91Ó°ÊÓ

First, we need to find the polar form of the complex number \(-3+3j\). The modulus can be calculated using the formula \(r = \sqrt{a^2 + b^2}\), where \(a\) and \(b\) represent the real and imaginary parts, respectively: \[r = \sqrt{(-3)^2 + (3)^2} = \sqrt{18}\] Now, we find the argument or the angle \(\theta\), which can be calculated using the formula \(\theta = tan^{-1}(\frac{b}{a})\) (and adjusting for the appropriate quadrant): \[\theta = tan^{-1}(\frac{3}{-3}) = tan^{-1}(-1)\] The angle lies in the second quadrant because both \(a\) and \(b\) are positive, so we should add \(\pi\) to the calculated angle: \[\theta = \pi + tan^{-1}(-1)\] Thus, the polar form of the complex number is: \[z = \sqrt{18}(\cos(\pi + tan^{-1}(-1)) + j\,sin(\pi + tan^{-1}(-1)))\]

To find the fifth roots of the complex number, we can apply De Moivre's theorem. For a given complex number \(z\) in polar form and a positive integer \(n\), the roots can be found using the following formula: \[z_k = r^{\frac{1}{n}}[\cos(\frac{\theta + 2k\pi}{n}) + j\,sin(\frac{\theta + 2k\pi}{n})]\] where \(k = 0, 1, 2, 3, \cdots, n - 1\). For our complex number and \(n = 5\), the fifth roots can be calculated as follows: \[z_k = (\sqrt{18})^{\frac{1}{5}}[\cos(\frac{\pi + tan^{-1}(-1) + 2k\pi}{5}) + j\,sin(\frac{\pi + tan^{-1}(-1) + 2k\pi}{5})]\] Now, we apply the formula for \(k = 0, 1, 2, 3, 4\): \[z_0 = (\sqrt{18})^{\frac{1}{5}}[\cos(\frac{\pi + tan^{-1}(-1)}{5}) + j\,sin(\frac{\pi + tan^{-1}(-1)}{5})]\] \[z_1 = (\sqrt{18})^{\frac{1}{5}}[\cos(\frac{\pi + tan^{-1}(-1) + 2\pi}{5}) + j\,sin(\frac{\pi + tan^{-1}(-1) + 2\pi}{5})]\] \[z_2 = (\sqrt{18})^{\frac{1}{5}}[\cos(\frac{\pi + tan^{-1}(-1) + 4\pi}{5}) + j\,sin(\frac{\pi + tan^{-1}(-1) + 4\pi}{5})]\] \[z_3 = (\sqrt{18})^{\frac{1}{5}}[\cos(\frac{\pi + tan^{-1}(-1) + 6\pi}{5}) + j\,sin(\frac{\pi + tan^{-1}(-1) + 6\pi}{5})]\] \[z_4 = (\sqrt{18})^{\frac{1}{5}}[\cos(\frac{\pi + tan^{-1}(-1) + 8\pi}{5}) + j\,sin(\frac{\pi + tan^{-1}(-1) + 8\pi}{5})]\]

Finally, we need to convert these fifth roots from polar form to exponential form: \[z_k = (\sqrt{18})^{\frac{1}{5}}e^{j\frac{\theta + 2k\pi}{5}}\] Thus, the fifth roots of \(-3 + 3j\) in exponential form are: \[z_0 = (\sqrt{18})^{\frac{1}{5}}e^{j\frac{\pi + tan^{-1}(-1)}{5}}\] \[z_1 = (\sqrt{18})^{\frac{1}{5}}e^{j\frac{\pi + tan^{-1}(-1) + 2\pi}{5}}\] \[z_2 = (\sqrt{18})^{\frac{1}{5}}e^{j\frac{\pi + tan^{-1}(-1) + 4\pi}{5}}\] \[z_3 = (\sqrt{18})^{\frac{1}{5}}e^{j\frac{\pi + tan^{-1}(-1) + 6\pi}{5}}\] \[z_4 = (\sqrt{18})^{\frac{1}{5}}e^{j\frac{\pi + tan^{-1}(-1) + 8\pi}{5}}\]