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February 17th, 2016, 10:20 AM   #1
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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
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February 17th, 2016, 12:58 PM   #2
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$\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} \\
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February 17th, 2016, 02:24 PM   #3
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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?

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February 18th, 2016, 05:23 AM   #4
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Originally Posted by greg1313 View Post
$\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} \\
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 $\displaystyle \frac{11- 13j}{10(-4- j)}\frac{-4+ j}{-4+ j}= \frac{-57- 41j}{10(17)}= \frac{-57- 41j}{170}$
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February 24th, 2016, 03:49 PM   #5
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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!
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