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.