Express each of the following complex numbers in polar form a. \( -\sqrt{6}-\sqrt{2} i \) b. \( -3 i \) c. \( 5+2 i \)

To express the complex numbers in polar form, we need to find their moduli and arguments. For \( -\sqrt{6}-\sqrt{2} i \): 1. **Modulus**: \[ r = \sqrt{(-\sqrt{6})^2 + (-\sqrt{2})^2} = \sqrt{6 + 2} = \sqrt{8} = 2\sqrt{2} \] 2. **Argument**: Since both components are negative, the complex number lies in the third quadrant. \[ \theta = \tan^{-1}\left(\frac{-\sqrt{2}}{-\sqrt{6}}\right) = \tan^{-1}\left(\frac{\sqrt{2}}{\sqrt{6}}\right) = \tan^{-1}\left(\frac{1}{\sqrt{3}}\right) = \frac{\pi}{6} \] Adding \(\pi\) for the third quadrant: \[ \theta = \frac{\pi}{6} + \pi = \frac{7\pi}{6} \] Thus, in polar form: \[ -\sqrt{6}-\sqrt{2} i = 2\sqrt{2} \text{cis} \frac{7\pi}{6} \] For \( -3 i \): 1. **Modulus**: \[ r = \sqrt{0^2 + (-3)^2} = \sqrt{9} = 3 \] 2. **Argument**: This number lies on the negative imaginary axis, so \(\theta = -\frac{\pi}{2}\) or \( \frac{3\pi}{2} \). Thus, in polar form: \[ -3 i = 3 \text{cis} \left(-\frac{\pi}{2}\right) \] For \( 5 + 2 i \): 1. **Modulus**: \[ r = \sqrt{5^2 + 2^2} = \sqrt{25 + 4} = \sqrt{29} \] 2. **Argument**: \[ \theta = \tan^{-1}\left(\frac{2}{5}\right) \] Thus, in polar form: \[ 5 + 2 i = \sqrt{29} \text{cis} \left(\tan^{-1}\left(\frac{2}{5}\right)\right) \] In summary: 1. \( -\sqrt{6}-\sqrt{2} i = 2\sqrt{2} \text{cis} \frac{7\pi}{6} \) 2. \( -3 i = 3 \text{cis} \left(-\frac{\pi}{2}\right) \) 3. \( 5 + 2 i = \sqrt{29} \text{cis} \left(\tan^{-1}\left(\frac{2}{5}\right)\right) \)