It seems like you're asking to evaluate $(\sqrt{8} - i)^8$, where $i$ is the imaginary unit. Let’s break this down step by step to simplify and compute the result.

Step 1: Express $\sqrt{8} - i$ in polar form

We start by converting $\sqrt{8} - i$ into polar form. Any complex number $a + bi$ can be expressed in polar form as: $r (\cos(\theta) + i\sin(\theta)) = r \text{cis}(\theta)$ where:

• $r = \sqrt{a^2 + b^2}$ is the modulus.
• $\theta = \tan^{-1} \left(\frac{b}{a}\right)$ is the argument (angle).

In our case: $a = \sqrt{8}, \quad b = -1$ So, $r = \sqrt{(\sqrt{8})^2 + (-1)^2} = \sqrt{8 + 1} = \sqrt{9} = 3$ Now, we find $\theta$: $\theta = \tan^{-1} \left(\frac{-1}{\sqrt{8}}\right) = \tan^{-1} \left(\frac{-1}{2\sqrt{2}}\right)$ Using a calculator or approximation, this is approximately $\theta = -\frac{\pi}{8}$.

So, we can write: $\sqrt{8} - i = 3 \text{cis}\left(-\frac{\pi}{8}\right)$

Step 2: Apply De Moivre’s Theorem

De Moivre’s theorem states that for any complex number in polar form $r \text{cis}(\theta)$, we have: $(r \text{cis}(\theta))^n = r^n \text{cis}(n\theta)$ For $(\sqrt{8} - i)^8$, we apply this theorem with $n = 8$: $(3 \text{cis}\left(-\frac{\pi}{8}\right))^8 = 3^8 \text{cis}\left(8 \times -\frac{\pi}{8}\right)$ This simplifies to: $3^8 \text{cis}(-\pi)$

Step 3: Simplify the result

Now, we know: $3^8 = 6561$ And: $\text{cis}(-\pi) = \cos(-\pi) + i\sin(-\pi) = -1 + 0i = -1$

Thus, the result is: $6561 \times (-1) = -6561$

Final Answer:

$(\sqrt{8} - i)^8 = -6561$

Would you like any further explanation or details? Here are some related questions:

1. How do you convert a complex number into polar form?
2. What is De Moivre's Theorem, and how does it apply to complex numbers?
3. How can you determine the argument of a complex number geometrically?
4. What happens if the angle is positive instead of negative in De Moivre’s Theorem?
5. Can this method be applied to higher powers or roots of complex numbers?

Tip: When working with powers of complex numbers, converting to polar form makes the process much easier using De Moivre’s Theorem.