For example, this is the rectangular form of the complex number whose absolute value is   and angle is  : PROBLEM 3.1Current  and

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

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

In the complex numbers, where 𝑖2=–1,2–𝑖–3+𝑖=?i 2 =–1, –3+i2–i​ =?

To solve these complex number expressions in both rectangular (Re + j Im) and polar/exponential (Ae^(jθ)) forms, we need to follow these steps: 1. **Rectangular Form**: Simplify the expression to the form \( a + bj \). 2. **Polar Form**: Convert the rectangular form to polar form using the magnitude \( A = \sqrt{a^2 + b^2} \) and the angle \( \theta = \tan^{-1}(b/a) \). Let's solve each part step by step. ### Part (a) Given: \( 6 - j\sqrt{3} \) **Rectangular Form**: \( 6 - j\sqrt{3} \) **Polar Form**: - Magnitude: \( A = \sqrt{6^2 + (-\sqrt{3})^2} = \sqrt{36 + 3} = \sqrt{39} \) - Angle: \( \theta = \tan^{-1}\left(\frac{-\sqrt{3}}{6}\right) = \tan^{-1}\left(-\frac{\sqrt{3}}{6}\right) \) So, the polar form is \( \sqrt{39} e^{j \tan^{-1}(-\sqrt{3}/6)} \). ### Part (b) Given: \( -8 \) **Rectangular Form**: \( -8 \) **Polar Form**: - Magnitude: \( A = |-8| = 8 \) - Angle: \( \theta = \pi \) (since it's on the negative real axis) So, the polar form is \( 8 e^{j\pi} \). ### Part (c) Given: \( \frac{-2}{\sqrt{5} - 3j} \) **Rectangular Form**: - Multiply numerator and denominator by the conjugate of the denominator: \[ \frac{-2}{\sqrt{5} - 3j} \cdot \frac{\sqrt{5} + 3j}{\sqrt{5} + 3j} = \frac{-2(\sqrt{5} + 3j)}{(\sqrt{5})^2 + (3j)^2} = \frac{-2\sqrt{5} - 6j}{5 + 9} = \frac{-2\sqrt{5} - 6j}{14} = -\frac{\sqrt{5}}{7} - \frac{3j}{7} \] **Polar Form**: - Magnitude: \( A = \sqrt{\left(-\frac{\sqrt{5}}{7}\right)^2 + \left(-\frac{3}{7}\right)^2} = \sqrt{\frac{5}{49} + \frac{9}{49}} = \sqrt{\frac{14}{49}} = \frac{\sqrt{14}}{7} \) - Angle: \( \theta = \tan^{-1}\left(\frac{-3/7}{-\sqrt{5}/7}\right) = \tan^{-1}\left(\frac{3}{\sqrt{5}}\right) \) So, the polar form is \( \frac{\sqrt{14}}{7} e^{j \tan^{-1}(3/\sqrt{5})} \). ### Part (d) Given: \( -3j \) **Rectangular Form**: \( -3j \) **Polar Form**: - Magnitude: \( A = |-3j| = 3 \) - Angle: \( \theta = -\frac{\pi}{2} \) (since it's on the negative imaginary axis) So, the polar form is \( 3 e^{-j\pi/2} \). ### Part (e) Given: \( \frac{5}{(\sqrt{2} - 3j)^2} \) **Rectangular Form**: - First, find \( (\sqrt{2} - 3j)^2 \): \[ (\sqrt{2} - 3j)^2 = (\sqrt{2})^2 - 2(\sqrt{2})(3j) + (3j)^2 = 2 - 6\sqrt{2}j - 9 = -7 - 6\sqrt{2}j \] - Then, \( \frac{5}{-7 - 6\sqrt{2}j} \): \[ \frac{5}{-7 - 6\sqrt{2}j} \cdot \frac{-7 + 6\sqrt{2}j}{-7 + 6\sqrt{2}j} = \frac{5(-7 + 6\sqrt