The problem asks to express complex numbers in their standard form.

The expressions are:

a) $(1 + i\sqrt{3})^9$
b) $\sqrt[5]{1 - i\sqrt{3}}$
c) $(2 + i\sqrt{12})^5 (\sqrt{3} - i)^{11}$

To solve these types of problems, we generally apply polar (trigonometric) form for complex numbers, which involves converting the complex number to polar coordinates and then using De Moivre’s Theorem for powers and roots. Let's go through each part:

a) $(1 + i\sqrt{3})^9$

1. Convert the complex number to polar form:
   A complex number $z = a + bi$ has polar form $z = r(\cos\theta + i\sin\theta)$, where $r = \sqrt{a^2 + b^2}$ and $\theta = \tan^{-1}\left(\frac{b}{a}\right)$.

   • Here, $a = 1$ and $b = \sqrt{3}$.
     So, $r = \sqrt{1^2 + (\sqrt{3})^2} = \sqrt{4} = 2$.
     Also, $\theta = \tan^{-1}\left(\frac{\sqrt{3}}{1}\right) = \frac{\pi}{3}$.

2. Apply De Moivre’s Theorem:
   For powers, $(r(\cos\theta + i\sin\theta))^n = r^n(\cos(n\theta) + i\sin(n\theta))$.
   Here, $n = 9$, so we compute:

   • $r^9 = 2^9 = 512$,
   • $n\theta = 9 \times \frac{\pi}{3} = 3\pi$.

3. Simplify:

   • $\cos(3\pi) = -1$,
   • $\sin(3\pi) = 0$.

   Thus, $(1 + i\sqrt{3})^9 = 512(-1 + 0i) = -512$.

b) $\sqrt[5]{1 - i\sqrt{3}}$

1. Convert to polar form:

   • $a = 1$, $b = -\sqrt{3}$.
     So, $r = \sqrt{1^2 + (-\sqrt{3})^2} = \sqrt{4} = 2$,
     $\theta = \tan^{-1}\left(\frac{-\sqrt{3}}{1}\right) = -\frac{\pi}{3}$.

2. Apply the formula for roots of complex numbers:
   The 5th root of a complex number is given by: $z_k = r^{1/5} \left( \cos\left(\frac{\theta + 2k\pi}{5}\right) + i\sin\left(\frac{\theta + 2k\pi}{5}\right) \right), \quad k = 0, 1, 2, 3, 4.$ Here, $r^{1/5} = 2^{1/5}$.

   For $k = 0$: $z_0 = 2^{1/5} \left( \cos\left(\frac{-\pi}{15}\right) + i\sin\left(\frac{-\pi}{15}\right) \right).$

   This will give one of the five roots. You can compute other values of $k$ for the other roots.

c) $(2 + i\sqrt{12})^5 (\sqrt{3} - i)^{11}$

1. For $(2 + i\sqrt{12})^5$:

   • Convert to polar form:
     $r = \sqrt{2^2 + (\sqrt{12})^2} = \sqrt{4 + 12} = \sqrt{16} = 4$,
     $\theta = \tan^{-1}\left(\frac{\sqrt{12}}{2}\right) = \tan^{-1}(2\sqrt{3})$.

   Apply De Moivre’s Theorem for powers.

2. For $(\sqrt{3} - i)^{11}$:
   Similar process as above, converting to polar form and applying De Moivre’s Theorem.

Would you like detailed steps for any specific part?

Here are 5 related questions:

1. How do you convert a complex number from rectangular form to polar form?
2. What is De Moivre’s Theorem, and how is it used in finding powers of complex numbers?
3. How can you find the nth roots of a complex number?
4. What are the applications of complex numbers in solving geometric problems?
5. How do you simplify trigonometric expressions involving multiples of angles?

Tip: When working with powers or roots of complex numbers, converting to polar form simplifies the calculations, especially when using De Moivre's Theorem.