My Math Forum cartesian and polar complex numbers help please
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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: 7,923 Thanks: 1123 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: 2,092 Thanks: 852 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

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