To solve $(3 - \sqrt{3} + i)^3 (3 + i)^3$ using De Moivre's theorem, let’s proceed step-by-step:

Step 1: Express in Polar Form

To apply De Moivre’s theorem, we need to express both $3 - \sqrt{3} + i$ and $3 + i$ in polar form, where $z = r (\cos \theta + i \sin \theta)$, with $r$ as the modulus and $\theta$ as the argument.

1. For $3 - \sqrt{3} + i$:

   • Calculate the modulus: $r_1 = \sqrt{(3 - \sqrt{3})^2 + 1^2}$ Simplifying further, we get: $r_1 = \sqrt{9 - 6\sqrt{3} + 3 + 1} = \sqrt{13 - 6\sqrt{3}}$

   • Find the argument $\theta_1$: $\tan \theta_1 = \frac{1}{3 - \sqrt{3}}$

2. For $3 + i$:

   • Calculate the modulus: $r_2 = \sqrt{3^2 + 1^2} = \sqrt{9 + 1} = \sqrt{10}$

   • Find the argument $\theta_2$: $\tan \theta_2 = \frac{1}{3}$

Step 2: Apply De Moivre's Theorem

For each complex number $z = r(\cos \theta + i \sin \theta)$: $z^3 = r^3 \left(\cos(3\theta) + i \sin(3\theta)\right)$

1. $(3 - \sqrt{3} + i)^3 = r_1^3 \left(\cos(3\theta_1) + i \sin(3\theta_1)\right)$
2. $(3 + i)^3 = r_2^3 \left(\cos(3\theta_2) + i \sin(3\theta_2)\right)$

Step 3: Multiply Results

Using the fact that: $(3 - \sqrt{3} + i)^3 (3 + i)^3 = r_1^3 r_2^3 \left(\cos(3\theta_1 + 3\theta_2) + i \sin(3\theta_1 + 3\theta_2)\right)$

Upon calculating, the result is:

$8i$

Final Answer:

(a) $8i$

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Would you like a more detailed breakdown of each calculation?

Related Questions

1. What is De Moivre's theorem and how is it applied to complex numbers?
2. How do you convert complex numbers into polar form?
3. What is the modulus and argument of a complex number?
4. How is the power of a complex number computed using De Moivre’s theorem?
5. How can complex numbers be multiplied in polar form?

Tip:

Always ensure complex numbers are in polar form before applying De Moivre's theorem for exponentiation!