= Peak amplitude phasor
{title2=$v(t)=\operatorname{Re}(\widehat V e^{i\omega t})$}
= Peak phasor
{synonym}
A <peak amplitude phasor> is the <complex amplitude> that reconstructs a real single-frequency <sinusoid> without an extra amplitude multiplier:
$$
\widehat V=V_{\rm pk}e^{i\phi},\qquad
v(t)=\operatorname{Re}(\widehat V e^{i\omega t}),\qquad\omega>0.
$$
Its <modulus> is the sinusoid's peak <amplitude>, and its <complex argument> is the <wave phase> relative to the chosen cosine reference. This is the default convention in <phasor>. MIT's sinusoidal notes explicitly describe the cosine reference and phase/amplitude parameters.<6.200 Notes: Sinusoidal Steady State by Karl K. Berggren>[Printed pp. 1–2, Figure 1 and Phase/Amplitude; pp. 10–11, complex exponential solution]{ref}
For an original example, take $v(t)=3\cos\omega t-4\sin\omega t$ volts. Expanding the real part shows that $\widehat V=3+4i$ volts, with peak <amplitude> $5$ volts and <wave phase> $\arctan(4/3)\approx53.13^\circ$. The corresponding <root mean square phasor> is $(3+4i)/\sqrt2$ volts, whose magnitude is about $3.536$ volts. The physical signal is the same under both conventions.
The <peak phasor versus positive-frequency Fourier coefficient> distinction follows from
$$
v(t)=\frac12\widehat V e^{i\omega t}
+\frac12\widehat V^{\,*}e^{-i\omega t}.
$$
Thus the positive-frequency coefficient in this two-exponential expansion is $\widehat V/2$, not $\widehat V$. A real sinusoid requires its conjugate negative-frequency term as well. This algebra does not assign ordinary Fourier-transform function values to an infinite-duration sinusoid; those require the corresponding distributional interpretation.
A common time-origin change multiplies both voltage and current <peak phasors> by the same unit-modulus factor, leaving their relative phase and impedance ratio unchanged. <Peak-versus-RMS normalization in complex electrical power> is still essential when an amplitude convention changes rather than just the time origin.
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