To find the complex cube roots of $2 + 2i$ and express the answer in exponential form, let’s follow these steps:

1. Represent the complex number in polar form:

The given complex number is $z = 2 + 2i$. Its magnitude ($r$) and argument ($\theta$) are computed as follows:

• Magnitude:
  $r = \sqrt{(2)^2 + (2)^2} = \sqrt{4 + 4} = \sqrt{8} = 2\sqrt{2}.$

• Argument ($\theta$):
  $\tan(\theta) = \frac{\text{Imaginary part}}{\text{Real part}} = \frac{2}{2} = 1.$ Thus, $\theta = \tan^{-1}(1) = \frac{\pi}{4}$ (in radians).

So, the polar form of $z$ is: $z = 2\sqrt{2} \text{cis} \frac{\pi}{4},$ where $\text{cis} \, \theta = \cos \theta + i \sin \theta$.

2. Find the cube roots:

The cube roots of a complex number $z = r \text{cis} \, \theta$ are given by: $z_k = r^{1/3} \text{cis} \left(\frac{\theta + 2k\pi}{3}\right),$ for $k = 0, 1, 2$.

• $r^{1/3} = (2\sqrt{2})^{1/3} = 2^{1/3} \cdot 2^{1/6} = 2^{2/3}.$

Now calculate the arguments for $k = 0, 1, 2$:

1. For $k = 0$:
   $\theta_0 = \frac{\frac{\pi}{4} + 2(0)\pi}{3} = \frac{\pi}{12}.$ So: $z_0 = 2^{2/3} \text{cis} \frac{\pi}{12}.$

2. For $k = 1$:
   $\theta_1 = \frac{\frac{\pi}{4} + 2(1)\pi}{3} = \frac{\pi}{4} + \frac{2\pi}{3} = \frac{9\pi}{12} = \frac{3\pi}{4}.$ So: $z_1 = 2^{2/3} \text{cis} \frac{3\pi}{4}.$

3. For $k = 2$:
   $\theta_2 = \frac{\frac{\pi}{4} + 2(2)\pi}{3} = \frac{\pi}{4} + \frac{4\pi}{3} = \frac{19\pi}{12}.$ So: $z_2 = 2^{2/3} \text{cis} \frac{19\pi}{12}.$

3. Final Answer:

The complex cube roots of $2 + 2i$ in exponential form are: $2^{2/3} \text{cis} \frac{\pi}{12}, \, 2^{2/3} \text{cis} \frac{3\pi}{4}, \, 2^{2/3} \text{cis} \frac{19\pi}{12}.$

Let me know if you need further clarification or a breakdown of any step!

Related Questions:

1. What is the geometric interpretation of cube roots in the complex plane?
2. How do you compute higher-order roots of complex numbers?
3. Can you convert the roots into rectangular form (real + imaginary)?
4. Why do the cube roots form a symmetric pattern on the Argand plane?
5. What is the role of the modulus and argument in determining the roots?

Tip: Always simplify the argument ($\theta$) to lie between $0$ and $2\pi$ for clarity in polar form.