Math Problem Statement

回答图片中选择题的第一和第二题。

Solution

Let's solve the first and second multiple-choice questions from the provided image.

Question 1:

z=1i1+i+2iz = \frac{1-i}{1+i} + 2i,则 z=|z| =

To solve this, we first simplify 1i1+i\frac{1-i}{1+i}.

1i1+i=(1i)(1i)(1+i)(1i)=12i+i21i2=12i11(1)=2i2=i\frac{1-i}{1+i} = \frac{(1-i)(1-i)}{(1+i)(1-i)} = \frac{1 - 2i + i^2}{1 - i^2} = \frac{1 - 2i - 1}{1 - (-1)} = \frac{-2i}{2} = -i

Thus, z=i+2i=iz = -i + 2i = i

Now, we need to find the magnitude of zz:

z=i=1|z| = |i| = 1

The correct answer is: C. 1

Question 2:

已知集合 A={xx2x2>0}A = \{x | x^2 - x - 2 > 0\},则 δRA=\delta_R A =

First, solve the inequality x2x2>0x^2 - x - 2 > 0.

x2x2=(x2)(x+1)>0x^2 - x - 2 = (x-2)(x+1) > 0

The solution for the inequality is found by considering the critical points x=2x = 2 and x=1x = -1:

  • For (x2)>0(x-2) > 0 and (x+1)>0(x+1) > 0, the solution is x>2x > 2.
  • For (x2)<0(x-2) < 0 and (x+1)<0(x+1) < 0, the solution is x<1x < -1.

So, the solution set is: A={xx<1 or x>2}A = \{x | x < -1 \text{ or } x > 2\}

Thus, the complement of AA with respect to the real numbers RR is: δRA={x1x2}\delta_R A = \{x | -1 \leq x \leq 2\}

The correct answer is: D. {x1x2}\{x | -1 \leq x \leq 2\}

Would you like more details on these solutions or have any further questions?

Here are 5 questions you might ask next:

  1. How do you solve a different type of complex number problem?
  2. Can you explain the steps to solving quadratic inequalities in more detail?
  3. What are the properties of complex numbers?
  4. How can you use the unit circle to understand complex numbers?
  5. What are some common methods for solving inequalities?

Tip: When solving complex number problems, always remember to simplify the expression first before finding magnitudes or arguments.

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Math Problem Analysis

Mathematical Concepts

Complex Numbers
Inequalities

Formulas

Formula for magnitude of a complex number

Theorems

Properties of complex numbers
Solving quadratic inequalities

Suitable Grade Level

High School