Electronic – Resistor math in complex circuit

This is only possible if \$|s|=|s_0|=1\$. Squaring the original inequality gives

$$|s-s_0|^2=|s|^2+|s_0|^2-2\cos(\Delta\phi)\tag{1}\le\epsilon^2$$

where \$\Delta\phi=\arg\{s\}-\arg\{s_0\}\$ is the phase difference between \$s\$ and \$s_0\$. If \$|s|=|s_0|=1\$ is satisfied we get from (1)

$$2(1-\cos(\Delta\phi))\le\epsilon^2$$

or, equivalently,

$$\cos(\Delta\phi)\ge1-\frac{\epsilon^2}{2}\tag{2}$$

From this inequality it follows that

$$|\Delta\phi|\le\arccos\left(1-\epsilon^2/2\right)$$

which is equivalent to the condition in your question.

The formula to convert from the phasor representation \$V=r+jx\$ to its time-domain equivalent is this:

\$ V(t) = Re\{V \, e^{\,j\omega t}\} \$

Using the complex identity:

\$ e^{\,j\omega t} = \cos(\omega t) + j\,\sin(\omega t) \$

You obtain:

\begin{align*} V(t) &= Re\{V \, e^{\,j\omega t}\} = Re\{ (r+jx) \cdot [\cos(\omega t) + j\,\sin(\omega t)] \} = \\[1em] &= Re\{ r \cos(\omega t) - x \sin(\omega t) + j[...] \} = \\[1 em] &= r \cos(\omega t) - x \sin(\omega t) \end{align*}