Difference between revisions of "1.49 (a,b,f,i) (ECE301Summer2008asan)" - Rhea

Latest revision as of 07:09, 5 January 2009

Express each of the following complex numbers in polar form, and plot them in the complex plane, indicating the magnitude and angle of each number.

A) $ 1 + j\sqrt{3} $

$ r = \sqrt{1^2 + \sqrt{3}^2} = \sqrt{4} = 2 $

$ \theta = arctan(\sqrt{3}/1) = arctan(\sqrt{3}) = \frac{\pi}{3} $

Therefore the polar form of this complex number is: $ 2e^{j\frac{\pi}{3}} $

B) $ -5 $

$ r = 5 $

$ \theta = \pi $

Therefore the polar form of this complex number is: $ 5e^{j\pi} $

F) $ (1 + j)^{5} $

$ r = \sqrt{1^2 + 1^2} = \sqrt{2} $

$ \theta = \frac{\pi}{4} $

$ (1 + j) = \sqrt{2}e^{j\frac{\pi}{4}} $

$ (1 + j)^{5} = (\sqrt{2}e^{j\frac{\pi}{4}})^{5} = 2^{\frac{5}{2}}e^{j\frac{5\pi}{4}} = 4\sqrt{2}e^{j(\pi + \frac{\pi}{4})} =4\sqrt{2}e^{j\pi}e^{j\frac{\pi}{4}} = -4(\sqrt{2}e^{j\frac{\pi}{4}}) = -4(1 + j) $

Therefore the polar form of this complex number is: $ -4(\sqrt{2}e^{j\frac{\pi}{4}}) $

I) $ \frac{1 + j\sqrt{3}}{\sqrt{3} + j} $

$ r = 2 $

$ Equation 1 = 1 + j\sqrt{3} => \theta_{1} = \frac{\pi}{3} $

$ Equation 2 = \sqrt{3} + j => \theta_{2} = \frac{\pi}{6} $

$ \frac{2e^{j\frac{\pi}{3}}}{2e^{j\frac{\pi}{6}}} = \frac{e^{j\frac{\pi}{3}}}{e^{j\frac{\pi}{6}}} = e^{j(\frac{\pi}{3} - \frac{\pi}{6})} = e^{j\frac{\pi}{6}} $

Therefore the polar form of this complex number is: $ e^{j\frac{\pi}{6}} $