Given total impedance find impedances for parallel RC

Question

How do you find impedances for parallel RC circuit given a total impedance?

More specifically, given this RC parallel combination:

$$Z = R\ ||\ \frac{1}{j\omega C} = \frac{R}{(1 + j\omega RC)}$$

And total impedance:

$$Z = 5.993 - j1.356\ M\Omega$$

How do you end up with these solutions for \$R\$ and \$C\$?

$$R = \frac{\Re(Z)^2 + \Im(Z)^2}{\Re(Z)} = \frac{5.993^2 + 1.356^2}{5.993} = 6.3\text{ M}\Omega$$

$$C = \frac{\Im(Z)^2}{\omega[\Re(Z)^2 + \Im(Z)^2]} = \frac{1.356}{ 2\pi f(5.993^2 + 1.356^2)} = 2.85\mu\text{F}\ \ (f=2kHz)$$

I don't understand transition in between those statements.

Answer 1

It's fairly simple complex number algebra.
Let \$x = Real[Z_x]\$
and \$y = Imag[Z_x]\$
then \$x+jy = R/(1+j\omega RC)\$
multiply through by \$(1+jwRC)\$
\$x+jy+jx\omega RC-\omega yRC=R\$
Then collect real and imaginary parts
\$x-\omega yRC-R=0 - (1)\$
and dividing by j
\$y+x\omega RC=0 - (2)\$
Hence from (1)
\$R=x/(\omega yC-1) - (3)\$
and substituting this in (2) gives
\$y + x^2\omega C/(\omega yC-1)=0\$
Multiplying by \$(\omega yC-1)\$
\$\omega y^2C-y+x^2\omega C=0\$
And hence \$C=y/w(x^2+y^2)\$
Substituting C back into (3) gives
\$R=x/(\omega y.y/\omega (x^2+y^2)-1)\$
Which simplifies to \$R=(x^2+y^2)/x\$
QED.