To determine all the cube roots of the equation $z^3 = i + 1$, we proceed by finding the roots in the complex plane.

Step-by-Step Solution

1. Express $i + 1$ in Polar Form: We need to convert $i + 1$ into polar form. First, write it in terms of real and imaginary components: $i + 1 = 1 + i$

   • The magnitude (modulus) $r$ of $1 + i$ is: $r = \sqrt{1^2 + 1^2} = \sqrt{2}$
   • The argument $\theta$ of $1 + i$ (angle with the positive real axis) is: $\theta = \tan^{-1} \left(\frac{1}{1}\right) = \frac{\pi}{4}$ Therefore, in polar form: $i + 1 = \sqrt{2} \left(\cos \frac{\pi}{4} + i \sin \frac{\pi}{4}\right)$

2. Use De Moivre's Theorem for Cube Roots: We want to find $z$ such that $z^3 = i + 1$. Let $z = r_z \left(\cos \phi + i \sin \phi \right)$, where $r_z$ is the modulus of $z$ and $\phi$ is the argument.

   Then, $z^3 = i + 1 = \sqrt{2} \left( \cos \frac{\pi}{4} + i \sin \frac{\pi}{4} \right)$.

   To satisfy this, the modulus $r_z$ of $z$ must be: $r_z^3 = \sqrt{2} \Rightarrow r_z = \sqrt[3]{\sqrt{2}} = 2^{1/6}$

   For the argument $\phi$, we have: $3\phi = \frac{\pi}{4} + 2k\pi, \quad k = 0, 1, 2$ Thus: $\phi = \frac{\pi}{12} + \frac{2k\pi}{3}$

3. Find the Three Cube Roots: For each integer $k = 0, 1, 2$, we get a distinct root $z_k$:

   • For $k = 0$: $z_0 = 2^{1/6} \left( \cos \frac{\pi}{12} + i \sin \frac{\pi}{12} \right)$

   • For $k = 1$: $z_1 = 2^{1/6} \left( \cos \frac{3\pi}{4} + i \sin \frac{3\pi}{4} \right)$

   • For $k = 2$: $z_2 = 2^{1/6} \left( \cos \frac{17\pi}{12} + i \sin \frac{17\pi}{12} \right)$

These are the three cube roots of $i + 1$, expressed in polar form.

Would you like further details on any specific part of this process?

Related Questions:

1. How would you find the fourth roots of a complex number?
2. What is De Moivre's theorem, and how does it apply to finding roots of complex numbers?
3. How can you convert a complex number from Cartesian form to polar form?
4. How does the argument of a complex number change when finding roots?
5. What is the geometric representation of complex roots on the Argand plane?

Tip:

When finding roots of complex numbers, always express the given number in polar form for easier calculation and use of De Moivre’s theorem.