To solve for $Z^{51}$ where $Z = \frac{\sqrt{3}}{2} - \frac{1}{2} i$, let’s proceed step-by-step.

Step 1: Express $Z$ in Polar Form

The complex number $Z = \frac{\sqrt{3}}{2} - \frac{1}{2} i$ can be represented in polar form as $Z = r \cdot e^{i \theta}$, where:

1. $r = |Z|$, the modulus of $Z$.
2. $\theta = \arg(Z)$, the argument of $Z$.

Calculate the Modulus $r$

The modulus $r$ of $Z$ is: $r = \sqrt{\left(\frac{\sqrt{3}}{2}\right)^2 + \left(-\frac{1}{2}\right)^2} = \sqrt{\frac{3}{4} + \frac{1}{4}} = \sqrt{1} = 1$

So, $|Z| = 1$.

Calculate the Argument $\theta$

To find $\theta$, we use the formula: $\theta = \tan^{-1}\left(\frac{-\frac{1}{2}}{\frac{\sqrt{3}}{2}}\right) = \tan^{-1}(-\frac{1}{\sqrt{3}})$ This angle corresponds to $-\frac{\pi}{6}$ (or equivalently, $11\pi/6$ if we consider it in the range $[0, 2\pi)$).

Therefore, $Z = e^{-i \frac{\pi}{6}}$.

Step 2: Calculate $Z^{51}$

Since $Z = e^{-i \frac{\pi}{6}}$, we have: $Z^{51} = \left(e^{-i \frac{\pi}{6}}\right)^{51} = e^{-i \frac{51 \pi}{6}}$

Step 3: Simplify the Exponent

Now, simplify $\frac{51 \pi}{6}$ modulo $2\pi$ to find the equivalent angle in the primary circle.

$\frac{51 \pi}{6} = 8 \pi + \frac{3 \pi}{6} = 8 \pi + \frac{\pi}{2}$ Since $8\pi$ is a multiple of $2\pi$, it can be removed, leaving: $Z^{51} = e^{-i \frac{\pi}{2}}$

Step 4: Express in Rectangular Form

The angle $-\frac{\pi}{2}$ corresponds to the point $0 - i$ on the unit circle. Therefore: $Z^{51} = -i$

Final Answer

$\text{Odgovor: } -i$

Would you like a more detailed explanation or have any further questions?

Here are five related questions for further exploration:

1. How can we express other powers of $Z$ in a similar way?
2. What would $Z^{100}$ be for the same $Z$?
3. How does the modulus of a complex number affect its powers?
4. Why is polar form helpful for calculating high powers of complex numbers?
5. How do we convert from polar to rectangular form for any complex power?

Tip: For high powers of complex numbers, using De Moivre's theorem is very effective, as it leverages both the modulus and argument in polar form.