How does the conversion between admittance and impedance affect phase angle?

You can analyze phasor notation in a quasi-2D Cartesian fashion. The real part is the "x", and the complex part is the "y".

So given a phasor magnitude M with angle Theta,

Using trig: \begin{equation} R = M \cos(\theta)\\ X = M \sin(\theta) \end{equation}

We now have the complex impedance R + Xj

To invert, you can multiply by the complex conjugate (R - Xj) to both the numerator and denominator.

\begin{equation} Y = \frac{R - Xj}{(R + Xj)(R - Xj)} = \frac{R - Xj}{R^2 + X^2} \end{equation}

To compute the magnitude of the admittance, use the distance formula:

\begin{equation} M_Y = \sqrt{\left(\frac{R}{R^2 + X^2}\right)^2 + \left(\frac{-X}{R^2 + X^2}\right)^2} \end{equation}

And the phase of the admittance:

\begin{equation} \theta_Y = \tan^{-1}\left(\frac{-X}{R}\right) \end{equation}

Note that tangent is a bit finicky for computing the phasor angle as you have to be careful about the quadrant. If you're using a computer, they often times have an "atan2" function which takes the x and y coordinates directly and computes the CCW angle from the positive X axis.

A closer look at the phase angle mapping, and it looks like the admittance phase angle is just the reflection of the impedance phase angle about the real/X axis.

For example, an impedance phase angle of 45 degrees is equal to an admittance phase angle of -45 degrees.

And this makes sense if I had used some identities above:

\begin{equation} \theta_Y = -\tan^{-1}\left(\frac{X}{R}\right) = -\theta_X \end{equation}

If I remember correctly, the phase angle just switches the sign and that intensity decreases. So if you had impedance of 10 ohms at 45 degrees, you'd get admittance of around 0.1 siemens at -45 degrees.

We keep in mind that \$ j = \sqrt {-1}\$.

Let's see if I can derive that:

\$ Y=Z^{-1}=\dfrac{1}{R+jX}=\dfrac{1}{R+jX} \dfrac{R-jX}{R-jX}=\dfrac{R-jX}{R^2 + RjX -RjX - j^2X^2}=\dfrac{R-jX}{R^2+X^2}=\dfrac{R}{R^2+X^2}+j\dfrac{-X}{R^2+X^2}=G+jB \$

So the angle switched because the sign in front of the imaginary part switched. The intensity decreased because we got the \$R^2+X^2\$ component. There was no change in angle's absolute value because we decreased both real and imaginary parts by same amount so their ratio stayed the same. Phase angle for admittance is \$ \arctan \left(\frac{B}{G}\right)\$ and since we divided both components by same number, the absolute value of ratio remained constant.

A "quick" way to get the \$R^2+X^2\$ part would be to calculate sine and cosine of the angle multiplied by intensity. So if we have \$Z=\vert Z\vert e^{\alpha}\$ we'd use \$R^2= [\vert Z \vert \cos(\alpha)]^2, X^2=[\vert Z \vert \sin(\alpha)]^2 \$ which should be easy enough to do with a simple calculator. Then we'd add those two together, divide intensity using them and flip the angle so we get:

\$Y= \dfrac{1}{ \vert Z \vert\ e^{a}}= \dfrac { \vert Z \vert}{[\vert Z \vert \cos(\alpha)]^2+[\vert Z \vert \sin(\alpha)]^2}e^{-\alpha} =\dfrac{1}{\vert Z \vert [\cos^2(\alpha)+sin^2(\alpha)]}e^{- \alpha}=\dfrac{1}{\vert Z \vert}e^{- \alpha}\$

which should have been obvious to me from the start.

So final formula for quick conversion is: \$Y= \dfrac{1}{ \vert Z \vert\ e^{a}}=\dfrac{1}{\vert Z \vert}e^{- \alpha}\$