question, m2-3m = 0m2-4m+3 = 0m2.

∴ m = 0 or m = 3m = 3 or m = 1 | m | < 10,

When m = 3, the original inequality holds.

16. If the point corresponding to the complex number z = (m2+m-1)+(4m2-8m+3) i (m ∈ r) is in the first quadrant, then the value range of the realistic number m is determined.

[Resolution] ∫z =(m2+m- 1)+(4 m2-8m+3)I,

∴z=(m2+m- 1)-(4m2-8m+3)i.

M2+M- 1 > 04m 2-8m+3 & lt； 0, the solution is-1+52.

That is, the range of the real number m is-1+52.

17. Given a∈R, in which quadrant does the complex number Z = (A2-2A+4)-(A2-2A+2) I correspond? What is the locus of the corresponding point of complex number Z?

[Resolution] A2-2A+4 = (A- 1) 2+3 ≥ 3,-(A2-2A+2) =-(a-1) 2-1.

The real part of z is positive and the imaginary part is negative.

The corresponding point of the complex number z is in the fourth quadrant.

Let z = x+yi (x, y∈R), then x = a2-2a+4y =-(a2-2a+2).

Y =-x+2 (x ≥ 3) By eliminating A2-2a,

∴ The locus of the corresponding point of complex number Z is a ray, and its equation is y =-x+2 (x ≥ 3).

18. The known complex numbers Z 1 = 3-i and Z 2 =- 12+32i.

(1) Find the values of |z 1| and |z2| and compare their sizes;

(2) Let z∈C, what is the locus of the point Z that satisfies the condition |z2|≤|z|≤|z 1|?

[Resolution] (1) | z 1 | = | 3+I |

=(3)2+ 12=2

|z2|=- 12-32i= 1。