https://r2.erweima.ai/mathbot/image/1718586202625lwoczmqk.png

回答图片中选择题的第一和第二题。

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

### Question 1:
设 \( z = \frac{1-i}{1+i} + 2i \)，则 \( |z| = \)

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

\[
\frac{1-i}{1+i} = \frac{(1-i)(1-i)}{(1+i)(1-i)} = \frac{1 - 2i + i^2}{1 - i^2} = \frac{1 - 2i - 1}{1 - (-1)} = \frac{-2i}{2} = -i
\]

Thus,
\[
z = -i + 2i = i
\]

Now, we need to find the magnitude of \( z \):

\[
|z| = |i| = 1
\]

The correct answer is:
**C. 1**

### Question 2:
已知集合 \( A = \{x | x^2 - x - 2 > 0\} \)，则 \( \delta_R A = \)

First, solve the inequality \( x^2 - x - 2 > 0 \).

\[
x^2 - x - 2 = (x-2)(x+1) > 0
\]

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

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

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

Thus, the complement of \( A \) with respect to the real numbers \( R \) is:
\[
\delta_R A = \{x | -1 \leq x \leq 2\}
\]

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

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.

Learn how to solve complex number problems involving magnitudes and inequalities. This page covers the properties of complex numbers, formulas for magnitude calculations, and detailed steps to solve quadratic inequalities. Suitable for high school students.

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