Unit 5

Parametric, Polar & Complex

Three new coordinate systems, all invented because \(y = f(x)\) is a bad description of a great many things. A circle is not a function. A rotating phasor is not a function. For an electrical engineer this unit is where impedance, Nyquist plots, radiation patterns and the DFT all get their language.

1 · Parametric curves

Instead of relating \(y\) to \(x\) directly, give both as functions of a third variable: \(x = f(t)\), \(y = g(t)\). The parameter \(t\) is very often time, and that is the point — the curve now carries information about when and how fast, not just where.

Curve	Parametrisation	Range
Circle, radius \(a\)	\(x = a\cos t,\; y = a\sin t\)	\(0 \le t \le 2\pi\)
Ellipse	\(x = a\cos t,\; y = b\sin t\)	\(0 \le t \le 2\pi\)
Line through \((x_0,y_0)\)	\(x = x_0 + at,\; y = y_0 + bt\)	all \(t\)
Cycloid	\(x = a(t - \sin t),\; y = a(1 - \cos t)\)	rolling circle
Any function	\(x = t,\; y = f(t)\)	the trivial case

Eliminating the parameter

Solve one equation for \(t\) and substitute, or use an identity. For the circle, square and add: \(x^2 + y^2 = a^2(\cos^2 + \sin^2) = a^2\). But note what is lost — direction of travel, speed, and how many times the curve is retraced. The Cartesian equation is a shadow of the parametrisation.

EE: Lissajous figures

Drive an oscilloscope's X input with one sinusoid and its Y input with another: \(x = A\sin(\omega_1 t)\), \(y = B\sin(\omega_2 t + \phi)\). The trace is a parametric curve, and its shape reveals the frequency ratio and the phase difference. Equal frequencies in phase give a diagonal line; equal frequencies 90° apart give a circle; a 2:1 ratio gives a figure eight. Before frequency counters were cheap, this was how you measured a frequency against a reference.

2 · Calculus with parametric equations

\[ \frac{dy}{dx} = \frac{dy/dt}{dx/dt}, \qquad \frac{dx}{dt} \neq 0 \] Chain rule, rearranged. The \(dt\)'s cancel — which is the whole idea.

\[ \frac{d^2y}{dx^2} = \frac{\dfrac{d}{dt}\!\left(\dfrac{dy}{dx}\right)}{\dfrac{dx}{dt}} \] Not \(\dfrac{d^2y/dt^2}{d^2x/dt^2}\). Differentiate the first derivative with respect to \(t\), then divide by \(dx/dt\) again. This is the most-missed formula in the unit.

\[ L = \int_{\alpha}^{\beta} \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2}\;dt \] Speed integrated over time. If \(t\) really is time, the integrand is the particle's speed and the integral is distance travelled.

Arc length counts every pass

If the parametrisation traverses part of the curve twice, the arc-length integral counts it twice. For \(x = \cos t,\; y = \sin t\) with \(0 \le t \le 4\pi\), the answer is \(4\pi\), not \(2\pi\) — you went around twice. Check the range before integrating.

3 · Polar coordinates

Locate a point by distance \(r\) from the origin and angle \(\theta\) from the positive \(x\)-axis.

\[ x = r\cos\theta, \qquad y = r\sin\theta \] \[ r^2 = x^2 + y^2, \qquad \tan\theta = \frac{y}{x} \] Going back to polar, check the quadrant. \(\arctan(y/x)\) always returns an angle in \((-\pi/2, \pi/2)\), so points in quadrants II and III need \(\pi\) added. This is exactly why programming languages provide atan2(y,x).

Polar coordinates are not unique

The point \((1, 0°)\) is also \((1, 360°)\) and \((-1, 180°)\). A negative \(r\) means "go backwards through the origin." This non-uniqueness is why finding where two polar curves intersect is genuinely tricky — solving the equations simultaneously can miss intersections that occur at different \((r,\theta)\) labels for the same point. Always sketch.

Cardioid \(r = 1 + \cos\theta\). The shape of a cardioid microphone's pickup pattern — maximum sensitivity forward, a null at the rear.

Three-petal rose \(r = \cos 3\theta\). Rule: \(r = \cos(n\theta)\) gives \(n\) petals for odd \(n\), and \(2n\) petals for even \(n\).

EE: polar plots are everywhere

Antenna radiation patterns are polar plots of gain versus angle — a dipole's figure-eight is literally \(r = |\cos\theta|\). Nyquist plots trace a transfer function in the complex plane as frequency sweeps. Microphone polar patterns (cardioid, hypercardioid, figure-eight) are named after the curves you are drawing here. When you meet one later, you will already know how to compute the area of a lobe.

4 · Polar area and arc length

\[ A = \frac12\int_{\alpha}^{\beta} r^2\,d\theta \] Slice into thin sectors, not rectangles. A sector of radius \(r\) and angle \(d\theta\) has area \(\tfrac12 r^2 d\theta\) — that is where the one-half comes from, and forgetting it is the classic error.

\[ L = \int_{\alpha}^{\beta} \sqrt{r^2 + \left(\frac{dr}{d\theta}\right)^2}\;d\theta \]

For the area between two polar curves, integrate \(\tfrac12\big(r_{\text{outer}}^2 - r_{\text{inner}}^2\big)\) — the same "difference of squares, not square of difference" rule as washers in Unit 2.

Getting the limits right

The hardest part of a polar area problem is almost never the integral — it is \(\theta_{\text{start}}\) and \(\theta_{\text{end}}\). For one petal of a rose, find where \(r = 0\) on either side of the petal. For a closed curve traced once, do not integrate over \(0\) to \(2\pi\) if the curve already closed at \(\pi\); you will double the answer.

5 · Complex numbers

Three equivalent notations for the same object. Fluency means switching without thinking.

Form	Written	Best for
Rectangular	\(z = a + bi\)	Adding and subtracting
Polar / trig	\(z = r(\cos\theta + i\sin\theta)\)	Seeing the geometry
Exponential	\(z = re^{i\theta}\)	Multiplying, dividing, powers, roots

Modulus \(r = |z|\) is length; argument \(\theta = \arg z\) is angle. The conjugate is the mirror image across the real axis, and \(z\bar z = a^2 + b^2 = |z|^2\) is always real.

Arithmetic

Rule of thumb: **add in rectangular, multiply in exponential.** Convert freely; it is nearly always cheaper than fighting the wrong form.
Operation	Rectangular	Exponential
Add	\((a+c) + (b+d)i\)	— (no nice form)
Multiply	\((ac-bd) + (ad+bc)i\)	\(r_1 r_2 e^{i(\theta_1+\theta_2)}\) — moduli multiply, angles add
Divide	Multiply top and bottom by the conjugate	\(\dfrac{r_1}{r_2}e^{i(\theta_1-\theta_2)}\)
Power	Painful	\(r^{n}e^{in\theta}\) — trivial

6 · De Moivre and roots of unity

\[ \left(re^{i\theta}\right)^{n} = r^{n}e^{in\theta} \implies (\cos\theta + i\sin\theta)^n = \cos n\theta + i\sin n\theta \]

Running it backwards gives roots. Every nonzero complex number has exactly \(n\) distinct \(n\)th roots, evenly spaced around a circle:

\[ z^{1/n} = r^{1/n}\,e^{i(\theta + 2\pi k)/n}, \qquad k = 0, 1, \ldots, n-1 \] Same modulus for all of them; the arguments differ by \(2\pi/n\). They sit at the vertices of a regular \(n\)-gon.

The six solutions of \(z^6 = 1\), spaced 60° apart on the unit circle. Their sum is zero — which is exactly why a balanced polyphase system carries no neutral current, and why DFT basis vectors are orthogonal.

EE: roots of unity are the DFT

The Discrete Fourier Transform is built entirely from \(W_N = e^{-2\pi i/N}\), a primitive \(N\)th root of unity. Its powers are the "twiddle factors" in an FFT, and the algorithm's speed comes from the symmetry visible in the picture above: \(W^{k+N/2} = -W^{k}\), so half the multiplications are the same numbers with a sign flip. Three-phase power is the \(n = 3\) case of the same diagram.

7 · Phasors and impedance

Here is the payoff. A sinusoidal steady-state signal \(v(t) = V_m\cos(\omega t + \phi)\) carries three numbers, but in a linear circuit at one frequency \(\omega\) is shared by everything. Discard it and keep the rest as a single complex number:

\[ V_m\cos(\omega t + \phi) \;\longleftrightarrow\; \mathbf{V} = V_m e^{j\phi} \] The phasor. Justified by Euler's formula from Unit 4: the cosine is the real part of \(V_m e^{j(\omega t + \phi)}\), and the \(e^{j\omega t}\) factor is common to every term, so it divides out.

Because \(\frac{d}{dt}e^{j\omega t} = j\omega\,e^{j\omega t}\), differentiation becomes multiplication by \(j\omega\). Every differential equation in a linear circuit collapses to algebra:

With impedances, Ohm's law \(\mathbf{V} = Z\mathbf{I}\) and the series/parallel rules work exactly as in DC — but with complex numbers. That is the entire reason this unit exists in an engineering syllabus.
Element	Time domain	Impedance \(Z\)	\(\|Z\|\)	Phase
Resistor	\(v = Ri\)	\(R\)	\(R\)	\(0°\)
Inductor	\(v = L\,di/dt\)	\(j\omega L\)	\(\omega L\)	\(+90°\)
Capacitor	\(i = C\,dv/dt\)	\(\dfrac{1}{j\omega C} = -\dfrac{j}{\omega C}\)	\(\dfrac{1}{\omega C}\)	\(-90°\)

The phase column is the physical content: inductor voltage leads current by 90°, capacitor voltage lags by 90°. Multiplying by \(j\) is a \(+90°\) rotation in the complex plane — the geometry and the physics are the same statement.

8 · Worked examples

1. Tangent line to a parametric curve

For \(x = t^2 - 2t\), \(y = t^3 - 3t\), find \(dy/dx\) at \(t = 2\), and locate any vertical tangents.

\[ \frac{dx}{dt} = 2t - 2, \qquad \frac{dy}{dt} = 3t^2 - 3 \] \[ \frac{dy}{dx} = \frac{3t^2-3}{2t-2} = \frac{3(t-1)(t+1)}{2(t-1)} = \frac{3(t+1)}{2}, \quad t \neq 1 \]

At \(t = 2\): \(dy/dx = 9/2\).

Vertical tangents need \(dx/dt = 0\) with \(dy/dt \neq 0\). Here \(dx/dt = 0\) only at \(t = 1\) — but \(dy/dt = 0\) there too, so it is a \(0/0\) case, not a vertical tangent. The cancellation above shows the slope actually approaches \(3\).

That point matters: both derivatives vanishing signals a possible cusp or a smooth point in disguise, and only the limit distinguishes them. Never declare a vertical tangent from \(dx/dt = 0\) alone.

2. Parametric arc length

Find the length of \(x = t^2\), \(y = \tfrac23 t^3\) for \(0 \le t \le 2\).

\[ \frac{dx}{dt} = 2t, \qquad \frac{dy}{dt} = 2t^2 \] \[ L = \int_0^2 \sqrt{4t^2 + 4t^4}\;dt = \int_0^2 2t\sqrt{1+t^2}\;dt \] Factor \(4t^2\) out of the radical: \(\sqrt{4t^2(1+t^2)} = 2|t|\sqrt{1+t^2}\), and \(t \ge 0\) here.

Substitute \(u = 1+t^2\), \(du = 2t\,dt\):

\[ = \int_1^5 \sqrt{u}\;du = \tfrac23\Big[u^{3/2}\Big]_1^5 = \tfrac23\left(5\sqrt5 - 1\right) \approx 6.79 \]

3. Area of one petal of a rose

Find the area enclosed by one petal of \(r = \cos 3\theta\).

Limits first. The petal starts and ends where \(r = 0\): \(\cos 3\theta = 0\) at \(3\theta = \pm\pi/2\), so \(\theta = \pm\pi/6\).

\[ A = \frac12\int_{-\pi/6}^{\pi/6}\cos^2(3\theta)\,d\theta = \frac12\int_{-\pi/6}^{\pi/6}\frac{1+\cos 6\theta}{2}\,d\theta \] \[ = \frac14\left[\theta + \frac{\sin 6\theta}{6}\right]_{-\pi/6}^{\pi/6} = \frac14\left(\frac{\pi}{6} + \frac{\pi}{6}\right) = \frac{\pi}{12} \approx 0.262 \]

At \(\theta = \pi/6\), \(\sin 6\theta = \sin\pi = 0\); likewise at the lower limit, so only the \(\theta\) terms survive.

Sanity check: the petal fits inside a quarter-circle of radius 1 (area \(\pi/4 \approx 0.785\)) and is clearly much thinner. ✓ All three petals together give \(\pi/4\).

4. Complex arithmetic in both forms

Let \(z_1 = 3+4i\) and \(z_2 = 1-i\). Find \(z_1 z_2\) and \(z_1/z_2\).

Product, rectangular:

\[ (3+4i)(1-i) = 3 - 3i + 4i - 4i^2 = 3 + i + 4 = 7 + i \] \(-4i^2 = +4\), since \(i^2 = -1\). That sign flip is where most errors live.

Quotient — multiply top and bottom by the conjugate of the denominator:

\[ \frac{3+4i}{1-i}\cdot\frac{1+i}{1+i} = \frac{3 + 3i + 4i + 4i^2}{1 - i^2} = \frac{-1+7i}{2} = -0.5 + 3.5i \]

Check with the exponential form. \(|z_1| = 5\), \(\arg = \arctan(4/3) = 53.13°\). \(|z_2| = \sqrt2\), \(\arg = -45°\).

Product: modulus \(5\sqrt2 \approx 7.07\), argument \(8.13°\). And \(|7+i| = \sqrt{50} \approx 7.07\), \(\arg = \arctan(1/7) = 8.13°\) ✓
Quotient: modulus \(5/\sqrt2 \approx 3.54\), argument \(98.13°\). And \(|-0.5+3.5i| = \sqrt{12.5} \approx 3.54\) ✓

Note the quotient lands in quadrant II, and its argument is \(98.13°\) — not the \(-81.87°\) a naive \(\arctan(3.5/-0.5)\) would return. Always check the quadrant.

5. De Moivre and roots

Compute \((1+i)^8\), then find all cube roots of \(8i\).

Power. Convert first: \(|1+i| = \sqrt2\), \(\arg = \pi/4\).

\[ (1+i)^8 = (\sqrt2)^8 e^{i\cdot 8\pi/4} = 16\,e^{2\pi i} = 16 \]

Expanding by the binomial theorem would take nine terms for the same answer. This is what "multiply in exponential form" buys you.

Cube roots of \(8i\). Write \(8i = 8e^{i\pi/2}\). Modulus of each root is \(8^{1/3} = 2\):

\[ z_k = 2\,e^{i(\pi/2 + 2\pi k)/3}, \qquad k = 0,1,2 \]

\(k\)	Angle	Rectangular
0	30°	\(\sqrt3 + i\)
1	150°	\(-\sqrt3 + i\)
2	270°	\(-2i\)

Check \(k=2\): \((-2i)^3 = -8i^3 = -8(-i) = 8i\) ✓

The three roots are 120° apart on a circle of radius 2 — a perfect equilateral triangle, which the general formula guarantees.

6. EE — series RLC impedance

A series circuit has \(R = 50\,\Omega\), \(L = 10\) mH, \(C = 1\) µF, driven at \(f = 1000\) Hz. Find the total impedance in rectangular and polar form, and say whether the circuit is inductive or capacitive.

Angular frequency: \(\omega = 2\pi(1000) = 6283\) rad/s.

\[ Z_L = j\omega L = j(6283)(0.010) = j62.8\ \Omega \] \[ Z_C = \frac{-j}{\omega C} = \frac{-j}{(6283)(10^{-6})} = -j159.2\ \Omega \]

Series means add — and addition is why we keep rectangular form:

\[ Z = 50 + j62.8 - j159.2 = 50 - j96.4\ \Omega \] \[ |Z| = \sqrt{50^2 + 96.4^2} = \sqrt{11793} = 108.6\ \Omega \] \[ \theta = \arctan\!\left(\frac{-96.4}{50}\right) = -62.6° \]

The reactance is negative, so the circuit is capacitive at this frequency: current leads voltage by 62.6°. We are below resonance.

Where is resonance? Where the two reactances cancel:

\[ \omega_0 = \frac{1}{\sqrt{LC}} = \frac{1}{\sqrt{10^{-8}}} = 10^{4}\ \text{rad/s} \implies f_0 \approx 1592\ \text{Hz} \]

We are driving below that, so the capacitor dominates — consistent with the negative angle we computed. Two independent routes to the same conclusion.

Checklist

I compute \(d^2y/dx^2\) by differentiating \(dy/dx\) with respect to \(t\), then dividing by \(dx/dt\).
I check the parameter range so arc length does not double-count.
I never forget the \(\tfrac12\) in the polar area formula.
I find polar limits by locating where \(r = 0\) or where curves meet, and I sketch first.
I add in rectangular form and multiply in exponential form.
I check the quadrant before trusting an arctangent.
I can find all \(n\) roots of a complex number and know they are evenly spaced.
I know \(Z_L = j\omega L\) and \(Z_C = 1/(j\omega C)\), and what the sign of the reactance means.