What the polar form is for. Establishes that multiplication multiplies moduli and adds arguments, that division divides and subtracts, and that powers follow by induction as DeMoivre's Theorem. Reads the product rule geometrically as a stretch followed by a rotation, then solves the root equation to show that every non-zero complex number has exactly n distinct nth roots, evenly spaced around a circle at the vertices of a regular polygon.
Subject: Trigonometry · 65 slides · symbolic lesson
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Title
Trigonometry · Chapter 11 — Applications of Trigonometry
§11.7 Polar Form of Complex Numbers, pp. 997-1004
Objectives
Five outcomes, and the fourth is where a whole class of equations becomes solvable.
Stitz & Zeager, College Trigonometry, Ch. 11 Applications of Trigonometry — §11.7 Polar Form of Complex Numbers §11.7, pp. 997-1004 — the pages these objectives are drawn from
Warm-up
The previous lesson assembled the polar form but never used it for anything.
Discussion prompt
Try to compute the fifth power of negative 1 plus i root 3 by expanding the binomial. How far do you get before it becomes unpleasant?
Hint: How many terms does a fifth power have, and what are the powers of i?
Answer:
Six terms, each with a binomial coefficient, a power of root three, and a power of i that cycles through four values with sign changes. It is doable and it is thoroughly unpleasant, and a tenth power would be worse still.
In polar form the same number is 2 cis of two thirds of pi, and its fifth power is 32 cis of ten thirds of pi — one real number raised to the fifth, and one angle multiplied by five.
That is the payoff. The polar form was built so that multiplication would be easy, and this lesson establishes why it is and what follows once it is.
Concept
Writing two complex numbers in polar form and multiplying, the sum identities collapse the four cross terms into a single cis of the sum. Multiplication of complex numbers becomes multiplication of two real numbers and addition of two angles.
Theorem 11.16 — For complex numbers in polar form: the product has modulus the product of the moduli and argument the sum of the arguments; the nth power has modulus the nth power and argument n times the argument; and the quotient has modulus the quotient and argument the difference, provided the divisor is non-zero.
\[ zw = |z||w|\,\text{cis}(\alpha + \beta) \]
The proof of the product rule is where the sum identities from Chapter 10 finally do something that could not be done without them. Everything else in the theorem follows from it.
Figure (svg): The three rules of Theorem 11.16, each shown as an operation on the modulus paired with an operation on the argument
Stitz & Zeager, College Trigonometry, Ch. 11 Applications of Trigonometry — §11.7 Polar Form of Complex Numbers §11.7, pp. 997-998
Section
Section 1
Concept
Expand the product of two cis expressions. The four terms regroup into a real part that is the cosine of the sum and an imaginary part that is the sine of the sum, by the identities from Chapter 10.
The quotient rule is proved the same way, multiplying above and below by the conjugate. The denominator collapses to 1 by the Pythagorean Identity and the numerator becomes cis of the difference by the difference identities.
Figure (svg): The three rules of Theorem 11.16, each shown as an operation on the modulus paired with an operation on the argument
Stitz & Zeager, College Trigonometry, Ch. 11 Applications of Trigonometry — §11.7 Polar Form of Complex Numbers §11.7, pp. 997-998
Picture it
Each turns one hard operation into two easy ones.
Figure (svg): The three rules of Theorem 11.16, each shown as an operation on the modulus paired with an operation on the argument
Note what is absent: there is no rule for sums. Addition is the operation rectangular form is good at, and neither form is good at both.
Worked example
Where the sum identities earn their place.
\[ \text{Prove } zw = |z||w|\,\text{cis}(\alpha + \beta). \]
Write both in polar form and multiply
Why: The moduli come straight out front.
Expand the bracket
Why: Four terms, one with i squared.
Group the real and imaginary parts
Why: Using i squared equal to negative 1.
\[ (\cos \cos - \sin \sin) + i(\sin \cos + \cos \sin) \]
Apply the sum identities
Why: Both brackets are recognisable.
\[ \operatorname{cis}(\alpha + \beta) \]
Figure (svg): The solution to Worked example proving the product rule shown as a ladder of expressions, one row per legal move
\[ \left[\cos\alpha + i\sin\alpha\right]\left[\cos\beta + i\sin\beta\right] = \text{cis}(\alpha + \beta) \]
Verify: test with a simple case
Why: Take z and w both equal to i, which is cis of pi over 2. The rule gives cis of pi, which is negative 1. And indeed i times i is negative 1. The rule reproduces the defining property of the imaginary unit.
Stitz & Zeager, College Trigonometry, Ch. 11 Applications of Trigonometry — §11.7 Polar Form of Complex Numbers §11.7, pp. 997-997
Faded example
Multiply 3 cis of pi over 4 by 5 cis of pi over 6.
Fill in the blanks
(3)(5)\,\text15\left(\tfrac5___ + \tfrac______\right) = ___\,\text___\left(\tfrac___\pi}___\right)
Why: The moduli multiply to 15 and the arguments add. Over a common denominator of 12, the angles are three twelfths and two twelfths of pi, summing to five twelfths. Neither operation required expanding anything.
Worked example
Example 11.7.3, parts 1 and 3.
\[ \text{With } z = 4\,\text{cis}\left(\tfrac{\pi}{6}\right) \text{ and } w = 2\,\text{cis}\left(\tfrac{2\pi}{3}\right), \text{ find } zw \text{ and } \tfrac{z}{w}. \]
Multiply moduli, add arguments
Why: Four times 2, and the two angles.
\[ 8 \operatorname{cis}(5 \pi / 6) \]
Convert to rectangular form
Why: Evaluate at five sixths of pi.
\[ -4 \sqrt{3} + 4 i \]
Divide moduli, subtract arguments
Why: Four over 2, and the difference.
\[ 2 \operatorname{cis}(-\frac{\pi}{2}) \]
Convert that too
Why: Negative a quarter turn from the real axis.
\[ -2 i \]
Figure (svg): The solution to Worked example a product and a quotient shown as a ladder of expressions, one row per legal move
\[ zw = 8\,\text{cis}\left(\tfrac{5\pi}{6}\right) = -4\sqrt{3} + 4i, \qquad \frac{z}{w} = 2\,\text{cis}\left(-\tfrac{\pi}{2}\right) = -2i \]
Verify: check the moduli
Why: The modulus of the product should be 8, and the root of 48 plus 16 is 8. The modulus of the quotient should be 2, and negative 2i has modulus 2. Both check, and the modulus check is the fastest way to catch an arithmetic slip in this kind of problem.
Stitz & Zeager, College Trigonometry, Ch. 11 Applications of Trigonometry — §11.7 Polar Form of Complex Numbers §11.7, pp. 999-999
Trap
\[ z + w = \left(|z| + |w|\right)\text{cis}(\alpha + \beta) \]
Extend the pattern of the product rule to addition
Why: Products and quotients both had clean rules, so sums should too.
But adding 1 and negative 1, both of modulus 1, would give a modulus of 2 for a sum that is zero. No such rule exists, and the polar form is genuinely bad at addition.
Convert to rectangular form to add. Adding is adding the real parts and adding the imaginary parts, which is the one thing rectangular form does effortlessly.
Then convert back if the result is needed in polar form. That round trip is the cost of mixing operations.
The general shape of it is worth stating: each form is good at one operation and bad at the other, and knowing which is which is what makes a computation short rather than long.
Prediction
Two complex numbers have arguments pi over 3 and negative pi over 3.
Predict first
What is the argument of their product?
Correct: 0.
Why: Arguments add, and the two given arguments are negatives of one another, so their sum is zero. A product with argument zero lies on the positive real axis, which means it is a positive real number. This is exactly what happens when a complex number is multiplied by its conjugate, and it explains why that product is always real.
Sorting
Each form is good at one thing.
Sort into buckets
Sort each operation.
Socratic
The quotient rule is proved by multiplying above and below by a conjugate.
Discussion prompt
What does the denominator become, and which identity makes it happen?
Hint: What is the product of a cis and its conjugate?
Answer:
The denominator becomes the cosine squared plus the sine squared of beta, since the two cross terms cancel and the i squared turns a minus into a plus. By the Pythagorean Identity that is 1.
So the denominator disappears entirely, which is a stronger outcome than usual for the conjugate trick — normally it leaves a real number to divide by, and here that number is 1.
It happens because cis of beta always has modulus 1, and a number times its conjugate is the square of its modulus. So the denominator was bound to be 1 before any expansion was done, and the algebra is confirming something the geometry already guaranteed.
Section
Section 2
Concept
Applying the product rule repeatedly gives the power rule: the modulus is raised to the power and the argument is multiplied by it. The proof is an induction whose inductive step is one application of the product rule.
DeMoivre's Theorem — For a complex number in polar form and any natural number n, the nth power has modulus the nth power of the modulus and argument n times the argument.
\[ z^n = |z|^n\,\text{cis}(n\theta) \]
The proof is structurally identical to the proof of the power rule for the modulus in the previous lesson. Both are inductions built on their respective product rules, which is worth noticing as a pattern rather than meeting twice as a coincidence.
Figure (svg): Two columns comparing the work of raising a complex number to the fifth power by binomial expansion against doing it in polar form
Stitz & Zeager, College Trigonometry, Ch. 11 Applications of Trigonometry — §11.7 Polar Form of Complex Numbers §11.7, pp. 997-998
Picture it
The comparison that justifies the whole apparatus.
Figure (svg): Two columns comparing the work of raising a complex number to the fifth power by binomial expansion against doing it in polar form
For a single product the rectangular route is arguably easier, since no conversion is needed. The advantage appears at the second power and becomes decisive by the fifth.
Worked example
Example 11.7.3, part 2. The computation the warmup abandoned.
\[ \text{Find } w^5 \text{ where } w = -1 + i\sqrt{3}. \]
Convert to polar form
Why: Modulus 2, argument two thirds of pi.
\[ 2 \operatorname{cis}(2 \pi / 3) \]
Apply DeMoivre
Why: Fifth power of the modulus, five times the argument.
\[ 32 \operatorname{cis}(10 \pi / 3) \]
Reduce the argument
Why: Ten thirds of pi exceeds a full turn.
\[ \text{coterminal with } 4 \pi / 3 \]
Convert back
Why: Evaluate at four thirds of pi.
\[ -16 - 16 i \sqrt{3} \]
Figure (svg): The solution to Worked example a fifth power shown as a ladder of expressions, one row per legal move
\[ w^5 = 32\,\text{cis}\left(\tfrac{4\pi}{3}\right) = -16 - 16i\sqrt{3} \]
Verify: check the modulus
Why: The modulus of the answer should be 2 to the fifth, namely 32. The root of 256 plus 768 is the root of 1024, which is 32. The check passes, and it would have caught any slip in the final trigonometric evaluation.
Stitz & Zeager, College Trigonometry, Ch. 11 Applications of Trigonometry — §11.7 Polar Form of Complex Numbers §11.7, pp. 999-999
Faded example
Compute the sixth power of 2 cis of pi over 4.
Fill in the blanks
\left(2\,\text64\tfrac3___\right)^6 = ___\,\text___\left(\tfrac______\right) = ___\,\text___\left(\tfrac___\pi}___\right)
Why: The modulus 2 is raised to the sixth, giving 64, and the argument is multiplied by 6, giving six quarters of pi, which simplifies to three halves. That angle points straight down, so the answer in rectangular form is negative 64i.
Worked example
The proof, which is one line of real content.
\[ \text{Prove DeMoivre's Theorem by induction on } n. \]
Base case
Why: The first power is the number as given.
\[ n = 1\text{ holds} \]
Assume for k
Why: The induction hypothesis.
\[ z ^{k} = | z | ^{k} \operatorname{cis}(k \theta) \]
Split the next power
Why: One factor peeled off.
\[ z ^{k + 1} = z ^{k} \times z \]
Apply the product rule
Why: Moduli multiply, arguments add.
\[ | z | ^{k + 1} \operatorname{cis}((k + 1) \theta) \]
Figure (svg): The solution to Worked example the induction shown as a ladder of expressions, one row per legal move
\[ z^{k+1} = \left(|z|^k|z|\right)\text{cis}(k\theta + \theta) = |z|^{k+1}\text{cis}((k+1)\theta) \]
Verify: note what the proof used
Why: Only the product rule and the properties of exponents. Nothing about complex numbers specifically enters after the product rule is available, which is why the same argument proved the power rule for the modulus in the previous lesson.
Stitz & Zeager, College Trigonometry, Ch. 11 Applications of Trigonometry — §11.7 Polar Form of Complex Numbers §11.7, pp. 997-998
Error analysis
A student applies DeMoivre to a fourth power.
Annotate
On: \( \left(3\,\text{cis}\left(\tfrac{\pi}{5}\right)\right)^4 = 12\,\text{cis}\left(\tfrac{4\pi}{5}\right) \)
The rule is asymmetric on purpose: the modulus is exponentiated and the argument is multiplied. Doing the same thing to both is the most common slip, and the modulus check catches it immediately.
Prediction
A complex number has modulus exactly 1.
Predict first
What happens to its powers?
Correct: They stay on the unit circle.
Why: DeMoivre raises the modulus to the power, and 1 to any power is 1. So every power has modulus 1 and lies on the unit circle, with only the argument advancing. This is why the numbers of modulus 1 are so important: they are the ones multiplication rotates without stretching, and they are exactly where the roots of unity live.
Sorting
The rule treats modulus and argument differently.
Sort into buckets
Sort each operation by which coordinate it acts on and how.
Edge cases
The theorem is stated for natural numbers n.
Discussion prompt
Does it still hold for a negative exponent, and what about a fractional one?
Hint: What is a negative power, and what would a fractional one mean?
Answer:
For a negative exponent it holds, since a negative power is the reciprocal of a positive one and the quotient rule handles reciprocals. So the modulus is raised to the negative power and the argument multiplied by it, exactly as stated.
For a fractional exponent it fails as stated, and the failure is instructive. Taking n equal to one half, the formula gives one answer, but there are two square roots and the formula names only one of them.
That is precisely the gap the rest of this lesson fills. The nth root of a complex number is not a single number, so no formula of DeMoivre's shape can give it; a list of n numbers is needed instead. Recognising that the theorem cannot simply be extended is what motivates the next theorem.
Section
Section 3
Concept
The product rule read geometrically says that multiplying by a number magnifies distances by its modulus and rotates by its argument. Multiplication by a fixed complex number is a rigid rotation combined with a uniform scaling.
This is why complex numbers appear wherever rotation matters. The whole apparatus of rotating a plane is packaged into a single multiplication.
Figure (svg): The product of two complex numbers shown as a two-step geometric process: first stretching the first number by the modulus of the second, then rotating it by the argument of the second
Stitz & Zeager, College Trigonometry, Ch. 11 Applications of Trigonometry — §11.7 Polar Form of Complex Numbers §11.7, pp. 999-1000
Picture it
First the magnification, then the rotation.
Figure (svg): The product of two complex numbers shown as a two-step geometric process: first stretching the first number by the modulus of the second, then rotating it by the argument of the second
The order does not matter: scaling and rotating about the same centre commute, which is the geometric shadow of multiplication being commutative.
Worked example
The simplest non-trivial case, and the most used.
\[ \text{Describe the effect of multiplying any complex number by } i. \]
Write i in polar form
Why: It sits one unit up the imaginary axis.
\[ i = \operatorname{cis}(\frac{\pi}{2}) \]
Read its modulus
Why: One, so no stretching.
\[ | i | = 1 \]
Read its argument
Why: A quarter turn.
\[ a r g = \frac{\pi}{2} \]
Apply the product rule
Why: Modulus unchanged, argument advanced.
\[ \text{rotation by } \frac{\pi}{2} \]
Figure (svg): The solution to Worked example multiplication by i shown as a ladder of expressions, one row per legal move
\[ iz = \text{cis}\left(\tfrac{\pi}{2}\right)\cdot z \;\Longrightarrow\; \text{rotate } z \text{ by } 90^\circ \]
Verify: check the defining property
Why: Multiplying by i twice is two quarter turns, a half turn, which sends every number to its negative. And indeed i squared is negative 1. The geometric picture reproduces the algebraic definition of the imaginary unit, which is a strong sign the picture is right.
Stitz & Zeager, College Trigonometry, Ch. 11 Applications of Trigonometry — §11.7 Polar Form of Complex Numbers §11.7, pp. 999-1000
Matching
Read the modulus for the stretch and the argument for the turn.
Match the pairs
Why: The first, second and fourth all have modulus 1, so they rotate without stretching; their arguments are a quarter turn, a half turn, and a negative quarter turn. The third has argument zero, so it stretches without rotating. Every complex multiplier is some combination of the two effects.
Worked example
Example 11.7.3, part 1, seen as a motion.
\[ \text{Describe } zw \text{ geometrically for } z = 4\,\text{cis}\left(\tfrac{\pi}{6}\right), \; w = 2\,\text{cis}\left(\tfrac{2\pi}{3}\right). \]
Plot z
Why: Four units out at 30 degrees.
Magnify by the modulus of w
Why: Double the distance.
\[ 8\text{ units out at } 30 ^\circ \]
Rotate by the argument of w
Why: A further 120 degrees counter-clockwise.
\[ \text{now at } 150 ^\circ \]
Read the result
Why: Eight units out at 150 degrees.
\[ 8 \operatorname{cis}(5 \pi / 6) \]
Figure (svg): The solution to Worked example reading a product geometrically shown as a ladder of expressions, one row per legal move
\[ zw = 8\,\text{cis}\left(\tfrac{5\pi}{6}\right) \]
Verify: check against the algebra
Why: The algebra gave 8 cis of five sixths of pi, and five sixths of pi is 150 degrees. The two agree exactly, which they must — the geometric description is only the product rule read aloud.
Stitz & Zeager, College Trigonometry, Ch. 11 Applications of Trigonometry — §11.7 Polar Form of Complex Numbers §11.7, pp. 999-1000
Trap
\[ z \cdot (2 + 3i) \;\Longrightarrow\; \text{move } z \text{ right } 2 \text{ and up } 3 \]
Read the factor's parts as a displacement
Why: Two and three look like an amount to move by.
But that describes addition, not multiplication. Adding 2 plus 3i translates; multiplying by it stretches by the modulus, about 3.6, and rotates by the argument, about 56 degrees.
Addition translates. Multiplication stretches and rotates. The two operations have completely different geometric characters.
A quick way to keep them apart: addition moves the origin's image away from the origin, while multiplication always fixes the origin — zero times anything is zero.
That fixed point is the signature. A transformation fixing the origin and preserving angles is a rotation with a scaling, and that is exactly what multiplication by a complex number is.
Prediction
You multiply a complex number by a number of modulus one half.
Predict first
What happens to its distance from the origin?
Correct: It halves.
Why: Moduli multiply, so the new distance is the old one times one half. The argument of the multiplier affects only the direction, not the distance, so the answer is the same whatever it happens to be. Repeated multiplication by such a number spirals points inward towards the origin, which is how a decaying oscillation is modelled.
Faded example
Describe multiplication by 2 cis of pi over 3.
Fill in the blanks
Every point is moved to 2 times its distance from the origin, and rotated counter-clockwise by 60 degrees.
Why: The modulus 2 doubles every distance and the argument pi over 3, which is 60 degrees, supplies the rotation. Applying this six times would rotate a full turn and multiply distances by 64, which is 2 to the sixth — DeMoivre read as a repeated motion.
Explain it to yourself
Multiplication by a complex number is a stretch composed with a rotation.
Discussion prompt
Explain why the two steps can be done in either order, and what that corresponds to algebraically.
Hint: What is the modulus of a pure rotation, and the argument of a pure stretch?
Answer:
Write the multiplier as a positive real number times a number of modulus 1. The first factor stretches with no rotation and the second rotates with no stretch, so the multiplication factors into the two steps.
They can be done in either order because multiplication of complex numbers is commutative, so the two factors can be applied in either sequence with the same result.
Geometrically that says a scaling about the origin and a rotation about the origin commute, which is true because they act on independent coordinates: the scaling changes only the distance and the rotation only the angle. The polar form is what makes those two effects separable, and their independence is exactly why the product rule has one operation for each coordinate.
Section
Section 4
Concept
To find the nth roots of a number, write the unknown in polar form and apply DeMoivre. Matching moduli gives one real root; matching arguments gives an angle determined only up to full turns, and that ambiguity produces exactly n answers.
Theorem 11.17 — A non-zero complex number with polar form r cis theta has exactly n distinct nth roots, given by the formula above for k running from 0 to n minus 1.
\[ w_k = \sqrt[n]{r}\,\text{cis}\left(\frac{\theta}{n} + \frac{2\pi}{n}k\right) \]
The two halves of the proof are that each formula value really is a root, checked by DeMoivre, and that no two of them coincide, checked by showing their arguments cannot be coterminal.
Figure (svg): The formula for the nth roots of a complex number, with each part of the expression annotated
Stitz & Zeager, College Trigonometry, Ch. 11 Applications of Trigonometry — §11.7 Polar Form of Complex Numbers §11.7, pp. 1001-1002
Picture it
One real root, one division, and a repeated step.
Figure (svg): The formula for the nth roots of a complex number, with each part of the expression annotated
The only genuinely new element is the k, and it is there because an argument is determined only up to full turns — the same non-uniqueness that has run through the whole chapter.
Worked example
A case checkable by factoring, which is why the book starts there.
\[ \text{Find all complex cube roots of } 8. \]
Write 8 in polar form
Why: Eight units along the positive real axis.
\[ 8 \operatorname{cis}(0) \]
Match the moduli
Why: The cube of the unknown modulus is 8.
\[ | w | = 2 \]
Match the arguments up to full turns
Why: Three times the argument is a multiple of 2 pi.
\[ \alpha = 2 \pi k / 3 \]
List the three distinct values
Why: k equal to 0, 1 and 2.
\[ 0, 2 \pi / 3, 4 \pi / 3 \]
Figure (svg): The solution to Worked example the cube roots of 8 shown as a ladder of expressions, one row per legal move
\[ 2\,\text{cis}(0) = 2, \quad 2\,\text{cis}\left(\tfrac{2\pi}{3}\right) = -1 + i\sqrt{3}, \quad 2\,\text{cis}\left(\tfrac{4\pi}{3}\right) = -1 - i\sqrt{3} \]
Verify: check against factoring
Why: The polynomial w cubed minus 8 factors as w minus 2 times w squared plus 2w plus 4, and the quadratic formula on the second factor gives negative 1 plus or minus i root three. The two methods agree, and the polynomial has degree 3 so there are no further roots.
Stitz & Zeager, College Trigonometry, Ch. 11 Applications of Trigonometry — §11.7 Polar Form of Complex Numbers §11.7, pp. 1001-1001
Faded example
Find the two square roots of 4 cis of two thirds of pi.
Fill in the blanks
w_0 = 2\,\text4\left(\tfrac______\right), \qquad w_1 = ___\,\text___\left(\tfrac______ + \pi\right) = ___\,\text___\left(\tfrac___\pi}___\right)
Why: The real square root of 4 is 2, and half of two thirds of pi is pi over 3. The step is 2 pi over 2, namely pi, so the second root is at pi over 3 plus pi, which is four thirds of pi. In rectangular form the two are 1 plus i root three and its negative.
Worked example
Example 11.7.4, part 2. A case factoring would not reach.
\[ \text{Find the four fourth roots of } -16. \]
Write in polar form
Why: Sixteen units along the negative real axis.
\[ 16 \operatorname{cis}(\pi) \]
Take the real fourth root of the modulus
Why: Two to the fourth is 16.
\[ | w | = 2 \]
Divide the argument by 4
Why: Giving the first root's argument.
\[ \frac{\pi}{4} \]
Add quarter turns three times
Why: Steps of 2 pi over 4.
\[ \frac{\pi}{4}, 3 \pi / 4, 5 \pi / 4, 7 \pi / 4 \]
Figure (svg): The solution to Worked example the four fourth roots of -16 shown as a ladder of expressions, one row per legal move
\[ 2\,\text{cis}\left(\tfrac{\pi}{4}\right), \; 2\,\text{cis}\left(\tfrac{3\pi}{4}\right), \; 2\,\text{cis}\left(\tfrac{5\pi}{4}\right), \; 2\,\text{cis}\left(\tfrac{7\pi}{4}\right) \]
Verify: check one by DeMoivre
Why: Raising the first to the fourth gives 2 to the fourth, namely 16, with argument four times pi over 4, namely pi. That is 16 cis of pi, which is negative 16. Correct. Note also that none of the four roots is real, which is right — no real number has a negative fourth power.
Stitz & Zeager, College Trigonometry, Ch. 11 Applications of Trigonometry — §11.7 Polar Form of Complex Numbers §11.7, pp. 1002-1002
Trap
\[ w^2 = -2 + 2i\sqrt{3} = 4\,\text{cis}\left(\tfrac{2\pi}{3}\right) \;\Longrightarrow\; w = 2\,\text{cis}\left(\tfrac{\pi}{3}\right) \]
Halve the argument and take the real square root of the modulus
Why: That is the formula with k equal to zero, and it is a genuine root.
But there are two square roots, and the second is obtained with k equal to 1, adding a half turn to the argument. Reporting one is answering half the question.
\[ w_0 = 2\,\text{cis}\left(\tfrac{\pi}{3}\right), \qquad w_1 = 2\,\text{cis}\left(\tfrac{4\pi}{3}\right) \]
Run k from 0 through n minus 1
Why: Every value gives a distinct root.
This is the same shape of error as taking only the acute angle from an arcsine back in Lesson 11.2. A root is not a function of the number, and expecting one answer is importing an assumption from real arithmetic that does not survive.
Prediction
You are finding the seventh roots of a complex number.
Predict first
How many are there, and how are their arguments related?
Correct: Seven, with arguments differing by 2 pi / 7.
Why: The formula gives one root for each value of k from 0 to 6, and consecutive roots differ in argument by the step 2 pi over 7. After seven steps the argument has advanced a full turn and the roots begin repeating, which is exactly why there are seven and not more. All seven share the same modulus, the real seventh root of the original modulus.
Sorting
Every non-zero complex number has exactly n distinct nth roots.
Sort into buckets
Sort each by how many of its roots are real.
Two truths and a lie
Three of these are true and one is false.
Eliminate the wrong options
One of these statements about nth roots is wrong.
Survives elimination: B
Why: Statement B is false. Taking k equal to n adds a full turn to the argument, which returns the root with k equal to zero. So the values repeat with period n and there are exactly n distinct roots, which is the second half of the theorem's proof.
Section
Section 5
Concept
All n roots share a modulus, so they lie on one circle. Consecutive arguments differ by an equal step, so they are equally spaced round it. The roots are the vertices of a regular n-gon centred at the origin.
This is the strongest available check on a root computation: if the answers are not equally spaced on a circle, something has gone wrong, and it is visible without recomputing anything.
Figure (svg): The four fourth roots of negative sixteen plotted in the complex plane, evenly spaced a quarter turn apart on a circle of radius two
Stitz & Zeager, College Trigonometry, Ch. 11 Applications of Trigonometry — §11.7 Polar Form of Complex Numbers §11.7, pp. 1003-1003
Picture it
The cleanest case: modulus 1, one root at the number 1, the rest spaced round.
Figure (svg): The five fifth roots of one plotted on the unit circle, at the vertices of a regular pentagon with one vertex at the number one
For the fifth roots of 1 only one is real, and the other four come in conjugate pairs. That pattern holds for every odd n, and for even n there are two real roots instead.
Worked example
Example 11.7.4, part 4. The roots of unity.
\[ \text{Find the five fifth roots of } 1. \]
Write 1 in polar form
Why: One unit along the positive real axis.
\[ 1 \operatorname{cis}(0) \]
Take the real fifth root of the modulus
Why: One to any power is 1.
\[ | w | = 1 \]
The first root's argument is zero
Why: Zero divided by 5.
\[ w 0 = 1 \]
Step by 2 pi over 5 four times
Why: Round the unit circle.
\[ 2 \pi / 5, 4 \pi / 5, 6 \pi / 5, 8 \pi / 5 \]
Figure (svg): The solution to Worked example the five fifth roots of 1 shown as a ladder of expressions, one row per legal move
\[ 1, \; \text{cis}\left(\tfrac{2\pi}{5}\right), \; \text{cis}\left(\tfrac{4\pi}{5}\right), \; \text{cis}\left(\tfrac{6\pi}{5}\right), \; \text{cis}\left(\tfrac{8\pi}{5}\right) \]
Verify: note what cannot be simplified
Why: No identity developed in this course gives exact values for the cosine or sine of two fifths of pi, so these roots cannot be written in a nicer rectangular form by the available means. That is a real limitation and worth stating rather than hiding — a decimal approximation is the honest answer if one is needed.
Stitz & Zeager, College Trigonometry, Ch. 11 Applications of Trigonometry — §11.7 Polar Form of Complex Numbers §11.7, pp. 1003-1003
Prediction
You compute the six sixth roots of a complex number and plot them.
Predict first
What shape do they form?
Correct: A regular hexagon.
Why: All six roots have the same modulus so they lie on one circle, and their arguments are equally spaced by 2 pi over 6. Equally spaced points on a circle are the vertices of a regular polygon, here a hexagon. This holds for every n, and it is the fastest possible check on a root computation: if the plotted answers are not a regular polygon, something is wrong.
Worked example
Example 11.7.4, part 3. The formula is easy; the conversion is not.
\[ \text{Find the three cube roots of } \sqrt{2} + i\sqrt{2}. \]
Convert to polar form
Why: Modulus 2, argument pi over 4.
\[ 2 \operatorname{cis}(\frac{\pi}{4}) \]
Take the real cube root of the modulus
Why: The cube root of 2.
\[ | w | = 2 ^{\frac{1}{3}} \]
Divide the argument by 3
Why: Pi over 12.
\[ \text{first argument } \frac{\pi}{12} \]
Step by 2 pi over 3 twice
Why: Adding eight twelfths of pi each time.
\[ \frac{\pi}{12}, 3 \pi / 4, 17 \pi / 12 \]
Figure (svg): The solution to Worked example three cube roots, awkwardly placed shown as a ladder of expressions, one row per legal move
\[ \sqrt[3]{2}\,\text{cis}\left(\tfrac{\pi}{12}\right), \; \sqrt[3]{2}\,\text{cis}\left(\tfrac{3\pi}{4}\right), \; \sqrt[3]{2}\,\text{cis}\left(\tfrac{17\pi}{12}\right) \]
Verify: check the spacing
Why: The three arguments are pi over 12, nine twelfths of pi and seventeen twelfths of pi, differing by eight twelfths each time — which is two thirds of pi, the required step. Converting to rectangular form would need the half-angle or difference identities for the first and third, which is why the book leaves it as an exercise.
Stitz & Zeager, College Trigonometry, Ch. 11 Applications of Trigonometry — §11.7 Polar Form of Complex Numbers §11.7, pp. 1003-1003
Trap
\[ \text{cis}\left(\tfrac{2\pi}{5}\right) = \text{some exact surd expression} \]
Convert every root to rectangular form
Why: Every previous example converted cleanly, so this one should too.
But no identity in this course produces the cosine of two fifths of pi. The angle is not built from the standard ones by sums, differences or halvings, so the polar form is as far as the available tools reach.
Leave it in polar form, or give a decimal approximation if a numerical answer is wanted.
The polar form is a complete and exact answer. Nothing is being left undone by not converting; there is simply no better expression available.
Worth knowing: an exact surd expression for that cosine does exist and involves the golden ratio, but deriving it needs machinery beyond this course. Being unable to simplify is not the same as the answer being incomplete, and saying which of the two you are in is part of a good answer.
Faded example
For the eight eighth roots of a number, find the angular step.
Fill in the blanks
\text8 = \frac4___} = \frac______}
Why: The step is always a full turn divided by n, here 2 pi over 8, which simplifies to pi over 4 — forty-five degrees. Eight steps of 45 degrees is 360, which is why the ninth value would coincide with the first.
Sorting
Across the n roots of a single number.
Sort into buckets
Sort each quantity.
Real world
A digital signal processor computes a discrete Fourier transform. At the heart of the algorithm is a set of numbers called twiddle factors, which are precisely the nth roots of 1.
Discussion prompt
Explain why the roots of unity appear there, and what property of them the fast algorithm exploits.
Hint: What does a root of unity do when you raise it to a power?
Answer:
The transform decomposes a signal into sinusoids, and a sinusoid sampled at n equally spaced instants is exactly a walk round the n roots of unity. Each root is one sample of one frequency, so the roots are not an implementation detail — they are what the transform is made of.
The property the fast algorithm exploits is that the roots of unity have enormous internal structure. The squares of the nth roots are the roots of order n over 2, so a transform of size n can be built from two of half the size, and that halving is what turns an algorithm needing n squared operations into one needing n log n.
That speedup is not a small matter. The fast Fourier transform is what makes digital audio, image compression and wireless communication practical, and the fact it rests on is the one this section proves: that the nth roots of a complex number are evenly spaced points on a circle.
It is a fair illustration of how the chapter's themes compound. Polar form made multiplication easy; easy multiplication made powers easy; easy powers made roots findable; and the structure of those roots is what a great deal of modern computation runs on.
Comparison
Fill the blanks from memory. Each row says which form to reach for.
Comparison matrix
| Operation | Rectangular | Polar |
|---|---|---|
| addition | add the parts | no useful rule |
| multiplication | four products and a sign flip | multiply moduli, add arguments |
| nth power | a binomial expansion | DeMoivre: one real power, one multiplied angle |
| nth roots | factoring, and only for small n | the root formula, for every n |
The last row is the decisive one. Factoring finds the cube roots of 8 with effort and the fifth roots of 32 not at all, while the formula handles both without noticing the difference.
Pattern
Five moves, covering products, powers and roots.
Plot the roots as a check: n roots of one number are always the vertices of a regular n-gon centred at the origin.
OpenStax Algebra and Trigonometry 2e, §10.5 Polar Form of Complex Numbers §10.5
Check
The product rule.
Check your understanding
What is the product of 5 cis of pi over 3 and 4 cis of pi over 6?
Answer: B
Why: Moduli multiply, giving 20, and arguments add, giving pi over 3 plus pi over 6, which is pi over 2. The answer is 20 cis of pi over 2, which in rectangular form is 20i.
Check
DeMoivre.
Check your understanding
What is the fourth power of 2 cis of pi over 3?
Answer: C
Why: The modulus is raised to the fourth power, giving 16, and the argument is multiplied by 4, giving four thirds of pi. That angle is already under a full turn so no reduction is needed.
Check
Roots.
Check your understanding
How many distinct sixth roots does the number 64 have, and what is their common modulus?
Answer: C
Why: Every non-zero complex number has exactly n distinct nth roots, so 64 has six sixth roots. Their common modulus is the real sixth root of 64, which is 2. They lie at the vertices of a regular hexagon of radius 2, two of them real and four not.
Real world
A three-phase electrical supply delivers power on three conductors whose voltages are equal in size but offset in phase by a third of a cycle. An engineer analysing it represents each phase as a complex number.
Discussion prompt
Explain what the three numbers are, in the language of this lesson, and why the sum of the three phase voltages is zero.
Hint: Equal magnitudes, arguments a third of a turn apart.
Answer:
The three voltages have equal moduli and arguments differing by 2 pi over 3. That is precisely the description of the three cube roots of a single complex number — they are the original voltage times the three cube roots of unity.
The sum is zero because the three cube roots of unity sum to zero, which follows from the symmetry of the equilateral triangle they form: three equal vectors at 120 degrees cancel exactly. The same is true of the n roots of unity for every n above 1.
This is not an incidental fact. It is why a balanced three-phase system needs no return conductor — the currents cancel, so the neutral carries nothing, and the cable can be lighter and cheaper than three separate single-phase circuits would need.
So a theorem about the vertices of a regular polygon shows up as a saving in copper. The geometry of the roots is the engineering constraint, and the reason three phases were chosen rather than two or four comes down to the same arithmetic.
Commit first
Answer, then rate your confidence honestly. Confident and wrong is the combination worth finding.
Predict first
How many distinct complex fifth roots does the number 32 have?
Correct: Five.
\[ w_k = 2\,\text{cis}\left(\tfrac{2\pi}{5}k\right), \quad k = 0, 1, 2, 3, 4 \]
Why: Every non-zero complex number has exactly n distinct nth roots, so 32 has five fifth roots. Only one of them is real, namely 2, which is why real arithmetic reports a single answer. The other four are complex and lie with 2 at the vertices of a regular pentagon of radius 2 centred at the origin.
Explain it
They have been told that 8 has three cube roots and object that the cube root of 8 is obviously 2 and nothing else.
Discussion prompt
In no more than five sentences, explain what they are missing without telling them they are wrong.
Hint: What question is the symbol answering, and what question was asked?
Answer:
They are right that the cube root of 8 is 2 — the symbol denotes the principal root, and that is a single number by convention. But the question asked is different: which numbers cubed give 8, and that is an equation with a degree of 3.
A cubic equation has three solutions, and here two of them are complex: negative 1 plus i root three and its conjugate. Cubing either really does give 8, which is worth checking once by hand to believe.
Plot all three and they sit at the corners of an equilateral triangle on a circle of radius 2. Only one corner happens to land on the real axis, which is the only reason real arithmetic sees just one answer.
Exit ticket
Name the weakest spot before you close the deck. That is the one worth ten minutes tonight.
Predict first
Which of these would you least want to be handed cold on a quiz tomorrow?
Correct: Whichever you picked is the right answer — and each one has a specific fix.
Why: Products and quotients are fixed by remembering that moduli multiply or divide while arguments add or subtract. Powers are fixed by exponentiating the modulus and multiplying the argument, never the same operation on both. Roots are fixed by running k from 0 to n minus 1 and checking the answers form a regular polygon. The geometry is fixed by the phrase stretch by the modulus, turn by the argument. Pick yours and do five of that kind tonight rather than twenty mixed ones.
Connect it up
Do this on paper. It is worth more than rereading the slides.
Draw it
At the top of the page write the three rules of Theorem 11.16, and beside each note what happens to the modulus and what happens to the argument. Underneath, write the four-line proof of the product rule, marking the step where the sum identities are used. In the middle of the page draw two complex plane diagrams showing a product as a stretch followed by a rotation, labelling the stretch factor and the turn. In the bottom left, write the root formula and use it to find all four fourth roots of negative 16, listing them in polar form. In the bottom right, plot those four roots and draw the polygon they form, noting its radius and the angle between consecutive vertices. Finally, circle the one feature of the plotted roots that would let you spot a computational error without redoing any arithmetic.
The circled feature is the even spacing. If the plotted roots are not the vertices of a regular polygon centred at the origin, something is wrong — and that check costs a glance rather than a recomputation.
Recap
Five things, and the fourth solves equations that were previously out of reach.
| If the question says | Your first move is |
|---|---|
| Multiply these complex numbers | Convert to polar; multiply moduli, add arguments |
| Find a high power | DeMoivre, then reduce the argument |
| Find all the nth roots | Real nth root of the modulus, then step by 2 pi over n |
| How many roots are there? | Exactly n, for any non-zero number |
| Check your roots | Plot them; they must form a regular polygon |
Complex numbers are now a complete arithmetic with a geometry attached. The chapter turns next to vectors, where a quantity with a magnitude and a direction gets its own algebra — closely related to this one, and deliberately kept separate from it.
Stitz & Zeager, College Trigonometry, Ch. 11 Applications of Trigonometry — §11.7 Polar Form of Complex Numbers §11.7, pp. 997-1004 — everything on these slides traces back here
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