Applications: The Complex Plane

The modulus of the impedance is its length on the Argand diagram and the argument is its angle with the real axis. In polar form a division is quick: the moduli are divided and the arguments are subtracted, so the size and the phase of the current come out separately.

1. Plot Z = 12 + 5i on an Argand diagram: 12 along the real axis for the resistance and 5 up for the reactance. By Pythagoras its modulus is |Z| = √12² + 5² = √169 = 13 ohms.
2. Z lies in the first quadrant, so its argument is arg Z = tan⁻¹ 512 = 22.6^° to 1 decimal place.
3. (a) In polar form, Z = 13(cos 22.6^° + i sin 22.6^°) ohms.
4. To divide in polar form, divide the moduli and subtract the arguments. The voltage has modulus 260 and argument 0, so |I| = 26013 = 20 and arg I = 0 − 22.6^° = −22.6^°.
5. (b) The current is 20 amperes, and it lags behind the voltage by 22.6^°. Check: cos 22.6^° = 1213 and sin 22.6^° = 513, so I = 20(1213 − 513i), and I × Z = 2013(12 − 5i)(12 + 5i) = 2013 × 169 = 260 volts.

answer(a) Z = 13(cos 22.6^° + i sin 22.6^°) ohms; (b) 20 amperes, lagging behind the voltage by 22.6^°

techniqueDividing in Polar Form · The Modulus of a Complex Number · The Argument of a Complex Number

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Polar form gives the range as the modulus r and the direction as the argument θ, measured counterclockwise from east. The real part is r cos θ and the imaginary part is r sin θ; going the other way, the modulus comes from Pythagoras and the argument from the inverse tangent, with the quadrant read off the signs.

1. For A the modulus is 10 km and the argument is 120^°. The real part is 10 cos 120^° = 10 × (−12) = −5.
2. The imaginary part is 10 sin 120^° = 10 × √32 = 5√3 ≈ 8.66. (a) So A = −5 + 8.66i: A is 5 km west and 8.66 km north of the radar.
3. For B = 6 − 8i the modulus is √6² + 8² = √100 = 10 km.
4. B has a positive real part and a negative imaginary part, so it is in the fourth quadrant, and its argument is −tan⁻¹ 86 = −53.1^°. (b) B = 10(cos(−53.1^°) + i sin(−53.1^°)).
5. Check: 10 cos 53.1^° = 6.00 and 10 sin 53.1^° = 8.00, which gives back 6 − 8i. Both contacts have modulus 10, so both lie on the radar's 10 km range ring.

answer(a) 5 km west and 5√3 ≈ 8.66 km north; (b) B = 10(cos(−53.1^°) + i sin(−53.1^°))

techniquePolar Form · The Form r(cos θ + i sin θ) · The Argand Diagram

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After n stops the car is at 20wⁿ. De Moivre's theorem says (cos θ + i sin θ)ⁿ = cos nθ + i sin nθ, so the power is a turn through n times the angle. The imaginary part is the height above the hub.

1. After n stops the car is at 20wⁿ, and by De Moivre's theorem wⁿ = cos nπ6 + i sin nπ6.
2. For 5 stops, w⁵ = cos 5π6 + i sin 5π6 = −√32 + 12i.
3. (a) The car is at 20w⁵ = −10√3 + 10i ≈ −17.32 + 10i. It is 10 m above the hub, so 22 + 10 = 32 m above the ground.
4. For 16 stops the angle is 16π6 = 8π3 = 2π + 2π3. A whole turn of 2π brings the car back to where it was, so w¹⁶ = cos 2π3 + i sin 2π3 = −12 + √32i.
5. (b) The car is at 20w¹⁶ = −10 + 10√3i ≈ −10 + 17.32i, so it is 22 + 17.32 = 39.3 m above the ground. Check: 12 stops make one full turn, and 16 − 12 = 4 stops of π6 make 2π3, the same angle.

answer(a) 32 m above the ground; (b) 22 + 10√3 ≈ 39.3 m above the ground

techniqueDe Moivre’s Theorem · Polar Form

The solutions of z⁶ = 1 are the sixth roots of unity. They have modulus 1 and arguments 0, π3, 2π3, …, a sixth of a turn apart, so they are the vertices of a regular hexagon. Multiplying by 60 scales the hexagon to the bolt circle.

1. Write z = cos θ + i sin θ. By De Moivre's theorem z⁶ = cos 6θ + i sin 6θ, which is 1 when 6θ = 2kπ, so θ = kπ3 for k = 0, 1, 2, 3, 4, 5.
2. The roots are 1, 12 + √32i, −12 + √32i, −1, −12 − √32i and 12 − √32i. Multiply each by 60, with 60 × √32 = 30√3 ≈ 51.96.
3. (a) The holes are at (60, 0), (30, 52.0), (−30, 52.0), (−60, 0), (−30, −52.0) and (30, −52.0), in millimeters from the center of the pipe.
4. Neighboring holes are the ends of one side of the hexagon. From 60 to 30 + 30√3i the distance is √30² + (30√3)² = √900 + 2700 = √3600.
5. (b) The centers of neighboring holes are 60 mm apart, the same as the radius, as a regular hexagon's side always is. Check: each root raised to the sixth power is cos 2kπ + i sin 2kπ = 1.

answer(a) (60, 0), (30, 52.0), (−30, 52.0), (−60, 0), (−30, −52.0), (30, −52.0) in mm; (b) 60 mm

techniqueRegular Polygons from the Roots of Unity · The Roots of Unity

In exponential form, z = re^iθ gives z³ = r³e^3iθ, so the modulus is cubed and the argument is tripled. Going backwards, take the cube root of the modulus and a third of the argument, remembering that the argument of 8i can have any whole number of turns added, which gives three different roots a third of a turn apart.

1. 8i has modulus 8 and argument π2, so 8i = 8e^{i(π2 + 2kπ)} for any whole number k.
2. Let z = re^iθ, so z³ = r³e^3iθ. Then r³ = 8, so r = 2, and 3θ = π2 + 2kπ, so θ = π6 + 2kπ3.
3. k = 0, 1, 2 give θ = π6, 5π6, 3π2; k = 3 would give π6 plus a whole turn, the first root again.
4. (a) 2e^iπ/6 = √3 + i, 2e^5iπ/6 = −√3 + i and 2e^3iπ/2 = −2i: the antennas are at about 1.73 + i, −1.73 + i and −2i.
5. (b) The first two differ by 2√3 along the real axis, so they are 2√3 ≈ 3.46 m apart. The roots are equally spaced round a circle, so the triangle is equilateral and every pair is 3.46 m apart. Check: (√3 + i)² = 2 + 2√3i, and (2 + 2√3i)(√3 + i) = 2√3 + 2i + 6i − 2√3 = 8i.

answer(a) √3 + i, −√3 + i and −2i, or about 1.73 + i, −1.73 + i and −2i meters; (b) 2√3 ≈ 3.46 m

techniqueThe nth Roots of a Complex Number · The Exponential Form of a Complex Number · The Roots of Unity

De Moivre's theorem gives cos 3θ + i sin 3θ as (cos θ + i sin θ)³. Expanding the cube by the binomial theorem and matching the real parts writes cos 3θ in powers of cos θ, which turns the cubic in the manual into one cosine.

1. Write c = cosθ and s = sinθ. By De Moivre's theorem, cos 3θ + i sin 3θ = (c + is)³.
2. By the binomial theorem, (c + is)³ = c³ + 3c²(is) + 3c(is)² + (is)³ = (c³ − 3cs²) + i(3c²s − s³), since i² = −1 and i³ = −i.
3. The real parts are equal, so cos 3θ = c³ − 3cs² = c³ − 3c(1 − c²) = 4c³ − 3c.
4. (a) Then 8c³ − 6c = 2(4c³ − 3c) = 2cos 3θ, so h = 2cos 3θ + 2.
5. (b) A lift of 3 cm needs 2cos 3θ + 2 = 3, so cos 3θ = 12. The smallest positive solution is 3θ = π3, so θ = π9 radians, which is 20^°. Check: cos 20^° = 0.9397, and 8(0.9397)³ − 6(0.9397) + 2 = 6.638 − 5.638 + 2 = 3.00 cm.

answer(a) h = 2cos 3θ + 2, since cos 3θ = 4cos³θ − 3cosθ; (b) θ = π9 radians, 20^°

techniqueMultiple Angles by De Moivre · De Moivre’s Theorem

With z = cosθ + i sinθ, De Moivre's theorem gives zⁿ = cos nθ + i sin nθ and 1zⁿ = cos nθ − i sin nθ, so zⁿ + 1zⁿ = 2cos nθ. A power of cosθ becomes a power of z + 1z, and the terms of its expansion pair up into cosines of multiple angles.

1. Adding zⁿ = cos nθ + i sin nθ and 1zⁿ = cos nθ − i sin nθ gives zⁿ + 1zⁿ = 2cos nθ; with n = 1, z + 1z = 2cosθ.
2. By the binomial theorem, (2cosθ)⁴ = (z + 1z)⁴ = z⁴ + 4z² + 6 + 4z² + 1z⁴.
3. Pair each power with its reciprocal: (z⁴ + 1z⁴) + 4(z² + 1z²) + 6 = 2cos 4θ + 8cos 2θ + 6.
4. (a) So 16cos⁴θ = 2cos 4θ + 8cos 2θ + 6; dividing by 16, cos⁴θ = 18(cos 4θ + 4cos 2θ + 3).
5. (b) At θ = π6: 18(cos 2π3 + 4cos π3 + 3) = 18(−12 + 2 + 3) = 916, so 916 of the brightness, about 56%, reaches that point. Check: cos π6 = √32, and (√32)⁴ = 916.

answer(a) cos⁴θ = 18(cos 4θ + 4cos 2θ + 3); (b) 916 of the center's brightness

techniqueThe Identity z + 1/z = 2 cos θ · Multiple Angles by De Moivre

The four phasors are a geometric series with common ratio w = e^iπ/6, so the sum has a formula. Writing e^iα − 1 as e^iα/2 × 2i sinα2, by taking out half the angle, turns the top and the bottom of that formula into a sine times an exponential, and the modulus and the argument can be read off.

1. With w = e^iπ/6 the sum is S = 1 + w + w² + w³, a geometric series of four terms, so S = w⁴ − 1w − 1 = e^2iπ/3 − 1e^iπ/6 − 1.
2. Take out half the angle: e^iα − 1 = e^iα/2(e^iα/2 − e^−iα/2) = e^iα/2 × 2i sinα2. So the top is e^iπ/3 × 2i sinπ3 and the bottom is e^iπ/12 × 2i sinπ12.
3. Dividing, the 2i cancels and the exponentials divide by subtracting their angles: S = e^{i(π3 − π12)} × sin(π/3)sin(π/12) = e^iπ/4 × sin(π/3)sin(π/12).
4. (a) The amplitude is |S| = sin(π/3)sin(π/12) = 0.86600.2588 ≈ 3.35 units, a little less than the 4 units the speakers would give in step.
5. (b) The phase is arg S = π4, halfway between the first wave and the last. Check by adding the four directly: the real parts are 1 + 0.866 + 0.5 + 0 = 2.366 and the imaginary parts 0 + 0.5 + 0.866 + 1 = 2.366. The two parts are equal, so the angle is π4, and √2.366² + 2.366² = 3.35.

answer(a) |S| = sin(π/3)sin(π/12) ≈ 3.35 units; (b) arg S = π4 radians ahead of the first wave

techniqueSumming a Series with De Moivre · The Exponential Form of a Complex Number · The Modulus of a Complex Number

examsH2

|z − a| is the distance from z to a. So |z − a| = r is a circle of radius r about a, and |z − a| = |z − b| is the set of points equally far from a and b, the perpendicular bisector of the line joining them. Writing z = x + yi turns each into an equation in x and y.

1. With z = x + yi, |z − (3 + 4i)| = √(x − 3)² + (y − 4)². The edge of the zone is where this is 5: (x − 3)² + (y − 4)² = 25, a circle of radius 5 about 3 + 4i.
2. For the path, square both sides: (x − 1)² + (y − 1)² = (x − 5)² + (y − 5)². The x² and y² cancel, leaving −2x − 2y + 2 = −10x − 10y + 50, so 8x + 8y = 48.
3. (a) The edge is (x − 3)² + (y − 4)² = 25 and the path is x + y = 6. The path passes through 3 + 3i, the midpoint of the masts, as a perpendicular bisector must.
4. Substitute y = 6 − x into the circle: (x − 3)² + (2 − x)² = 25, so 2x² − 10x + 13 = 25, which is x² − 5x − 6 = 0, or (x − 6)(x + 1) = 0. So x = 6, y = 0, or x = −1, y = 7.
5. (b) The path meets the edge at 6 and at −1 + 7i, and the drone can fly √7² + 7² = 7√2 ≈ 9.90 km between them. Check: 6 is √25 + 1 = √26 km from each mast, and |6 − (3 + 4i)| = √9 + 16 = 5, on the edge.

answer(a) the edge (x − 3)² + (y − 4)² = 25 and the path x + y = 6; (b) at 6 and −1 + 7i, a flight of 7√2 ≈ 9.90 km

techniqueCircles and Bisectors on the Argand Diagram · The Modulus of a Complex Number

examsH2

|z − 1| is the distance from the searchlight and arg(z − 1) is the direction from the searchlight, measured from east. So the region is a sector: every point within 4 km of 1 whose direction from 1 is between east and north-east. A boat is lit only if it passes both tests.

1. Move the origin to the searchlight by writing w = z − 1. The region is a sector of radius 4 with its vertex at 1, between the ray going east, arg w = 0, and the ray going north-east, arg w = π4 ≈ 0.785.
2. For P, w = (3 + i) − 1 = 2 + i: its distance is |w| = √5 ≈ 2.24 ≤ 4, and its direction is arg w = tan⁻¹12 ≈ 0.46 ≤ 0.785. P passes both tests.
3. For Q, w = (2 + 3i) − 1 = 1 + 3i: its distance is √10 ≈ 3.16 ≤ 4, but its direction is tan⁻¹ 3 ≈ 1.25, which is more than π4.
4. (a) P is lit. Q is within range but outside the beam, north of the north-east ray, so it is not lit.
5. (b) The sector has radius 4 and angle π4, so its area is 12r²θ = 12 × 16 × π4 = 2π ≈ 6.28 km². Check: π4 is an eighth of a turn, and an eighth of the whole disc, 16π, is 2π.

answer(a) P is lit; Q is not, being outside the beam; (b) 2π ≈ 6.28 km²

techniqueRays and Regions on the Argand Diagram · The Argument of a Complex Number

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