We are given the equation:

$(\sqrt{3} - 2)^n = 2^n (\cos(n \pi / 6) - i \sin(n \pi / 6))$

This looks like an equation involving complex numbers and exponential functions, particularly invoking Euler's formula and polar form representations. The goal is to prove this identity for all $n \in \mathbb{Z}$.

Step 1: Analyze the right-hand side

The right-hand side of the equation, $2^n (\cos(n \pi / 6) - i \sin(n \pi / 6))$, looks like the polar form of a complex number. By Euler's formula, we know that:

$e^{i\theta} = \cos(\theta) + i\sin(\theta)$

Thus, the right-hand side can be rewritten as:

$2^n \cdot e^{-i n \pi / 6}$

Step 2: Express the left-hand side in polar form

Now, let's analyze the left-hand side, $(\sqrt{3} - 2)^n$. First, we recognize that $\sqrt{3} - 2$ can be expressed as a complex number in polar form.

The number $\sqrt{3} - 2$ can be written in the complex plane as $r \text{cis} \theta$, where $r$ is the modulus and $\theta$ is the argument.

Step 2.1: Find the modulus

The modulus $r$ of the complex number $\sqrt{3} - 2$ is given by:

$r = \sqrt{(\sqrt{3})^2 + (-2)^2} = \sqrt{3 + 4} = \sqrt{7}$

Step 2.2: Find the argument

The argument $\theta$ of $\sqrt{3} - 2$ is the angle it makes with the positive real axis. We can find it using the arctangent of the imaginary part over the real part:

$\theta = \text{arg}(\sqrt{3} - 2) = \tan^{-1}\left(\frac{-2}{\sqrt{3}}\right) = -\frac{\pi}{6}$

So, we can write:

$\sqrt{3} - 2 = \sqrt{7} \left( \cos\left(-\frac{\pi}{6}\right) + i\sin\left(-\frac{\pi}{6}\right) \right)$

Step 3: Use De Moivre's Theorem

De Moivre's Theorem states that for any complex number in polar form:

$(r (\cos \theta + i \sin \theta))^n = r^n (\cos(n \theta) + i \sin(n \theta))$

Using this theorem, we can raise both sides to the power $n$. On the left-hand side:

$(\sqrt{3} - 2)^n = (\sqrt{7})^n \left( \cos\left( n \cdot -\frac{\pi}{6} \right) + i \sin\left( n \cdot -\frac{\pi}{6} \right) \right)$

This simplifies to:

$(\sqrt{3} - 2)^n = 7^{n/2} \left( \cos\left( -\frac{n \pi}{6} \right) + i \sin\left( -\frac{n \pi}{6} \right) \right)$

Step 4: Compare both sides

Now compare both the left-hand side and right-hand side of the original equation. We have:

$(\sqrt{3} - 2)^n = 7^{n/2} \left( \cos\left( -\frac{n \pi}{6} \right) + i \sin\left( -\frac{n \pi}{6} \right) \right)$

and

$2^n \left( \cos\left( \frac{n \pi}{6} \right) - i \sin\left( \frac{n \pi}{6} \right) \right) = 2^n e^{-i n \pi / 6}$

For the equality to hold, $7^{n/2} = 2^n$, which is not true in general for all $n$.

Conclusion

The equation given does not seem to hold true for all $n$, because the moduli $7^{n/2}$ and $2^n$ do not match for general values of $n$. Therefore, the equation cannot be proven as stated, and it seems there is a discrepancy in the formulation.

Would you like to clarify or ask about specific steps?

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Here are 5 questions to expand understanding:

1. What is Euler's formula, and how is it used in complex analysis?
2. How do you convert a complex number into polar form?
3. What is De Moivre's Theorem and why is it useful in complex number problems?
4. Can complex exponentials always be simplified using trigonometric identities?
5. What are some typical errors that occur when manipulating complex equations?

Tip: Always ensure the moduli of both sides match when dealing with complex exponentials, or the equation will not hold.