To find the roots of the equation
$w^3 = 2i$ we express $2i$ in polar form.

Step 1: Express $2i$ in Polar Form

The complex number $2i$ can be written as
$2i = 2 e^{i\frac{\pi}{2}}$ where the modulus is $2$ and the argument is $\frac{\pi}{2}$.

Step 2: Apply De Moivre’s Theorem

The roots of $w^3 = 2 e^{i\frac{\pi}{2}}$ are given by
$w_k = \sqrt[3]{2} e^{i\frac{\left(\frac{\pi}{2} + 2k\pi\right)}{3}}, \quad k = 0,1,2.$

Step 3: Compute the Roots

We evaluate the three roots for $k = 0, 1, 2$:

1. For $k = 0$:
   [ w_0 = \sqrt[3]{2} e^{i\frac{\pi}{6}} = \sqrt[3]{2} \left(\cos \frac{\pi}{6} + i \sin \frac{\pi}{6} \right). ]

2. For $k = 1$:
   [ w_1 = \sqrt[3]{2} e^{i\frac{5\pi}{6}} = \sqrt[3]{2} \left(\cos \frac{5\pi}{6} + i \sin \frac{5\pi}{6} \right). ]

3. For $k = 2$:
   [ w_2 = \sqrt[3]{2} e^{i\frac{9\pi}{6}} = \sqrt[3]{2} e^{i\frac{3\pi}{2}} = \sqrt[3]{2} \left(\cos \frac{3\pi}{2} + i \sin \frac{3\pi}{2} \right). ]

Final Answer in Polar Form

[ w_0 = \sqrt[3]{2} e^{i\frac{\pi}{6}}, \quad w_1 = \sqrt[3]{2} e^{i\frac{5\pi}{6}}, \quad w_2 = \sqrt[3]{2} e^{i\frac{3\pi}{2}}. ]

Would you like a verification of these roots or a graphical representation of them on the complex plane?

Related Questions:

1. How can De Moivre's theorem be used for higher-order roots?
2. How do you convert from polar to Cartesian form for complex numbers?
3. What are the cube roots of unity, and how do they relate to this problem?
4. How would the roots change if the equation were $w^3 = -2i$?
5. Can this method be applied to solving general polynomial equations in the complex plane?

Tip:

For any equation of the form $w^n = z$, always express $z$ in polar form first, then use De Moivre’s theorem to find the roots.