Given the complex number, 0 − 2i, how would it be written in polar form?

When calculating "θ" using the arctan function to write the complex number, −2+3i−2+3𝑖 , in polar form, you must do what to find the actual "θ" that exists in the polar form?

To convert complex numbers to polar and rectangular forms, we need to understand the following: 1. **Rectangular Form**: \( z = a + bj \), where \( a \) is the real part and \( b \) is the imaginary part. 2. **Polar Form**: \( z = r(\cos \theta + j \sin \theta) \) or \( z = re^{j\theta} \), where \( r \) is the magnitude and \( \theta \) is the argument (angle). The magnitude \( r \) and the argument \( \theta \) can be found using: \[ r = \sqrt{a^2 + b^2} \] \[ \theta = \tan^{-1}\left(\frac{b}{a}\right) \] Let's convert each expression to both forms: ### f) \( (3 + \sqrt{5}j)(3 - \sqrt{5}j) \) First, simplify the expression: \[ (3 + \sqrt{5}j)(3 - \sqrt{5}j) = 3^2 - (\sqrt{5}j)^2 = 9 - 5j^2 = 9 - 5(-1) = 9 + 5 = 14 \] - **Rectangular Form**: \( 14 \) - **Polar Form**: \( 14(\cos 0 + j \sin 0) \) or \( 14e^{j0} \) ### g) \( \frac{1}{3 + \sqrt{5}j} \) To simplify, multiply the numerator and the denominator by the conjugate of the denominator: \[ \frac{1}{3 + \sqrt{5}j} \cdot \frac{3 - \sqrt{5}j}{3 - \sqrt{5}j} = \frac{3 - \sqrt{5}j}{(3 + \sqrt{5}j)(3 - \sqrt{5}j)} = \frac{3 - \sqrt{5}j}{14} = \frac{3}{14} - \frac{\sqrt{5}}{14}j \] - **Rectangular Form**: \( \frac{3}{14} - \frac{\sqrt{5}}{14}j \) - **Polar Form**: \[ r = \sqrt{\left(\frac{3}{14}\right)^2 + \left(\frac{\sqrt{5}}{14}\right)^2} = \frac{1}{\sqrt{14}} \] \[ \theta = \tan^{-1}\left(\frac{-\sqrt{5}/14}{3/14}\right) = \tan^{-1}\left(\frac{-\sqrt{5}}{3}\right) \] So, the polar form is \( \frac{1}{\sqrt{14}} \left( \cos \theta + j \sin \theta \right) \) or \( \frac{1}{\sqrt{14}} e^{j\theta} \) ### h) \( (1 + j\sqrt{3})(3 + 2j) \) First, expand the expression: \[ (1 + j\sqrt{3})(3 + 2j) = 1 \cdot 3 + 1 \cdot 2j + j\sqrt{3} \cdot 3 + j\sqrt{3} \cdot 2j \] \[ = 3 + 2j + 3j\sqrt{3} + 2j^2\sqrt{3} \] \[ = 3 + 2j + 3j\sqrt{3} + 2(-1)\sqrt{3} \] \[ = 3 + 2j + 3j\sqrt{3} - 2\sqrt{3} \] - **Rectangular Form**: \( 3 - 2\sqrt{3} + (2 + 3\sqrt{3})j \) - **Polar Form**: \[ r = \sqrt{(3 - 2\sqrt{3})^2 + (2 + 3\sqrt{3})^2} \] \[ \theta = \tan^{-1}\left(\frac{2 + 3\sqrt{3}}{3 - 2\sqrt{3}}\right) \] So, the polar form is \( r \left( \cos \theta + j \sin \theta \right) \) or \( r e^{j\theta} \) ### i) \( (1 - j\sqrt{3})(3 - 2j) \) First, expand the expression: \[ (

Express z = 2 + 2i√3 in polar form.

When converting a complex number from rectangular form to polar form, "θ" must always be expressed in radians.

The product of two complex numbers in polar form is found by​ __________ their moduli and​ __________ their arguments.Group of answer choicesadding; multiplyingmultiplying; addingdividing; subtractingsubtracting; dividing