Questions: Write the complex number in polar form with argument θ between 0 and 2π. -√6+√2 i

Solution Steps

The complex number is given as \(-\sqrt{6} + \sqrt{2}i\). Here, the real part is \(-\sqrt{6}\) and the imaginary part is \(\sqrt{2}\).

The modulus \(r\) of a complex number \(a + bi\) is given by: \[ r = \sqrt{a^2 + b^2} \] Substituting \(a = -\sqrt{6}\) and \(b = \sqrt{2}\): \[ r = \sqrt{(-\sqrt{6})^2 + (\sqrt{2})^2} = \sqrt{6 + 2} = \sqrt{8} = 2\sqrt{2} \]

The argument \(\theta\) is given by: \[ \theta = \arctan\left(\frac{b}{a}\right) \] Substituting \(a = -\sqrt{6}\) and \(b = \sqrt{2}\): \[ \theta = \arctan\left(\frac{\sqrt{2}}{-\sqrt{6}}\right) = \arctan\left(-\frac{1}{\sqrt{3}}\right) \] Since the real part is negative and the imaginary part is positive, the complex number lies in the second quadrant. Therefore: \[ \theta = \pi - \arctan\left(\frac{1}{\sqrt{3}}\right) = \pi - \frac{\pi}{6} = \frac{5\pi}{6} \]

The polar form of a complex number is given by: \[ z = r(\cos\theta + i\sin\theta) \] Substituting \(r = 2\sqrt{2}\) and \(\theta = \frac{5\pi}{6}\): \[ z = 2\sqrt{2}\left(\cos\left(\frac{5\pi}{6}\right) + i\sin\left(\frac{5\pi}{6}\right)\right) \]

Final Answer