(3) \( \frac{(-1+i)^{4}}{(\sqrt{3}-i)^{10}}+ \)

To simplify the expression \( \frac{(-1+i)^{4}}{(\sqrt{3}-i)^{10}} \), we will first simplify the numerator and the denominator separately. ### Step 1: Simplify the numerator \( (-1+i)^{4} \) We can express \( -1+i \) in polar form. The modulus \( r \) is calculated as follows: \[ r = | -1 + i | = \sqrt{(-1)^2 + 1^2} = \sqrt{1 + 1} = \sqrt{2} \] Next, we find the argument \( \theta \): \[ \theta = \tan^{-1}\left(\frac{1}{-1}\right) = \tan^{-1}(-1) = \frac{3\pi}{4} \quad (\text{since } -1+i \text{ is in the second quadrant}) \] Thus, we can write: \[ -1+i = \sqrt{2} \left( \cos\frac{3\pi}{4} + i \sin\frac{3\pi}{4} \right) \] Now, we raise this to the power of 4: \[ (-1+i)^{4} = \left( \sqrt{2} \right)^{4} \left( \cos\left(4 \cdot \frac{3\pi}{4}\right) + i \sin\left(4 \cdot \frac{3\pi}{4}\right) \right) \] Calculating \( \left( \sqrt{2} \right)^{4} \): \[ \left( \sqrt{2} \right)^{4} = 2^{2} = 4 \] Now, calculate the angle: \[ 4 \cdot \frac{3\pi}{4} = 3\pi \] Thus, we have: \[ \cos(3\pi) = -1, \quad \sin(3\pi) = 0 \] So, \[ (-1+i)^{4} = 4(-1 + 0i) = -4 \] ### Step 2: Simplify the denominator \( (\sqrt{3}-i)^{10} \) Next, we express \( \sqrt{3}-i \) in polar form. The modulus \( r \) is: \[ r = | \sqrt{3} - i | = \sqrt{(\sqrt{3})^2 + (-1)^2} = \sqrt{3 + 1} = \sqrt{4} = 2 \] Now, we find the argument \( \theta \): \[ \theta = \tan^{-1}\left(\frac{-1}{\sqrt{3}}\right) = -\frac{\pi}{6} \quad (\text{since } \sqrt{3}-i \text{ is in the fourth quadrant}) \] Thus, we can write: \[ \sqrt{3}-i = 2 \left( \cos\left(-\frac{\pi}{6}\right) + i \sin\left(-\frac{\pi}{6}\right) \right) \] Now, we raise this to the power of 10: \[ (\sqrt{3}-i)^{10} = \left( 2 \right)^{10} \left( \cos\left(10 \cdot -\frac{\pi}{6}\right) + i \sin\left(10 \cdot -\frac{\pi}{6}\right) \right) \] Calculating \( \left( 2 \right)^{10} \): \[ \left( 2 \right)^{10} = 1024 \] Now, calculate the angle: \[ 10 \cdot -\frac{\pi}{6} = -\frac{10\pi}{6} = -\frac{5\pi}{3} \] Since \( -\frac{5\pi}{3} \) is equivalent to \( \frac{\pi}{3} \) (adding \( 2\pi \)), we have: \[ \cos\left(-\frac{5\pi}{3}\right) = \cos\left(\frac{\pi}{3}\right) = \frac{1}{2}, \quad \sin\left(-\frac{5\pi}{3}\right) = \sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2} \] Thus, \[ (\sqrt{3}-i)^{10} = 1024 \left( \frac{1}{