Express \(z=-3+2 \mathrm{j}\) in polar form and hence find \(z^{6}\), converting your answer into cartesian form.

To convert the given complex number -\(3+2j\) into polar form, we can find the modulus and argument of the complex number. The modulus is the magnitude of the complex number, given by \(\rho = \sqrt{x^2 + y^2}\), where \(x=-3\) and \(y=2\) \[\rho = \sqrt{(-3)^2 + (2)^2} = \sqrt{9+4} = \sqrt{13}\] The argument \(\theta\) is the angle formed by the line connecting the origin \((0,0)\) and the point \((-3,2)\) in the complex plane. To find the argument, we can use the arctangent formula \(\theta = \arctan{\frac{y}{x}}\), where \(x=-3\) and \(y=2\). Keep in mind that since \(x\) is negative, we will need to add \(\pi\) radians (180°) to the angle. \[\theta = \arctan{\frac{2}{-3}} + \pi \] Now we can write the complex number in polar form as \(z = \rho (\cos{\theta} + j\sin{\theta})\).

De Moivre's theorem states that for any complex number \(z = \rho (\cos{\theta} + j\sin{\theta})\) and any integer \(n\), \[z^n = \rho^n (\cos{(n\theta)} + j\sin{(n\theta)})\] Using De Moivre's theorem, we can find \(z^6\) as follows: \[z^6 = (\sqrt{13})^6 (\cos{(6(\arctan{\frac{2}{-3}} + \pi))} + j\sin{(6(\arctan{\frac{2}{-3}} + \pi))})\]

To convert the complex number back to Cartesian form, we can use the formula \(z = x+ jy\), where \(x\) and \(y\) are given by: \[x = \rho^n \cos{(n\theta)}\] \[y = \rho^n \sin{(n\theta)}\] Therefore, we will compute \(x\) and \(y\) and then write the final result. \[x = (\sqrt{13})^6 \cos{(6(\arctan{\frac{2}{-3}} + \pi))}\] \[y = (\sqrt{13})^6 \sin{(6(\arctan{\frac{2}{-3}} + \pi))}\] Finally, the sixth power of the complex number in Cartesian form is \(z^6 = x+ jy\).