Complex Arithmetic · applications

Applications: Complex Arithmetic

10 question types · Pre-University · each worked step by step with a figure that follows the steps

H2

01

A Pump Motor on a Long Cable: The Impedance of Three Parts in Series, and the Capacitor That Cancels Its Reactance

methodWrite Each Part as R + iX, Add the Real Parts and the Imaginary Parts Separately, and Choose the Capacitor That Brings the Imaginary Part to Zero

In an AC circuit the impedance of each part is written Z = R + iX ohms, where R is its resistance and X is its reactance: positive for a coil and negative for a capacitor. A pump motor has impedance 5 + 9i ohms. It is fed through a cable of impedance 1 + i ohms and an ideal capacitor, which has no resistance, of impedance −2i ohms, all in series, and impedances in series add. (a) Find the total impedance of the circuit. (b) The electrician replaces the capacitor with another ideal capacitor that makes the total impedance a real number. What must the new capacitor's impedance be, and what is the total impedance then?

ReImmotorcable−2imotor 5 + 9i, cable 1 + i
Each impedance is an arrow on the Argand diagram, and a series circuit adds them head to tail.
Engineers write j for the imaginary unit, because i already stands for current; here it is i throughout, with i2 = −1. The three impedances are 5 + 9i, 1 + i and 0 − 2i ohms.
step 1 of 5

Impedances in series add like any complex numbers: the real parts together and the imaginary parts together. A total that is a real number has no imaginary part, so the new capacitor's reactance must cancel the reactance of everything else.

  1. Engineers write j for the imaginary unit, because i already stands for current; here it is i throughout, with i2 = −1. The three impedances are 5 + 9i, 1 + i and 0 − 2i ohms.
  2. Add the real parts, which are the resistances: 5 + 1 + 0 = 6 ohms.
  3. Add the imaginary parts, which are the reactances: 9 + 1 − 2 = 8 ohms.
  4. (a) The total impedance is Z = 6 + 8i ohms.
  5. (b) Without a capacitor, the motor and the cable give (5 + 1) + (9 + 1)i = 6 + 10i ohms. The new capacitor must add −10i to bring the imaginary part to 0, so its impedance is −10i ohms, and the total is then 6 + 10i − 10i = 6 ohms. Check: 5 + 1 + 0 = 6 and 9 + 1 − 10 = 0.

answer(a) 6 + 8i ohms; (b) −10i ohms, which makes the total 6 ohms

techniqueAdding Complex Numbers · The Imaginary Unit

examsH2

Common pitfalls

  • Adding every number as if it were one plain quantity, 5 + 9 + 1 + 1 − 2 = 14 ohms. Resistance and reactance are the real and imaginary parts of one complex number, and they are added separately.
  • Choosing −8i for the new capacitor because the total in (a) has imaginary part 8. That total already includes the old capacitor's −2i, which is taken out, so the new capacitor must cancel 9 + 1 = 10.
02

A Lamp and a Choke Coil on a 50-Volt Supply: The Current, and the Voltage Across Each Part

methodDivide the Voltage by the Impedance, Multiplying Top and Bottom by the Conjugate; Then Multiply the Current by Each Part's Impedance and Check That the Voltages Add to the Supply

A lamp of resistance 8 ohms is in series with a choke coil of impedance 6i ohms, on an AC supply of V = 50 volts. The current is I = VZ amperes, where Z is the total impedance. (a) Find the current I in the form a + bi. (b) The voltage across each part is I times that part's impedance. Find the voltage across the lamp and the voltage across the coil, and check that they add up to the supply.

ReImZ = 8 + 6iI = 50/(8 + 6i)
The total impedance is 8 + 6i ohms, and the current is 508 + 6i.
Impedances in series add, so Z = 8 + 6i ohms and I = 508 + 6i.
step 1 of 5

Dividing by a complex number is done by multiplying the top and the bottom by its conjugate, which makes the bottom a real number. The voltages across the parts are then products, and in series they must add up to the supply voltage.

  1. Impedances in series add, so Z = 8 + 6i ohms and I = 508 + 6i.
  2. Multiply the top and the bottom by the conjugate 8 − 6i. The bottom becomes (8 + 6i)(8 − 6i) = 64 − 36i2 = 64 + 36 = 100, a real number.
  3. (a) The top becomes 50(8 − 6i) = 400 − 300i, so I = 400 − 300i100 = 4 − 3i amperes.
  4. (b) Across the lamp the voltage is 8(4 − 3i) = 32 − 24i volts. Across the coil it is 6i(4 − 3i) = 24i − 18i2 = 18 + 24i volts.
  5. Check: (32 − 24i) + (18 + 24i) = 50 volts, which is the supply. The size of the current is √42 + 32 = 5 amperes.

answer(a) I = 4 − 3i amperes; (b) 32 − 24i volts across the lamp and 18 + 24i volts across the coil, which add to 50 volts

techniqueDividing Complex Numbers · Complex Conjugates · Multiplying Complex Numbers

examsH2

Common pitfalls

  • Splitting the fraction over the sum in its denominator, 508 + 506i. A fraction can be split over a sum in its numerator, never in its denominator; multiply by the conjugate instead.
  • Writing −18i2 as −18, which gives the coil −18 + 24i volts. Since i2 = −1, −18i2 = +18, and the two voltages then add to 50 as they must.
03

A Coil Branch and a Capacitor Branch in Parallel: The Combined Impedance

methodMultiply the Two Impedances, Add Them, and Divide the Product by the Sum Using the Conjugate; Check by Adding the Reciprocals

Two branches are connected in parallel across an AC supply. One branch, a coil with some resistance, has impedance Z1 = 2 + 4i ohms. The other, a capacitor with some resistance, has impedance Z2 = 4 − 2i ohms. The combined impedance is Z = Z1 Z2Z1 + Z2. (a) Find Z1 Z2 and Z1 + Z2. (b) Find Z in the form a + bi ohms, and check it with 1Z = 1Z1 + 1Z2.

2 + 4i4 − 2iZ₁Z₂(2 + 4i)(4 − 2i) = 16 + 12i
The product: (2 + 4i)(4 − 2i) = 8 − 4i + 16i − 8i2 = 16 + 12i.
Multiply out the product and replace i2 by −1: (2 + 4i)(4 − 2i) = 8 − 4i + 16i − 8i2 = 8 + 12i + 8 = 16 + 12i.
step 1 of 5

The product is found by multiplying out the brackets and replacing i2 by −1, and the sum by adding the real and imaginary parts. The division is done with the conjugate of the bottom. The reciprocals give a second route to the same answer.

  1. Multiply out the product and replace i2 by −1: (2 + 4i)(4 − 2i) = 8 − 4i + 16i − 8i2 = 8 + 12i + 8 = 16 + 12i.
  2. Add the two impedances: (2 + 4) + (4 − 2)i = 6 + 2i. (a) Z1 Z2 = 16 + 12i and Z1 + Z2 = 6 + 2i.
  3. To divide 16 + 12i by 6 + 2i, multiply the top and the bottom by the conjugate 6 − 2i. The bottom becomes 36 + 4 = 40.
  4. The top becomes (16 + 12i)(6 − 2i) = 96 − 32i + 72i − 24i2 = 120 + 40i. (b) Z = 120 + 40i40 = 3 + i ohms.
  5. Check: 12 + 4i = 2 − 4i20 and 14 − 2i = 4 + 2i20, which add to 6 − 2i20. Also 13 + i = 3 − i10 = 6 − 2i20, the same.

answer(a) Z1 Z2 = 16 + 12i and Z1 + Z2 = 6 + 2i; (b) Z = 3 + i ohms

techniqueDividing Complex Numbers · Multiplying Complex Numbers · Adding Complex Numbers

examsH2

Common pitfalls

  • Giving Z1 + Z2 = 6 + 2i as the answer, as if the branches were in series. In parallel the current has two paths, so the combined impedance is found from the product over the sum.
  • Dividing the real parts and the imaginary parts separately, 166 + 122i. Division by a complex number needs the conjugate, just as a surd in a denominator needs rationalizing.
04

A Marker Buoy After a Boat's Wake: What the Complex Roots Say About Its Bobbing

methodSolve the Characteristic Equation with the Square Root of a Negative Number, Then Read the Real Part as the Decay and the Imaginary Part as the Oscillation

After a boat's wake passes, the up-and-down motion of a marker buoy is governed by the characteristic equation 250r2 + 500r + 2500 = 0, from the buoy's mass of 250 kg, a damping by the water of 500 N s/m, and a stiffness of 2500 N/m from its buoyancy. When the roots are r = p ± qi, the motion is ept times an oscillation of q radians per second. (a) Solve the equation. (b) Say what the real part and the imaginary part tell you about the motion, and find the time from the top of one bob to the top of the next.

ReImr2+ 2r + 10 = 0, discriminant −36
Divided by 250, the equation is r2 + 2r + 10 = 0, with discriminant 4 − 40 = −36.
Divide by 250: r2 + 2r + 10 = 0. The discriminant is 22 − 4 × 10 = 4 − 40 = −36, which is negative, so there are no real roots.
step 1 of 5

A negative discriminant gives a pair of conjugate complex roots. The real part says whether the motion grows or dies away, and the imaginary part says how fast it swings.

  1. Divide by 250: r2 + 2r + 10 = 0. The discriminant is 22 − 4 × 10 = 4 − 40 = −36, which is negative, so there are no real roots.
  2. Use the quadratic formula with √−36 = 6i: r = −2 ± 6i2.
  3. (a) r = −1 + 3i or r = −1 − 3i, a pair of conjugates. Check: (−1 + 3i)2 = 1 − 6i − 9 = −8 − 6i, and −8 − 6i + 2(−1 + 3i) + 10 = 0.
  4. (b) The real part is −1, which is negative, so the size of the bobbing is multiplied by e−t and dies away. The imaginary part is 3, which is not zero, so the buoy does bob up and down, at 3 radians per second.
  5. From the top of one bob to the top of the next is one cycle of 2π radians, which takes 2π3 ≈ 2.09 seconds.

answer(a) r = −1 ± 3i; (b) the real part −1 is negative, so the bobbing dies away, and the imaginary part 3 makes it oscillate at 3 radians per second, from one top to the next every 2π3 ≈ 2.09 seconds

techniqueComplex Roots of a Quadratic · Roots of Negative Numbers

examsH2

Common pitfalls

  • Stopping at the negative discriminant and saying that the buoy does not move. A negative discriminant means complex roots, and complex roots mean an oscillation.
  • Reading the imaginary part as the decay and the real part as the bobbing. The real part gives the factor ept, which grows or shrinks; only the imaginary part makes the motion swing.
05

A Floor Robot That Turns a Quarter-Turn Left 2026 Times: Its Heading and Where It Ends Up

methodA Quarter-Turn Left Is Multiplication by i, So the Heading Is a Power of i; Use the Remainder on Dividing by 4, and Add the Moves, Which Cancel in Blocks of Four

A floor robot starts at the origin facing east. Each command makes it turn 90° to the left and then drive 1 m forward. On an Argand diagram east is 1 and north is i, and a quarter-turn to the left is multiplication by i, so after n commands the robot faces the direction in. (a) Which way does it face after 2026 commands? (b) Where does it end up after those 2026 commands?

EN1i−1−ii, −1, −i, 1, then i again
Each left turn multiplies the heading by i: east, north, west, south, and east again.
The powers of i repeat every four: i1 = i, i2 = −1, i3 = −i and i4 = 1, and then i5 = i again.
step 1 of 5

The powers of i repeat every four, so only the remainder of the power on dividing by 4 matters. The position is the sum of all the moves, and each block of four moves adds to zero.

  1. The powers of i repeat every four: i1 = i, i2 = −1, i3 = −i and i4 = 1, and then i5 = i again.
  2. Divide 2026 by 4: 2026 = 4 × 506 + 2, so i2026 = (i4)506 × i2 = 1 × (−1) = −1.
  3. (a) The robot faces the direction −1, which is west.
  4. The nth drive moves the robot 1 m in the direction in, so its position is i + i2 + i3 + ⋯ + i2026. Each block of four is i − 1 − i + 1 = 0: four commands bring it back to the start.
  5. (b) The first 2024 = 4 × 506 drives cancel, which leaves i2025 + i2026 = i + (−1) = −1 + i. The robot ends 1 m west and 1 m north of the origin. Check: after the first two commands it is at i − 1 as well.

answer(a) west, because i2026 = −1; (b) at −1 + i, which is 1 m west and 1 m north of the start

techniquePowers of i · Adding Complex Numbers

examsH2

Common pitfalls

  • Working out 2026 ÷ 4 = 506.5 and trying to use the .5. What decides the power is the remainder, 2, so i2026 = i2 = −1.
  • Giving i2026 = −1 as the position as well as the heading. The heading is the last power alone, but the position is the sum of every move the robot has made.
06

A Game Sprite Multiplied by 1 + i: Where One Corner Moves After One Power-Up and After Two

methodStore the Point as x + yi, Multiply Out the Brackets and Replace i Squared by −1; Two Multiplications by 1 + i Are One Multiplication by (1 + i) Squared

In a game, each point (x, y) of a sprite is stored as the complex number x + yi. A power-up multiplies every point of the sprite by 1 + i, which turns it and makes it larger. One corner of the sprite is at (3, 2). (a) Where is that corner after the power-up? (b) The player collects a second power-up, so the corner is multiplied by 1 + i again. Where is it now, and what single multiplication does the same as the two?

ReIm(3, 2)(3 + 2i)(1 + i) = 3 + 3i + 2i + 2i2
The corner (3, 2) is 3 + 2i; multiply out the brackets.
The corner is 3 + 2i. Multiply out: (3 + 2i)(1 + i) = 3 + 3i + 2i + 2i2.
step 1 of 5

Multiplying two complex numbers means multiplying every term of one bracket by every term of the other, then replacing i2 by −1. Doing it twice is the same as multiplying once by the square of 1 + i.

  1. The corner is 3 + 2i. Multiply out: (3 + 2i)(1 + i) = 3 + 3i + 2i + 2i2.
  2. Replace i2 by −1: 3 + 5i − 2 = 1 + 5i. (a) The corner moves to (1, 5).
  3. Multiply by 1 + i again: (1 + 5i)(1 + i) = 1 + i + 5i + 5i2 = 1 + 6i − 5 = −4 + 6i.
  4. (b) The corner is now at (−4, 6). The two power-ups together multiply by (1 + i)2 = 1 + 2i + i2 = 2i.
  5. Check: 2i(3 + 2i) = 6i + 4i2 = −4 + 6i, the same point. Multiplying by 2i is multiplying by 2 and then by i, which is a quarter-turn.

answer(a) (1, 5), from 1 + 5i; (b) (−4, 6), from −4 + 6i, and the two together multiply by 2i

techniqueMultiplying Complex Numbers · The Imaginary Unit

examsH2

Common pitfalls

  • Multiplying only the real parts together and the imaginary parts together, (3 × 1) + (2 × 1)i = 3 + 2i. Every term of one bracket multiplies every term of the other, and the term 2i × i is 2i2 = −2.
  • Leaving 2i2 as +2, which puts the corner at (5, 5). Since i2 = −1, 2i2 = −2.
07

A Fish Tank That Must Hold 100 Liters: Why Only One Size Works

methodForm the Cubic from the Volume, Find the Real Root by the Factor Theorem, Divide Out the Factor, and Solve the Quadratic Left Over

An aquarium maker wants a tank with a square base of side x dm and a height 1 dm less than the side, holding exactly 100 liters, where 1 liter is 1 dm3. (a) Show that x3 − x2 − 100 = 0, and find a real root by the factor theorem. (b) Find the other two roots, and explain why exactly one tank has this shape and this capacity.

xx2(x − 1) = 100, x3− x2− 100 = 0
The base area x2 times the height x − 1 is 100 liters.
The volume is the base area times the height: x2(x − 1) = 100, so x3 − x2 − 100 = 0.
step 1 of 5

A cubic has three roots. Once one real root is found, dividing out its factor leaves a quadratic, and a negative discriminant there means that the other two roots are a complex conjugate pair, which cannot be lengths.

  1. The volume is the base area times the height: x2(x − 1) = 100, so x3 − x2 − 100 = 0.
  2. Try factors of 100. At x = 4 the cubic is 64 − 16 − 100 = −52, and at x = 5 it is 125 − 25 − 100 = 0. (a) By the factor theorem (x − 5) is a factor, and x = 5 is a root.
  3. Divide by (x − 5): x3 − x2 − 100 = (x − 5)(x2 + 4x + 20).
  4. Solve the quadratic. The discriminant is 16 − 80 = −64 and √−64 = 8i, so x = −4 ± 8i2 = −2 ± 4i.
  5. (b) The other roots are −2 + 4i and −2 − 4i, a conjugate pair. A side must be a real, positive length, so only x = 5 gives a tank: 5 dm by 5 dm by 4 dm, which is 50 cm by 50 cm by 40 cm. Check: 52 × 4 = 100.

answer(a) x = 5; (b) x = −2 ± 4i, which are not lengths, so the only tank is 5 dm by 5 dm by 4 dm (50 cm by 50 cm by 40 cm)

techniqueComplex Roots of a Cubic · Complex Roots of a Quadratic · Roots of Negative Numbers

examsH2

Common pitfalls

  • Stopping at x = 5 without showing that no other root is real. A cubic can have three real roots; here the quadratic factor has a negative discriminant, and that is what makes the tank the only one.
  • Writing the quotient as x2 − 4x + 20. Multiplying back, (x − 5)(x2 − 4x + 20) = x3 − 9x2 + 40x − 100, which has −9x2 instead of −x2, so the middle term must be +4x.
08

A Fountain Jet Under a Glass Roof: The Roots of the Roof Equation, and the Greatest Height

methodSet the Height Equal to the Roof and Solve, Read What Kind of Roots Come Out, and Complete the Square for the Greatest Height

A fountain in a shopping center throws a jet whose height is h = 6x − x2 meters at a horizontal distance of x meters from the nozzle. The glass roof is 13 m above the nozzle. (a) Solve 6x − x2 = 13, and say what the roots tell the architect. (b) Find the greatest height of the jet, and how far below the roof it stays.

xroof 13 mx2− 6x + 13 = 0, discriminant −16
Setting the height equal to 13 gives x2 − 6x + 13 = 0, and its discriminant is −16.
Rearrange: x2 − 6x + 13 = 0. The discriminant is 36 − 52 = −16, which is negative.
step 1 of 5

A height is reached where the equation has a real root. Complex roots mean there is no such distance, so the water never gets there. Completing the square shows the greatest height and how the roots are linked to it.

  1. Rearrange: x2 − 6x + 13 = 0. The discriminant is 36 − 52 = −16, which is negative.
  2. Use √−16 = 4i in the quadratic formula: x = 6 ± 4i2 = 3 ± 2i.
  3. (a) The roots are 3 + 2i and 3 − 2i. No real distance gives a height of 13 m, so the water never reaches the roof. Check: (3 + 2i)2 − 6(3 + 2i) + 13 = 5 + 12i − 18 − 12i + 13 = 0.
  4. Complete the square: 6x − x2 = 9 − (x − 3)2, and (x − 3)2 is never negative when x is real.
  5. (b) The greatest height is 9 m, at x = 3, the real part of the roots. The jet stays 13 − 9 = 4 m below the roof, and 4 = 22 is the square of the imaginary part.

answer(a) x = 3 ± 2i: there is no real solution, so the jet never reaches the roof; (b) 9 m, at x = 3, which is 4 m below the roof

techniqueComplex Roots of a Quadratic · Roots of Negative Numbers

examsH2

Common pitfalls

  • Reading 3 ± 2i as distances of 3 m and 2 m. A root with an imaginary part is not a place on the floor: the equation has no real solution, and that is the answer to the architect's question.
  • Solving 6x − x2 = 0 instead. That gives where the jet comes down, x = 0 and x = 6, and says nothing about the roof.
09

A Third-Order Audio Filter: Rebuilding Its Polynomial from One Real Root and One Complex Root

methodNon-Real Roots of a Real Polynomial Come in Conjugate Pairs; Build the Quadratic Factor from the Sum and Product of the Pair, Then Multiply by the Linear Factor

An audio engineer designs a filter whose behavior is set by a cubic p(z) = z3 + az2 + bz + c. The coefficients must be real, because the filter is built from real resistors and capacitors. The design needs roots at z = −2 and z = −3 + 4i, in thousands per second. (a) What must the third root be, and why? (b) Find a, b and c.

ReIm−2−3 + 4ithe conjugate of −3 + 4i is −3 − 4i
A real polynomial has its non-real roots in conjugate pairs, mirror images in the real axis.
A polynomial with real coefficients has its non-real roots in conjugate pairs. The conjugate of −3 + 4i is −3 − 4i.
step 1 of 5

A polynomial with real coefficients has its non-real roots in conjugate pairs, so the third root is fixed. The pair gives a real quadratic factor, and the real root gives a linear factor.

  1. A polynomial with real coefficients has its non-real roots in conjugate pairs. The conjugate of −3 + 4i is −3 − 4i.
  2. (a) The third root must be −3 − 4i; without it the coefficients could not all be real.
  3. The conjugate pair gives a real quadratic factor. The sum of the pair is −6 and the product is (−3)2 + 42 = 25, so the factor is z2 + 6z + 25.
  4. Multiply by the factor (z + 2) for the real root: (z + 2)(z2 + 6z + 25) = z3 + 6z2 + 25z + 2z2 + 12z + 50 = z3 + 8z2 + 37z + 50.
  5. (b) a = 8, b = 37 and c = 50. Check: p(−2) = −8 + 32 − 74 + 50 = 0.

answer(a) −3 − 4i, because the non-real roots of a real polynomial come in conjugate pairs; (b) a = 8, b = 37, c = 50

techniqueComplex Roots of a Cubic · Complex Conjugates · Multiplying Complex Numbers

examsH2

Common pitfalls

  • Using only the two given roots, (z + 2)(z + 3 − 4i), which gives a quadratic with non-real coefficients. A real design needs the conjugate root as well, and a cubic has room for exactly three roots.
  • Writing the factor for the root −3 + 4i as (z − 3 + 4i). The factor for a root w is (z − w), which here is (z + 3 − 4i).
10

Coloring One Point of the Mandelbrot Set: Squaring and Adding from Zero

methodApply the Rule One Step at a Time, Squaring by Multiplying Out and Replacing i Squared by −1, and Watch Whether the Values Grow

A program that draws the Mandelbrot set tests each complex number c by starting at z0 = 0 and repeating zn+1 = zn2 + c. The point c is colored black if the values never grow without limit. (a) For c = i, find z1, z2, z3 and z4. (b) Explain why the point c = i is colored black.

ReIm0iz₁ = 0 + i = i
From z0 = 0, the first step gives z1 = i.
Start from z0 = 0: z1 = 02 + i = i.
step 1 of 5

Each value is the square of the one before, plus c. Squaring a complex number means multiplying it by itself and replacing i2 by −1. When a value comes back, everything after it repeats.

  1. Start from z0 = 0: z1 = 02 + i = i.
  2. z2 = i2 + i = −1 + i.
  3. Square first: (−1 + i)2 = 1 − 2i + i2 = −2i. Then z3 = −2i + i = −i.
  4. (a) z4 = (−i)2 + i = −1 + i. So z1 = i, z2 = −1 + i, z3 = −i and z4 = −1 + i.
  5. (b) z4 = z2, and each value depends only on the one before, so from z2 on the values go −1 + i, −i, −1 + i, −i, … without end. They never grow, so c = i is in the set and is colored black.

answer(a) z1 = i, z2 = −1 + i, z3 = −i, z4 = −1 + i; (b) z4 = z2, so the values repeat −1 + i, −i and never grow: c = i is colored black

techniqueMultiplying Complex Numbers · Adding Complex Numbers · The Imaginary Unit

examsH2

Common pitfalls

  • Squaring −1 + i as (−1)2 + i2 = 0, squaring each part and dropping the middle term. The square of a sum has the term 2ab, here 2(−1)(i) = −2i.
  • Adding c only once, at the start. The rule adds c = i at every step, so each square is followed by + i.
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