Apparent power in exponential form
The complex power \$S\$ is defined as

$$S = P + jQ = \frac{1}{2}\vec V \vec I^* = \tilde{V} \tilde{I^*}$$

where \$\vec V\$ is a peak phasor voltage and \$\tilde{V}\$ denotes a rms phasor voltage; the relationship between rms phasors and peak phasors being

$$\tilde{V} = \frac{\vec V}{\sqrt{2}} $$

Carefully note that the complex power is (proportional to) the product of the voltage phasor and the conjugate current phasor.

Now, Ohm's Law for phasors is

$$\vec V = \vec IZ$$

Thus, for a circuit element with impedance \$Z\$, the associated complex power is

$$S =  P + jQ = \frac{1}{2}\vec V \vec I^* = \frac{1}{2}\vec V \frac{\vec V^*}{Z^*} = \frac{|\vec V|^2}{2Z^*}$$

For \$\vec V = |V|e^0\$ and \$Z = |Z|e^{j\psi} \$, the complex power is

$$S =  P + jQ = \frac{|\vec V|^2}{2Z^*} = \frac{|V|^2}{2|Z|e^{-j\psi}} =  \frac{|V|^2}{2|Z|}e^{j\psi}$$

See that, when we conjugate the impedance, the sign of the phase changes from plus to minus.  Then, since the (conjugate) impedance is in the denominator, we factor out the phase by bringing to the numerator with yet another sign change from minus to plus.