Given: \( Z=(3-4 j)^{3} \) Convert \( Z \) to polar form. \( \theta \) may always be positive. Solve for \( x \) and \( y \) simultaneously using complex numbers if: \( x-j y=\frac{7-j^{9}}{1-j} \) Evaluate: \( j^{9}-j^{2} \)

To convert \( Z = (3 - 4j)^3 \) to polar form, we first need to find the modulus and argument of the complex number \( 3 - 4j \). 1. **Calculate the modulus**: \[ |3 - 4j| = \sqrt{3^2 + (-4)^2} = \sqrt{9 + 16} = \sqrt{25} = 5 \] 2. **Calculate the argument**: \[ \theta = \tan^{-1}\left(\frac{-4}{3}\right) \] Since the point \( (3, -4) \) is in the fourth quadrant, we can find the angle: \[ \theta = -\tan^{-1}\left(\frac{4}{3}\right) \] To express \( \theta \) as a positive angle, we can add \( 2\pi \): \[ \theta = 2\pi - \tan^{-1}\left(\frac{4}{3}\right) \] 3. **Convert to polar form**: The polar form of \( 3 - 4j \) is: \[ 5 \left( \cos(\theta) + j \sin(\theta) \right) \] Therefore, raising it to the power of 3: \[ Z = (5 \text{cis}(\theta))^3 = 5^3 \text{cis}(3\theta) = 125 \text{cis}(3\theta) \] 4. **Calculate \( 3\theta \)**: \[ 3\theta = 3 \left( 2\pi - \tan^{-1}\left(\frac{4}{3}\right) \right) = 6\pi - 3\tan^{-1}\left(\frac{4}{3}\right) \] Thus, the polar form of \( Z \) is: \[ Z = 125 \text{cis}(6\pi - 3\tan^{-1}\left(\frac{4}{3}\right)) \] Next, we solve for \( x \) and \( y \) in the equation: \[ x - jy = \frac{7 - j^9}{1 - j} \] 5. **Calculate \( j^9 \)**: Since \( j = e^{j\frac{\pi}{2}} \), we have: \[ j^9 = e^{j\frac{9\pi}{2}} = e^{j(4\pi + \frac{\pi}{2})} = e^{j\frac{\pi}{2}} = j \] Therefore: \[ 7 - j^9 = 7 - j \] 6. **Calculate \( 1 - j \)**: To simplify \( \frac{7 - j}{1 - j} \), we multiply the numerator and denominator by the conjugate of the denominator: \[ \frac{(7 - j)(1 + j)}{(1 - j)(1 + j)} = \frac{(7 - j + 7j - j^2)}{1 + 1} = \frac{(7 + 6j + 1)}{2} = \frac{8 + 6j}{2} = 4 + 3j \] 7. **Equate real and imaginary parts**: From \( x - jy = 4 + 3j \), we have: \[ x = 4 \quad \text{and} \quad -y = 3 \implies y = -3 \] Finally, we evaluate \( j^9 - j^2 \): 8. **Calculate \( j^2 \)**: \[ j^2 = -1 \] Therefore: \[ j^9 - j^2 = j - (-1) = j + 1 \] In summary, the results are: - Polar form of \( Z \): \( 125 \text{cis}(6\pi - 3\tan^{-1}\left(\frac{4}{3}\right)) \) - Values of \( x \) and \( y \): \( x = 4, y = -3 \) - Evaluation of \( j^9 - j^2 \): \( j + 1 \)