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 February 17th, 2016, 10:20 AM #1 Newbie   Joined: Feb 2016 From: uk Posts: 8 Thanks: 3 cartesian and polar complex numbers help please hi guys, i have a few questions and was wondering if you could help me out and explain the processes involved: the power of J: the expression (2+j5) . . . . . . . . . (3+j)(-4-j) can be expressed in a single number form (a + jb) as -0.335 - j0.241 but i would like to know the process of how you get to that answer. also i would like to know how to turn that equation into polar form any input would be much appreciated. thanks in advance February 17th, 2016, 12:58 PM #2 Global Moderator   Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,958 Thanks: 1146 Math Focus: Elementary mathematics and beyond \displaystyle \begin{align*}-\dfrac{2+5j}{(3+j)(4+j)}&=-\dfrac{2+5j}{11+7j}\cdot\dfrac{11-7j}{11-7j} \\ &=-\dfrac{(2+5j)(11-7j)}{170} \\ &=-\dfrac{57+41j}{170}\end{align*} Thanks from thundertiger February 17th, 2016, 02:24 PM #3 Math Team   Joined: May 2013 From: The Astral plane Posts: 2,256 Thanks: 926 Math Focus: Wibbly wobbly timey-wimey stuff. Given $\displaystyle z = a + ib$: Let $\displaystyle a + ib = re^{i \theta} = r~cos( \theta ) + ir~sin( \theta )$ Note that since $\displaystyle a = r~cos( \theta )$ and $\displaystyle b = r~sin( \theta )$ we know that $\displaystyle a^2 + b^2 = r^2$ and $\displaystyle tan( \theta ) = \frac{b}{a}$. So what does this give you for your problem? -Dan Thanks from greg1313 February 18th, 2016, 05:23 AM   #4
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Quote:
 Originally Posted by greg1313 \displaystyle \begin{align*}-\dfrac{2+5j}{(3+j)(4+j)}&=-\dfrac{2+5j}{11+7j}\cdot\dfrac{11-7j}{11-7j} \\ &=-\dfrac{(2+5j)(11-7j)}{170} \\ &=-\dfrac{57+41j}{170}\end{align*}
Thundertiger, do you see where greg1313 got "11+ 7j"?

You could also do this "one step at a time", rationalizing the denominator:
$\displaystyle \frac{2+ 5j}{(3+ j)(-4- j)}\frac{3- j}{3- j}= \frac{11+ 13j}{(10)(-4-j)}$

and then $\displaystyle \frac{11- 13j}{10(-4- j)}\frac{-4+ j}{-4+ j}= \frac{-57- 41j}{10(17)}= \frac{-57- 41j}{170}$ February 24th, 2016, 03:49 PM #5 Newbie   Joined: Feb 2016 From: uk Posts: 8 Thanks: 3 Thanks for the reply guys, gregs post was pretty useful, I did manage to get to the bottom of it all in the end. Much appreciated! Thanks from greg1313 Tags cartesian, complex, numbers, polar Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post chocolate Algebra 1 December 11th, 2012 06:03 AM chocolate Algebra 10 November 21st, 2012 10:16 AM ChloeG Calculus 1 March 6th, 2011 01:15 PM jakeward123 Complex Analysis 5 February 9th, 2011 08:15 AM blackobisk Algebra 1 April 16th, 2009 05:39 PM

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