Multiplying an imaginary number by a constant in AC circuit, Z impedance

Question

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.

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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

Answer 1

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 $5-j20$ $$ \eqalign{ & {{5 \cdot j20} \over {5 + j20}} = {{\left( {5 - j20} \right)j100} \over {\left( {5 - j20} \right)\left( {5 + j20} \right)}} = \cr & = {{\left( {5 - j20} \right)j100} \over {\left( {25 + 400} \right)}} = {{2000 + j500} \over {425}} = \cr & = {{80} \over {17}} + j{{20} \over {17}} \cr} $$

Answer 2

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

you can finish.