# Impedance equivs and complex numbers.

1. Jan 23, 2006

### seang

Hey, I'm finding equivalent impedances of circuits, and I always run into things like this:

1/(-j25) + 1/(600 +j900) = 1/Zeq

I don't know how to proceed from here. I know this is more of a math issue than anything else, but I appreciate your help

2. Jan 23, 2006

### Staff: Mentor

Just put them over a common denominator and add. Remember that j*j=-1 and you should be fine.

3. Jan 23, 2006

### phantom_photon

What works best for me is to rationalise the denominators (by multiplying numerator and denominator by the complex conjugate of the denominator) then simply adding real and imaginary parts. After rationalising, the denominator will be the square of the modulus while the numerator will be the conjugate. Also, remember that 1/j = -j.
So for your example you would go:
1/(-j25) + 1/(600 +j900) = j/25 + (600 - 900j)/(600^2 + 900^2) = 600/1170000 + (1/25 - 900/1170000)j = 1/1950 + j51/1300 = 1/Z
-> Z = 1/(1/1950 + j51/1300) = (1/1950 - j51/1300)/(1/1950^2 + (51/1300)^2) = 0.3331 - 25.49j
Alternatively, you could convert the cartesian impedances into polar admitances then add them (admitances in parallel add) and put the result into polar form and invert to obtain the equivalent impedance.
There are many ways of doing these calculations. I'm not sure which is quickest.