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I have a hard time calculating the equation shown in the image attached. I'm good with the denominator, but I'm confused on how to convert the numerator in polar.

See attached picture.

More precisely, what is $5\times j20$ in polar?

Could someone please explain the steps in obtaining result?

Thank you for your time. A person preparing for a test tomorrow

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    $\begingroup$ I am more used to seeing things written in the form $a+bi$ rather than $a+jb$ but assuming these mean the same thing, we would have $5\times j20=5\times j\times 20=j\times 100 = 0+j100$ $\endgroup$ – JMoravitz Aug 13 '17 at 20:58
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    $\begingroup$ Yesss!!! Thank you. That was the answer I was looking, and got the same. From this, I replaced this with the numerator, converted the fraction to polar, proceed the calculation and got the same answer as in the picture! Thank you so much $\endgroup$ – Emergeon Aug 13 '17 at 21:08
  • $\begingroup$ The topic of complex-values of impedance for RLC circuits has come up in a few previous Questions, such as this one. $\endgroup$ – hardmath Aug 17 '17 at 2:48
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In polar coordinates you have that
- the numerator is $5 \cdot j20=j100=100 \, e^{j \pi /2}$.
- the denominator is $5 + j20=\sqrt{5^2+20^2}\, e^{j \arctan{(20/4)}}$.

But to rationalize the formula you cite, you do not need to go through polar. Just multiply over and below the fraction for $5-j20$ $$ \eqalign{ & {{5 \cdot j20} \over {5 + j20}} = {{\left( {5 - j20} \right)j100} \over {\left( {5 - j20} \right)\left( {5 + j20} \right)}} = \cr & = {{\left( {5 - j20} \right)j100} \over {\left( {25 + 400} \right)}} = {{2000 + j500} \over {425}} = \cr & = {{80} \over {17}} + j{{20} \over {17}} \cr} $$

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  • $\begingroup$ Indeed no polar conversion is needed by looking at your solution. Gracias, Thank you! $\endgroup$ – Emergeon Aug 13 '17 at 21:26
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$$\frac {5\times 20j}{5+20j}=\frac {20j}{1+4j}=$$ $$\frac {20j}{1+4j}\frac {1-4j}{1-4j}=$$ $$\frac {20 (j+4)}{1^2+4^2}=\frac {80}{17}+j\frac {20}{17} $$

you can finish.

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  • $\begingroup$ That is a much simpler way of solving without using the polar! Thank you! $\endgroup$ – Emergeon Aug 13 '17 at 21:25
  • $\begingroup$ @Emergeon no problem . $\endgroup$ – hamam_Abdallah Aug 13 '17 at 21:33

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