A sinusoid has two independent properties, amplitude and phase. A single real number cannot carry both; a complex number carries exactly two.
Rectangular (\(Re + jIm\)) for addition and subtraction. Phasor (polar) (\(A\angle\phi\)) for multiplication and division.
\(A = \sqrt{Re^2+Im^2}\) and \(\phi = \tan^{-1}(Im/Re)\). Check the quadrant — \(\tan^{-1}\) cannot distinguish quadrant II from IV.
\(v(t) = V_m\cos(2\pi f t + \phi) \longleftrightarrow V = V_m\angle\phi\). The frequency never changes, so only magnitude and angle travel. A phasor carries whichever kind of magnitude the problem gives you — an amplitude or an RMS value — and the answer comes back in the same kind; \(\sqrt{2}\) only converts between them.
Capacitor: current leads by \(90^\circ\). Inductor: current lags by \(90^\circ\). Check it against the sign of \(Z_L = +j\omega L\).
\(Z_R = R\), \(Z_C = 1/(j\omega C) = -j/(\omega C)\), \(Z_L = j\omega L\), with \(\omega = 2\pi f\).
At DC, \(|Z_C| \to \infty\) — an open circuit. At high frequency, \(|Z_C| \to 0\) — a short circuit.
Exactly the reverse of a capacitor: a short at DC and an open at high frequency.
Nothing. Series impedances add, parallel is product over sum, and Ohm's Law and the divider rules apply unchanged. Only the arithmetic is complex. One Lesson 4 habit does not carry over: two opposite reactances in parallel can be larger than either branch.
Because \(Z_C\) and \(Z_L\) both contain \(\omega\). That frequency dependence is precisely what makes filters possible.
The angle unit — degrees versus radians. A calculator left in the wrong mode produces an answer that looks entirely reasonable and is wrong.
\(Z_\text{EQ} = 1000 - j1592 = 1880\angle{-}57.9^\circ\ \Omega\). The angle is negative, so the circuit is capacitive and the current leads: \(\phi = -57.9^\circ\) and \(\text{pf} = \cos\phi = 0.532\).