To find the cube roots of \( 1 + i \), we proceed step by step by converting the complex number into polar form and applying De Moivre's Theorem.

### Step 1: Convert \( 1 + i \) into polar form
A complex number in rectangular form \( z = x + yi \) can be written in polar form as \( r(\cos \theta + i\sin \theta) \), where:
- \( r = |z| = \sqrt{x^2 + y^2} \) is the modulus
- \( \theta = \arg(z) = \tan^{-1} \left( \frac{y}{x} \right) \) is the argument

For \( z = 1 + i \):
- \( x = 1 \), \( y = 1 \)
- Modulus: \( r = \sqrt{1^2 + 1^2} = \sqrt{2} \)
- Argument: \( \theta = \tan^{-1} \left( \frac{1}{1} \right) = \frac{\pi}{4} \)

Thus, in polar form:
\[
1 + i = \sqrt{2} \left( \cos \frac{\pi}{4} + i \sin \frac{\pi}{4} \right)
\]

### Step 2: Find the cube roots using De Moivre's Theorem
The cube roots of \( 1 + i \) can be found by taking the cube root of the modulus and dividing the argument by 3. Also, we need to consider the periodicity of the argument by adding \( 2k\pi \), where \( k = 0, 1, 2 \) for the three cube roots.

- Modulus of cube root: \( r^{1/3} = (\sqrt{2})^{1/3} = 2^{1/6} \)
- Argument for cube roots: \( \frac{\theta + 2k\pi}{3} \) for \( k = 0, 1, 2 \)

So, the cube roots are given by:
\[
z_k = 2^{1/6} \left( \cos \frac{\pi + 2k\pi}{12} + i \sin \frac{\pi + 2k\pi}{12} \right) \quad \text{for} \, k = 0, 1, 2
\]

### Step 3: Calculate the cube roots
Now, calculate the angles for each \( k \):
1. For \( k = 0 \): \( z_0 = 2^{1/6} \left( \cos \frac{\pi}{12} + i \sin \frac{\pi}{12} \right) \)
2. For \( k = 1 \): \( z_1 = 2^{1/6} \left( \cos \frac{5\pi}{12} + i \sin \frac{5\pi}{12} \right) \)
3. For \( k = 2 \): \( z_2 = 2^{1/6} \left( \cos \frac{9\pi}{12} + i \sin \frac{9\pi}{12} \right) = 2^{1/6} \left( \cos \frac{3\pi}{4} + i \sin \frac{3\pi}{4} \right) \)

Thus, the cube roots are:
\[
z_0 = 2^{1/6} \left( \cos \frac{\pi}{12} + i \sin \frac{\pi}{12} \right), \, z_1 = 2^{1/6} \left( \cos \frac{5\pi}{12} + i \sin \frac{5\pi}{12} \right), \, z_2 = 2^{1/6} \left( \cos \frac{3\pi}{4} + i \sin \frac{3\pi}{4} \right)
\]

Would you like further details on this or have any questions?

### Related Questions:
1. What are De Moivre's Theorem and its applications in complex numbers?
2. How do you calculate higher roots of complex numbers?
3. Can you find the square roots of \( 1 + i \) instead of the cube roots?
4. How does polar form simplify the process of root-finding for complex numbers?
5. What are some real-world applications of finding roots of complex numbers?

### Tip:
When working with roots of complex numbers, always remember that the argument has multiple values due to the periodicity of trigonometric functions, giving rise to multiple distinct roots.

Find the cube roots of 1+i by converting it to polar form and applying De Moivre's Theorem. This detailed step-by-step solution covers complex numbers, polar form, and trigonometric identities. Suitable for advanced high school and college students.

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