# Multiplying an imaginary number by a constant in AC circuit, Z impedance [closed]

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. .

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

• 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 $5\times j20=5\times j\times 20=j\times 100 = 0+j100$ – JMoravitz Aug 13 '17 at 20:58
• 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 – Emergeon Aug 13 '17 at 21:08
• The topic of complex-values of impedance for RLC circuits has come up in a few previous Questions, such as this one. – hardmath Aug 17 '17 at 2:48

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}

• Indeed no polar conversion is needed by looking at your solution. Gracias, Thank you! – Emergeon Aug 13 '17 at 21:26

$$\frac {5\times 20j}{5+20j}=\frac {20j}{1+4j}=$$ $$\frac {20j}{1+4j}\frac {1-4j}{1-4j}=$$ $$\frac {20 (j+4)}{1^2+4^2}=\frac {80}{17}+j\frac {20}{17}$$

you can finish.

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