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Si las raí­ces cuadradas de 256 son 16 y -16 entonces

The problem asks for the eighth roots of a number, based on the information that the square roots of 256 are 16 and -16. To solve this, we need to find all the eighth roots of 256, which is equivalent to solving the equation \( z^8 = 256 \).

### Step 1: Express 256 in exponential form
First, express 256 in exponential form:
\[
256 = 2^8
\]
So, we need to find all the complex numbers \( z \) such that:
\[
z^8 = 2^8
\]

### Step 2: General solution for the eighth roots
The general solution for \( z \) is given by:
\[
z = r \left( \cos\left(\frac{\theta + 2k\pi}{8}\right) + i\sin\left(\frac{\theta + 2k\pi}{8}\right) \right)
\]
where \( r = \sqrt[8]{256} = 2 \) and \( \theta = 0 \) (since \( 2^8 \) is a positive real number). \( k \) takes values from 0 to 7 to account for all the eighth roots.

### Step 3: Calculate the roots
For each \( k = 0, 1, 2, \dots, 7 \), calculate the corresponding roots:
\[
z_k = 2 \left( \cos\left(\frac{2k\pi}{8}\right) + i\sin\left(\frac{2k\pi}{8}\right) \right)
\]

These will result in:
- \( z_0 = 2 \left( 1 + i0 \right) = 2 \)
- \( z_1 = 2 \left( \cos\left(\frac{\pi}{4}\right) + i\sin\left(\frac{\pi}{4}\right) \right) = 2 \left( \frac{\sqrt{2}}{2} + i\frac{\sqrt{2}}{2} \right) = \sqrt{2} + i\sqrt{2} \)
- \( z_2 = 2 \left( 0 + i1 \right) = 2i \)
- \( z_3 = 2 \left( -\frac{\sqrt{2}}{2} + i\frac{\sqrt{2}}{2} \right) = -\sqrt{2} + i\sqrt{2} \)
- \( z_4 = 2 \left( -1 + i0 \right) = -2 \)
- \( z_5 = 2 \left( -\frac{\sqrt{2}}{2} - i\frac{\sqrt{2}}{2} \right) = -\sqrt{2} - i\sqrt{2} \)
- \( z_6 = 2 \left( 0 - i1 \right) = -2i \)
- \( z_7 = 2 \left( \frac{\sqrt{2}}{2} - i\frac{\sqrt{2}}{2} \right) = \sqrt{2} - i\sqrt{2} \)

### Step 4: Match the roots with the options given
Now, comparing these roots with the options given in the problem:
- \( \sqrt{2} + i\sqrt{2} \) matches with option \( \sqrt{2} + i\sqrt{2} \).
- \( -\sqrt{2} + i\sqrt{2} \) matches with option \( -\sqrt{2} + i\sqrt{2} \).
- \( 2i \) matches with option \( 2i \).
- \( -\sqrt{2} - i\sqrt{2} \) matches with option \( -\sqrt{2} - i\sqrt{2} \).
- \( -2i \) matches with option \( -2i \).
- \( 2 \) matches with option \( 2 \).
- \( -2 \) matches with option \( -2 \).
- \( \sqrt{2} - i\sqrt{2} \) matches with option \( \sqrt{2} - i\sqrt{2} \).

### Conclusion
The correct roots corresponding to the options in the image are:
- \( 2 \)
- \( \sqrt{2} + i\sqrt{2} \)
- \( 2i \)
- \( -\sqrt{2} + i\sqrt{2} \)
- \( -2 \)
- \( -\sqrt{2} - i\sqrt{2} \)
- \( -2i \)
- \( \sqrt{2} - i\sqrt{2} \)

All of these match the roots calculated above, so the solution is correct.

Would you like more details or have any questions?

### Related Questions
1. How do you find the fourth roots of a complex number?
2. What are the general steps to convert a complex number into polar form?
3. How would the eighth roots change if the original number was negative?
4. What is De Moivre's Theorem and how is it used in finding roots?
5. How can the geometric representation of these roots help in understanding their properties?

### Tip
When dealing with complex roots, it's helpful to remember that they are evenly spaced on the complex plane, forming a regular polygon.

Explore the solution to finding the eighth roots of 256, involving complex numbers and De Moivre's Theorem. Learn how to calculate these roots step-by-step with detailed explanations. Suitable for advanced high school students.

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