What is the value of `(-1+isqrt3)^(48)`?

Options

A. 1
B. 2
C. `2^(24)`
D. `2^(48)`

Explanation

To solve the problem of finding the value of \((-1 + i\sqrt{3})^{48}\), we can follow these steps:

### Step 1: Rewrite the expression
We start with the expression:
\[ (-1 + i\sqrt{3})^{48} \]

### Step 2: Identify the complex number in polar form
We can express \(-1 + i\sqrt{3}\) in polar form. First, we find the modulus \(r\) and the argument \(\theta\).

The modulus \(r\) is given by:
\[ r = \sqrt{(-1)^2 + (\sqrt{3})^2} = \sqrt{1 + 3} = \sqrt{4} = 2 \]

Next, we find the argument \(\theta\):
\[ \theta = \tan^{-1}\left(\frac{\sqrt{3}}{-1}\right) \]
Since the point \((-1, \sqrt{3})\) lies in the second quadrant, we have:
\[ \theta = \pi - \frac{\pi}{3} = \frac{2\pi}{3} \]

Thus, we can write:
\[ -1 + i\sqrt{3} = 2\left(\cos\frac{2\pi}{3} + i\sin\frac{2\pi}{3}\right) \]

### Step 3: Apply De Moivre's Theorem
Using De Moivre's Theorem, we can raise the polar form to the 48th power:
\[ (-1 + i\sqrt{3})^{48} = \left(2\right)^{48}\left(\cos\left(48 \cdot \frac{2\pi}{3}\right) + i\sin\left(48 \cdot \frac{2\pi}{3}\right)\right) \]

Calculating \(2^{48}\):
\[ 2^{48} = 2^{48} \]

Now, we calculate the angle:
\[ 48 \cdot \frac{2\pi}{3} = 32\pi \]
Since \(\cos\) and \(\sin\) are periodic with a period of \(2\pi\), we can reduce \(32\pi\) modulo \(2\pi\):
\[ 32\pi \mod 2\pi = 0 \]

### Step 4: Evaluate cosine and sine
Now we evaluate:
\[ \cos(0) = 1 \quad \text{and} \quad \sin(0) = 0 \]

### Step 5: Combine results
Putting it all together:
\[ (-1 + i\sqrt{3})^{48} = 2^{48} \cdot (1 + 0i) = 2^{48} \]

### Final Answer
Thus, the value of \((-1 + i\sqrt{3})^{48}\) is:
\[ \boxed{2^{48}} \]

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