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Complex Numbers and solving a parallel arrangement of impedances

I have the answer but I do not understand part of the working out (where it changes the sign and squares the bottom - after the third "="), someone please explain.(I posted the circuit and question in an attachment in the thread so can be read clearer)

1/Zt = 1/Z1 + 1/Z2 + 1/Z3 = 1/30-j20 + 1/40+j50 + 1/25 = (30+j20/30^2+20^) + (40-j50/40^2+50^2) + 1/25

=( 30/1300) + (j20/1300) + (40/4100) - (j50/4100) + 1/25

For the circuit shown, determine the current I flowing and its phase relative to the applied voltage.

Please help guys, really interested as to why this is!

(edited 5 years ago)

Reply 1

mercedes36

Attached the question/working out with the circuit here

Circuit.PNG39.9KB

Reply 2

username3249896

Original post by mercedes36

Attached the question/working out with the circuit here

It's multiplying top/bottom by conjugate of denominator to make the denominator real (similar to rationalising for surds)

so, noting that (a-bj)(a+bj) = a^2 + b^2, we have