If `(sqrt(3)+i)^(100)=2^(99)(a+ib)`. Then show that `a^(2)+b^(2)=4`

Answer

To solve the problem, we need to show that if \((\sqrt{3} + i)^{100} = 2^{99}(a + ib)\), then \(a^2 + b^2 = 4\).

### Step-by-step Solution:

1. **Express \(\sqrt{3} + i\) in polar form**:
The complex number \(\sqrt{3} + i\) can be expressed in polar form. First, we find the modulus \(r\) and the argument \(\theta\):
\[ r = \sqrt{(\sqrt{3})^2 + (1)^2} = \sqrt{3 + 1} = \sqrt{4} = 2 \]
The argument \(\theta\) can be calculated as:
\[ \theta = \tan^{-1}\left(\frac{1}{\sqrt{3}}\right) = \frac{\pi}{6} \]
Thus, we can write:
\[ \sqrt{3} + i = 2 \left(\cos\frac{\pi}{6} + i\sin\frac{\pi}{6}\right) = 2 \text{cis} \frac{\pi}{6} \]

2. **Raise to the power of 100**:
Using De Moivre's Theorem, we can raise this expression to the power of 100:
\[ (\sqrt{3} + i)^{100} = (2 \text{cis} \frac{\pi}{6})^{100} = 2^{100} \text{cis} \left(\frac{100\pi}{6}\right) = 2^{100} \text{cis} \left(\frac{50\pi}{3}\right) \]

3. **Simplify the argument**:
The angle \(\frac{50\pi}{3}\) can be simplified by subtracting multiples of \(2\pi\):
\[ \frac{50\pi}{3} - 16\pi = \frac{50\pi - 48\pi}{3} = \frac{2\pi}{3} \]
Therefore, we have:
\[ (\sqrt{3} + i)^{100} = 2^{100} \text{cis} \left(\frac{2\pi}{3}\right) \]

4. **Convert back to rectangular form**:
Now, we convert \(\text{cis} \left(\frac{2\pi}{3}\right)\) back to rectangular form:
\[ \text{cis} \left(\frac{2\pi}{3}\right) = \cos\left(\frac{2\pi}{3}\right) + i\sin\left(\frac{2\pi}{3}\right) = -\frac{1}{2} + i\frac{\sqrt{3}}{2} \]
Thus,
\[ (\sqrt{3} + i)^{100} = 2^{100} \left(-\frac{1}{2} + i\frac{\sqrt{3}}{2}\right) = 2^{99}(-1 + i\sqrt{3}) \]

5. **Compare with given expression**:
From the problem statement, we have:
\[ 2^{99}(a + ib) = 2^{99}(-1 + i\sqrt{3}) \]
By comparing the coefficients, we find:
\[ a = -1 \quad \text{and} \quad b = \sqrt{3} \]

6. **Calculate \(a^2 + b^2\)**:
Now, we compute \(a^2 + b^2\):
\[ a^2 + b^2 = (-1)^2 + (\sqrt{3})^2 = 1 + 3 = 4 \]

### Conclusion:
Thus, we have shown that \(a^2 + b^2 = 4\).