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?

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

Given the complex number, 0 − 2i, how would it be written in 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: \[ (

Write the complex number in trigonometric form, once using degrees and once using radians. Begin by sketching the graph to help find the argument 𝜃.−1 + i

Express z = 2 + 2i√3 in polar form.