# Proof of formula to convert from parallel to series impedance

So, I tried to find a proof to the formulas to convert from parallel to series reactive circuit, which I found in the ARRL Handbook:
$R_s=\frac{R_pX_p^2}{R_p^2+X_p^2}$ and $X_s=\frac{R_p^2X_p}{R_p^2+X_p^2}$
where $R_s$ and $X_s$ are the series resistance and reactance to match the impedance of a parallel circuit with $R_p$ and $X_p$ resistance and reactance.
The resistances are positive and real values and the reactances are signed real values (positive for inductive, and negative for capacitive).

I decided to use the complex representation of impedance (I use j for i because it seems to be the convention in electronics):
$Z_s=R_s+jX_s$
and the complex representation of admittance:
$Y_p=G_p+jB_p$
which gives: $\frac{1}{Z_p}=\frac{1}{R_p}+j\frac{1}{X_p}$, then, with some algebric manipulation:
$Z_p=\frac{R_pX_p}{X_p+jR_p}$
After that, I do $Z_s=Z_p$, meaning:
$R_s+jX_s=\frac{R_pX_p}{X_p+jR_p}$
To bring the complex term to the numerator, I multiply by the conjugate of the denominator:
$R_s+jX_s=\frac{R_pX_p}{X_p+jR_p}*\frac{X_p-jR_p}{X_p-jR_p}=\frac{R_pX_p^2-jR_p^2X_p}{R_p^2+X_p^2}=\frac{R_pX_p^2}{R_p^2+X_p^2}+j\frac{-R_p^2X_p}{R_p^2+X_p^2}$
Now, to set real and imaginary parts equal, we get:
$R_s=\frac{R_pX_p^2}{R_p^2+X_p^2}$ and $X_s=-\frac{R_p^2X_p}{R_p^2+X_p^2}$

The first formula is fine, but what is wrong with the second formula? Why is there that extra "-" in front?

It must be like $\dfrac{1}{Z_p} = \dfrac{1}{Rp} + \dfrac{1}{j.X_p}$. In your case, the imaginary $j$ is in numerator.