High School Math Questions on Complex Numbers 10

Hello classmates, I am Zhou Shuai, a teacher at New Oriental Youneng Learning Center, and I hope it will help you in the future.

This is a question about the basic concept of complex numbers. There needs to be a clear distinction to start learning.

A + bi is a plural number of general representations;

When b = 0, i.e., a real number;

b is not 0, a + bi is imaginary;

0 and b 0 > special one, which is called a pure imaginary number.

When its form appears in a + bidirectional, it is called the true part, and b is called the imaginary part.

equation, we know that this is when the discriminant is greater than or equal to zero, and the discriminant of the real root is less than zero, then both of these are fictitious, therefore, we must first satisfy this condition, 2-8 <0 of the two roots.

Denote. Basically similar to the previous real root, just need to add one me. (equivalent to -1 prescription).

In this topic, x = -a (a 2-8)] 2 (the subtraction of the negative root of the half party minus 8 times i).

I wish you the best of luck.

Because the complex number z has a real part and an imaginary part, your system of equations has a problem, you should set the real part and then the modulus length is the sum of squares of the real part and the open root, so that you can get the constraints, and just go ahead. Hope, thanks.

If you check it again, you should be able to find out.

Power operations on complex trigonometry. The power of the mold is multiplied by the power of the number of times. Hungry.

This is the answer I just made, landlord, please see**.

Solution: From the meaning of the question, it will be |z + root number 3 + i|Less than or equal to 1 is regarded as having (negative root number 3,1) as the center of the circle, and a circle c with a radius less than or equal to 1 is |z-root number 3|The maximum value of the corresponding point (root number 3, 0) is root number 13+1, and the minimum value is root number 13-1The same goes for it|z-2i|The maximum value is 2 roots 3+1, and the minimum value is 2 roots 3-1.

then |z-root number 3|Squared +|z-2i|The maximum value of the square is (root number 13 + 1) square + (2 root number 3 + 1) square = 27 + 2 root number 13 + 4 root number 3 The minimum value is the square of (root number 13-1) + (2 root number 3 + 1) square = 27-2 root number 13-4 root number 3 ps: I'm not very sure about this question.,I don't know if it will mislead you.,But I don't seem to like to see that kind of person.。。。 It's a fellow believer... Hehe.

Why don't you ask the teacher, just ask your classmates.

The solution is as follows: substituting the real number b into the equation x2-(6+i)x+9+ai=0b2-(6+i)b+9+bi=0 only if it is -ib+:ai=0 imaginary number, then the imaginary part of the imaginary number cancels out, and the real part is also zero b2-6b+9=0 to solve b=3:a=3

（1+i））^2=2i

1+i））^2005=(2i)^1002 × 1+i)=-2^1002 (1+i)

Root No. 2 (1+i)) 2005=- (1-i) [2 Hope the above answers are helpful to you

1.The system of equations m 2-2m-3=5 m 2-4m+3=3 is solved to give m=4

2.The inequality group m 2-2m-3≠5 and m 2-4m+3≠3 gives m≠-2m and ≠0 and m≠4

Ascend and multiply i

z=(i-√2i²)/i²

i+√2)/(-1)

The 2-i counterpart is (- 2, -1).

In the third quadrant, it should be C.

The numerator and denominator are multiplied by i, and according to i 2=-1, we get.

z= - root number 2-i, since both the real part and the imaginary part are negative, z belongs to the third quadrant.

1) Let Z=A+Bi, then (A+Bi) 2=K+2i, that is, A2+2Abi-B 2=K+2i, you can launch 2AB=2I, AB=1, and A 2+B 2=2, so A=B=1 or A=B=-1. There is an abbreviated multiplier sign between the middle trace caves, I hope you can understand it.

2) Z=1+i, Z 2=2i, Z-Z 2=1-i, the orange state is a: (1, 1), b: (0, 2), c: (1, -1), and the area is 1 2*1*2=1

or Z=-1-i, Z2=2i, Z-Z2=-1-3i, A:(-1,-1)B:(0,2)C:(-1..)-3), the area is 1 2*2*1=1

I haven't studied math for a long time, and I hope I'm doing it right