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Let's solve the second and third questions from the image provided.

### Question 2:
**Problem: Find the principal argument of the complex number \( e^{5+6i} \)**

**Solution:**

We start with the complex number in the form \( e^{a+bi} \). Here, \( a = 5 \) and \( b = 6 \).

1. Recall Euler's formula:
\[ e^{a+bi} = e^a (\cos(b) + i \sin(b)) \]

So,
\[ e^{5+6i} = e^5 (\cos(6) + i \sin(6)) \]

2. The principal argument (θ) of a complex number \( e^{bi} \) is \( b \mod 2\pi \).

Therefore,
\[ \text{Principal argument} = 6 \mod 2\pi \]

Since \( 6 \) is less than \( 2\pi \approx 6.2832 \), the principal argument is:

\[ \theta = 6 \]

### Question 3:
**Problem: Find the rotation angle of the transformation \( w = z^2 \) at \( z = 2i \)**

**Solution:**

1. First, express \( z = 2i \) in polar form:
\[ z = 2i = 2(\cos \frac{\pi}{2} + i \sin \frac{\pi}{2}) \]

2. The transformation \( w = z^2 \):
\[ w = (2i)^2 = 4i^2 = 4(-1) = -4 \]

3. Find the argument of \( w \):
\[ w = -4 = 4(\cos \pi + i \sin \pi) \]

So, the argument of \( w \) is \( \pi \).

4. The rotation angle for the transformation \( w = z^2 \) is twice the argument of \( z \):
\[ \text{Rotation angle} = 2 \times \frac{\pi}{2} = \pi \]

Therefore, the rotation angle is \( \pi \).

If you have any further questions or need additional details, please let me know.

Here are 5 questions you might consider next:
1. How do you find the magnitude of a complex number?
2. What is Euler's formula and how is it used in complex analysis?
3. How do you convert a complex number from rectangular form to polar form?
4. What is the principal argument of a complex number and how is it determined?
5. How do transformations affect the argument and magnitude of complex numbers?

**Tip:** When working with complex numbers, always try to convert them to polar form to make multiplication, division, and finding powers easier.

Explore how to find the principal argument of a complex number and the rotation angle of a transformation. This problem set covers complex numbers, Euler's formula, and polar form, demonstrating their application in advanced high school mathematics.

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