Complex numbers as points in a plane, and the polar description of those points. Defines the modulus as the distance from the origin and the argument as the set of angles that reach the number, singling out the principal argument in the interval from negative pi to pi. Establishes the properties of both, then assembles the polar form: the modulus times cosine theta plus i sine theta, abbreviated cis, and converts in both directions.
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. 991-997
Objectives
Five outcomes, and the second is the one where the notation earns its capital letter.
Stitz & Zeager, College Trigonometry, Ch. 11 Applications of Trigonometry — §11.7 Polar Form of Complex Numbers §11.7, pp. 991-997 — the pages these objectives are drawn from
Warm-up
Complex numbers were introduced as an algebraic device. They also have a geometry.
Discussion prompt
The complex number negative 4 plus 2i has two real numbers in it. What natural picture does that suggest, and what would the two axes be?
Hint: How many numbers does it take to locate a point in a plane?
Answer:
Two real numbers locate a point in a plane, so negative 4 plus 2i can be plotted at the point negative 4 comma 2. The horizontal axis carries the real part and the vertical axis the imaginary part.
Those axes get new names — the real axis and the imaginary axis — and the plane they determine is the complex plane. Nothing about the plane itself is new.
Which means every tool from the polar lessons applies immediately. A complex number has polar coordinates, and the whole of this lesson is working out what those coordinates are called and what they are good for.
Concept
Plot a complex number and give the resulting point polar coordinates with a non-negative first entry. The distance is called the modulus and the angle is called an argument.
Definition 11.2 — For a complex number z with a polar representation of its point having r non-negative: the modulus of z is r, written with absolute-value bars; the angle theta is an argument of z, and the set of all such angles is written arg z; and if z is non-zero, the unique argument in the interval from negative pi to pi is the principal argument, written Arg z.
\[ z = a + bi \;\longleftrightarrow\; (a, b) \;\longleftrightarrow\; (r, \theta) \]
The requirement that r be non-negative is what makes the modulus a single number. Nothing pins down the argument in the same way, which is why it is a set and needs a separate convention to produce one value.
Figure (svg): A complex number in the plane with its distance from the origin marked as the modulus and its angle from the positive real axis marked as an argument
Stitz & Zeager, College Trigonometry, Ch. 11 Applications of Trigonometry — §11.7 Polar Form of Complex Numbers §11.7, pp. 991-992
Section
Section 1
Concept
Associate the number with the ordered pair of its real and imaginary parts. The horizontal axis becomes the real axis and the vertical the imaginary axis.
The correspondence is one-to-one because two complex numbers are equal precisely when their real parts agree and their imaginary parts agree. That is what makes the real and imaginary parts well-defined functions.
Figure (svg): The complex plane with a real axis and an imaginary axis, and four complex numbers plotted as points
Stitz & Zeager, College Trigonometry, Ch. 11 Applications of Trigonometry — §11.7 Polar Form of Complex Numbers §11.7, pp. 991-991
Picture it
Real numbers on one axis, imaginary numbers on the other, everything else in between.
Figure (svg): The complex plane with a real axis and an imaginary axis, and four complex numbers plotted as points
The number negative 117 lies far out on the negative real axis, which is worth picturing when its modulus and argument are computed later.
Worked example
Example 11.7.1, the first two parts. Both need rewriting before reading.
\[ \text{Find the real and imaginary parts of } z = \sqrt{3} - i \text{ and of } z = 3i. \]
Rewrite the first in standard form
Why: Make the coefficient of i explicit.
\[ \sqrt{3} + (-1) i \]
Read off
Why: The coefficient of i is the imaginary part.
\[ R e = \sqrt{3}, I m = -1 \]
Rewrite the second
Why: Supply the missing real part.
\[ 0 + 3 i \]
Read off again
Why: The real part is zero.
\[ R e = 0, I m = 3 \]
Figure (svg): The solution to Worked example reading real and imaginary parts shown as a ladder of expressions, one row per legal move
\[ \text{Re}(\sqrt{3}-i) = \sqrt{3}, \; \text{Im} = -1; \qquad \text{Re}(3i) = 0, \; \text{Im} = 3 \]
Verify: check the imaginary part is real
Why: The imaginary part of a complex number is a real number, not an imaginary one — it is the coefficient of i, not the term. So the imaginary part of 3i is 3 and not 3i, which is a distinction worth being firm about.
Stitz & Zeager, College Trigonometry, Ch. 11 Applications of Trigonometry — §11.7 Polar Form of Complex Numbers §11.7, pp. 992-993
Sorting
Look at the signs of the two parts.
Sort into buckets
Sort each complex number.
Worked example
Example 11.7.1, all four, placed in the plane.
\[ \text{Plot } \sqrt{3}-i, \; -2+4i, \; 3i, \; -117. \]
First number
Why: Positive real, negative imaginary.
Second number
Why: Negative real, positive imaginary.
Third number
Why: Zero real part.
Fourth number
Why: Zero imaginary part, negative real.
Figure (svg): The solution to Worked example plotting four numbers shown as a ladder of expressions, one row per legal move
\[ \text{IV}, \quad \text{II}, \quad \text{on the } \text{Im} \text{ axis}, \quad \text{on the negative } \text{Re} \text{ axis} \]
Verify: check the last two are on axes
Why: Both have one part equal to zero, and a zero part means the point sits on the axis belonging to the other part. That will matter shortly, since the argument formula divides by the real part and so fails on the imaginary axis.
Stitz & Zeager, College Trigonometry, Ch. 11 Applications of Trigonometry — §11.7 Polar Form of Complex Numbers §11.7, pp. 993-993
Trap
\[ z = 3 + 4i \;\Longrightarrow\; \text{Im}(z) = 4i \]
Take the imaginary term as the imaginary part
Why: The term 4i is the part of the number that is imaginary, so the name seems to fit.
But the imaginary part is defined as the coefficient of i, which is 4 — a real number. That is what makes the real and imaginary parts a pair of real coordinates.
\[ \text{Re}(z) = 3, \quad \text{Im}(z) = 4 \]
Take the coefficient, not the term
Why: Both parts are real numbers.
The point of the definition is that a complex number is a pair of real numbers, which is what lets it be plotted at all. If the imaginary part were itself imaginary, there would be nothing to measure along the vertical axis.
Faded example
Write negative 5 in the standard form and identify its parts.
Fill in the blanks
-5 = -5 + 0 i, so the real part is -5 and the imaginary part is 0.
Why: Every real number is a complex number with imaginary part zero, and writing it out makes that explicit. The point lies on the real axis, five units to the left of the origin, which is why its argument will turn out to be pi.
Prediction
A complex number has real part zero and a positive imaginary part.
Predict first
Where is it in the plane?
Correct: On the positive imaginary axis.
Why: A zero real part means no horizontal displacement, so the point sits on the vertical axis, and a positive imaginary part puts it above the origin. This is precisely the case in which the tangent formula for the argument fails, since it would divide by the real part, and the argument has to be read off the picture instead.
Socratic
Every complex number gives a point, and every point gives a complex number.
Discussion prompt
What algebraic fact guarantees that no two different complex numbers land on the same point?
Hint: When are two complex numbers equal?
Answer:
Two complex numbers are equal exactly when their real parts agree and their imaginary parts agree. That is a fact from earlier algebra, not something the picture supplies.
It is what makes the real and imaginary parts well-defined: however a number is written, its two parts come out the same. Without it, one number could correspond to several points and the picture would be meaningless.
It is worth noticing how much rests on that small fact. The whole geometric interpretation of complex arithmetic depends on the correspondence being a genuine bijection, and the next lesson's results about multiplication as rotation would be unstateable without it.
Section
Section 2
Concept
Give the point polar coordinates with a non-negative distance. That distance is the modulus, unique because of the non-negativity requirement. The angle is an argument, and there are infinitely many of them.
The set of all arguments is written with a lower-case arg, and the single representative lying in the interval from negative pi to pi is the principal argument, written with a capital A.
Figure (svg): The set of all arguments of a complex number shown as infinitely many coterminal angles, with exactly one lying in the interval from negative pi to pi
Stitz & Zeager, College Trigonometry, Ch. 11 Applications of Trigonometry — §11.7 Polar Form of Complex Numbers §11.7, pp. 991-992
Picture it
One column has a convention that works; the other has one that works except at a single point.
Figure (svg): Two columns contrasting the modulus, which is a single well-defined number, with the argument, which is an infinite set
For zero, every angle is an argument and none lies distinguished among them, so the principal argument is simply left undefined there.
Worked example
Example 11.7.1, part 1. The whole calculation.
\[ \text{For } z = \sqrt{3} - i, \text{ find } |z|, \; \arg(z) \text{ and } \text{Arg}(z). \]
Compute the modulus
Why: Three plus one, then the root.
\[ | z | = 2 \]
Locate the point
Why: Positive real, negative imaginary.
Find the reference angle
Why: The tangent is negative one over root three.
\[ \frac{\pi}{6} \]
Place it and list the set
Why: Quadrant four, then add whole turns.
\[ -\frac{\pi}{6} + 2 \pi k \]
Figure (svg): The solution to Worked example a quadrant four number shown as a ladder of expressions, one row per legal move
\[ |z| = 2, \quad \arg(z) = \left\{-\tfrac{\pi}{6} + 2\pi k\right\}, \quad \text{Arg}(z) = -\tfrac{\pi}{6} \]
Verify: check the principal value is in range
Why: The interval is from negative pi to pi, including the right endpoint only. Negative pi over 6 is comfortably inside, and it is the only member of the set that is — the next ones are 11 pi over 6 and negative 13 pi over 6, both outside.
Stitz & Zeager, College Trigonometry, Ch. 11 Applications of Trigonometry — §11.7 Polar Form of Complex Numbers §11.7, pp. 992-992
Faded example
Find the modulus of negative 3 plus 4i.
Fill in the blanks
|z| = \sqrt25 = \sqrt5} = ___
Why: Nine plus 16 is 25, whose root is 5. Note that both parts are squared, so the sign of the real part makes no difference — the modulus is a distance and cannot be negative.
Worked example
Example 11.7.1, part 2. The angle is not a common one.
\[ \text{For } z = -2 + 4i, \text{ find } |z| \text{ and } \text{Arg}(z). \]
Compute the modulus
Why: Four plus 16, then the root.
\[ | z | = 2 \sqrt{5} \]
Locate the point
Why: Negative real, positive imaginary.
Compute the tangent
Why: Four over negative two.
\[ \tan \theta = -2 \]
Place the angle in quadrant two
Why: Pi minus the arctangent of 2.
\[ A r g = \pi - \arctan 2 \]
Figure (svg): The solution to Worked example a quadrant two number shown as a ladder of expressions, one row per legal move
\[ |z| = 2\sqrt{5}, \quad \text{Arg}(z) = \pi - \arctan(2) \approx 2.03 \]
Verify: check the quadrant
Why: The arctangent of 2 is about 1.107, so the principal argument is about 2.03 radians, or about 116 degrees. That is in quadrant two, matching the plotted point. Writing the arctangent alone would have given about negative 1.107, in quadrant four, which is wrong.
Stitz & Zeager, College Trigonometry, Ch. 11 Applications of Trigonometry — §11.7 Polar Form of Complex Numbers §11.7, pp. 992-992
Error analysis
A student writes down the arguments of a number.
Annotate
On: \( \text{Arg}(z) = \left\{-\tfrac{\pi}{6} + 2\pi k \;\middle|\; k \in \mathbb{Z}\right\} \)
The capitalisation is doing real work. One notation names an infinite set and the other names one of its elements, and equations mixing them are type errors rather than arithmetic errors.
Prediction
You are asked for the principal argument of the number 0.
Predict first
What is it?
Correct: Undefined.
Why: Zero is at the origin, which every angle reaches, so its set of arguments is all real numbers. Since every angle in the interval from negative pi to pi qualifies, no one of them is distinguished and there is no principal value to choose. The book leaves it undefined for exactly this reason, and it is the same exceptional point that had no polar angle back in Lesson 11.4a.
Sorting
One is a distance and the other is a direction.
Sort into buckets
Sort each statement by which it is about.
Edge cases
The principal argument lives in the interval from negative pi to pi.
Discussion prompt
Why is that interval half-open, and which endpoint is included?
Hint: What would happen if both endpoints were allowed?
Answer:
The interval includes pi and excludes negative pi. Those two angles are coterminal, so allowing both would give a negative real number two principal arguments and destroy uniqueness.
The choice of which endpoint to keep is a convention, and this book keeps the positive one. So the principal argument of negative 117 is pi rather than negative pi.
The same half-open device appeared for polar coordinates, where theta ran from 0 up to but not including 2 pi. Any interval of length one full turn needs exactly one endpoint removed, and which one is removed is arbitrary but must be stated. Here it is chosen so that a positive real number gets argument 0 sitting comfortably in the interior.
Section
Section 3
Concept
The modulus is a distance, so it is non-negative and vanishes only at zero. It also multiplies and divides cleanly, which is the property everything later depends on.
Theorem 11.14 — The modulus is the distance from the number to zero; it is non-negative and zero only at zero; it equals the square root of the sum of the squares of the real and imaginary parts; and it obeys the product, power and quotient rules.
\[ |zw| = |z||w|, \quad |z^n| = |z|^n, \quad \left|\frac{z}{w}\right| = \frac{|z|}{|w|} \]
There is deliberately no sum rule here. The modulus of a sum is not the sum of the moduli, any more than the length of a sum of two displacements is the sum of their lengths.
Figure (svg): The properties of the modulus, grouped by what each one says
Stitz & Zeager, College Trigonometry, Ch. 11 Applications of Trigonometry — §11.7 Polar Form of Complex Numbers §11.7, pp. 993-995
Picture it
Three describe what the modulus is, and three describe how it interacts with arithmetic.
Figure (svg): The properties of the modulus, grouped by what each one says
The three arithmetic rules all reduce to the product rule, so proving that one carefully is the whole work of the theorem.
Worked example
Pure algebra, with no geometry and no identities.
\[ \text{Prove } |zw| = |z||w|. \]
Multiply the two numbers out
Why: Real and imaginary parts of the product.
\[ (a c - b d) + (a d + b c) i \]
Write the modulus of the product
Why: Sum of the squares under a root.
\[ \sqrt{(a c - b d) ^{2} + (a d + b c) ^{2}} \]
Expand; the cross terms cancel
Why: Both contain the same product of four letters.
\[ a ^{2} c ^{2} + a ^{2} d ^{2} + b ^{2} c ^{2} + b ^{2} d ^{2} \]
Factor and split the root
Why: Group by the first pair.
\[ \sqrt{a ^{2} + b ^{2}} \sqrt{c ^{2} + d ^{2}} \]
Figure (svg): The solution to Worked example proving the product rule shown as a ladder of expressions, one row per legal move
\[ |zw| = \sqrt{a^2+b^2}\,\sqrt{c^2+d^2} = |z||w| \]
Verify: test on a case
Why: Take z equal to 3 plus 4i with modulus 5 and w equal to i with modulus 1. Their product is negative 4 plus 3i, whose modulus is also 5. Five times one is five, so the rule holds here.
Stitz & Zeager, College Trigonometry, Ch. 11 Applications of Trigonometry — §11.7 Polar Form of Complex Numbers §11.7, pp. 994-994
Prediction
Two complex numbers have moduli 3 and 5.
Predict first
What is the modulus of their product?
Correct: 15.
Why: The product rule says moduli multiply, so the modulus of the product is 3 times 5. This holds regardless of the arguments, which is what makes it so useful. Note the contrast with the sum, whose modulus is only bounded between 2 and 8 and genuinely depends on the arguments.
Worked example
The product rule doing all the work.
\[ \text{Prove } |z^n| = |z|^n \text{ for every natural number } n. \]
Check the base case
Why: The first power is the number itself.
\[ n = 1\text{ holds} \]
Assume the statement for k
Why: The induction hypothesis.
\[ | z ^{k} | = | z | ^{k} \]
Split the next power
Why: One factor of z peeled off.
\[ z ^{k + 1} = z ^{k} \times z \]
Apply the product rule then the hypothesis
Why: Twice.
\[ = | z | ^{k} | z | = | z | ^{k + 1} \]
Figure (svg): The solution to Worked example the power rule by induction shown as a ladder of expressions, one row per legal move
\[ |z^{k+1}| = |z^k|\,|z| = |z|^k|z| = |z|^{k+1} \]
Verify: sanity-check the shape of the argument
Why: The base case and the inductive step are both needed: the base case anchors the chain and the step propagates it. Note the step used only the product rule, so anything satisfying that rule satisfies this one — the proof never mentions complex numbers specifically.
Stitz & Zeager, College Trigonometry, Ch. 11 Applications of Trigonometry — §11.7 Polar Form of Complex Numbers §11.7, pp. 994-994
Trap
\[ |z + w| = |z| + |w| \]
Extend the product rule to sums
Why: The other operations all worked, so addition should too.
But take z equal to 1 and w equal to negative 1. Each has modulus 1, so the right side is 2, while the sum is 0 with modulus 0. The rule fails badly.
\[ |z + w| \le |z| + |w| \]
Only an inequality holds
Why: This is the triangle inequality.
Geometrically it is exactly the triangle inequality from Lesson 11.3: the direct route is no longer than going via a third point. Equality holds only when the two numbers point in the same direction, which is the degenerate flat triangle.
Faded example
Find the modulus of the fourth power of 1 plus i.
Fill in the blanks
|1 + i| = \sqrt4 \;\Longrightarrow\; |(1+i)^4| = \left(\sqrt4\right)^___} = ___
Why: The power rule turns a fourth power of a complex number into a fourth power of a single real number. Root two to the fourth is 4, obtained without ever expanding the binomial — which is the practical value of the rule.
Sorting
The modulus respects some operations and not others.
Sort into buckets
Sort each proposed rule.
Explain it to yourself
The absolute-value notation is reused for the modulus.
Discussion prompt
Explain why that reuse is justified rather than merely convenient.
Hint: What is the absolute value of a real number, geometrically?
Answer:
The absolute value of a real number is its distance from 0 on the number line, and the modulus of a complex number is its distance from 0 in the plane. Those are the same idea in one dimension and two.
More concretely, a real number is a complex number with imaginary part zero, and its modulus is the root of its square plus zero — which is its absolute value exactly. The new definition agrees with the old one wherever both apply, so the notation is unambiguous.
That is the standard for reusing notation, and it is worth insisting on. A symbol may be extended to a wider setting only if it keeps its old meaning on the old setting, and here it does.
Section
Section 4
Concept
For a number off the imaginary axis, the tangent of any argument is the imaginary part over the real part. On the imaginary axis that quotient is undefined, and the argument is read off the picture instead.
Theorem 11.15 — If the real part is non-zero, every argument satisfies the tangent relation. If the real part is zero, the arguments are pi over 2 plus full turns when the imaginary part is positive, and negative pi over 2 plus full turns when it is negative. If both parts are zero the number is zero and every angle is an argument.
\[ \tan(\theta) = \frac{\text{Im}(z)}{\text{Re}(z)}, \quad \text{Re}(z) \ne 0 \]
This is Theorem 11.7 from Lesson 11.4b in different clothing, including its exclusion. The arctangent still cannot see the quadrant, and the fix is still to plot the number first.
Stitz & Zeager, College Trigonometry, Ch. 11 Applications of Trigonometry — §11.7 Polar Form of Complex Numbers §11.7, pp. 995-995
Picture it
Every argument, and the one that gets a capital letter.
Figure (svg): The set of all arguments of a complex number shown as infinitely many coterminal angles, with exactly one lying in the interval from negative pi to pi
Listing the whole set and then selecting from it is more work than needed in practice, but doing it once makes clear what the capital letter is selecting from.
Worked example
Example 11.7.1, parts 3 and 4. No calculation needed.
\[ \text{Find } \text{Arg}(3i) \text{ and } \text{Arg}(-117). \]
Locate the first
Why: Three units up the imaginary axis.
Read its argument
Why: A quarter turn counter-clockwise.
\[ \frac{\pi}{2} \]
Locate the second
Why: One hundred seventeen units along the negative real axis.
Read its argument
Why: A half turn, and pi is in the interval.
Figure (svg): The solution to Worked example a number on an axis shown as a ladder of expressions, one row per legal move
\[ \text{Arg}(3i) = \tfrac{\pi}{2}, \quad \text{Arg}(-117) = \pi \]
Verify: check the interval for the second
Why: The set of arguments of negative 117 is the odd multiples of pi, namely pi, negative pi, 3 pi and so on. The interval from negative pi to pi includes its right endpoint and excludes its left, so pi qualifies and negative pi does not. The half-open convention decides this case and only this case.
Stitz & Zeager, College Trigonometry, Ch. 11 Applications of Trigonometry — §11.7 Polar Form of Complex Numbers §11.7, pp. 993-993
Matching
Plot it first, then place the reference angle.
Match the pairs
Why: All four have the same reference angle of pi over 4, since the two parts have equal magnitude in every case. Only the quadrant differs, and the principal argument places the angle in the interval from negative pi to pi — so the two lower quadrants take negative values rather than values above pi.
Worked example
Two numbers with identical tangents in opposite quadrants.
\[ \text{Find } \text{Arg}(1 + i) \text{ and } \text{Arg}(-1 - i). \]
Compute both tangents
Why: One over one, and negative one over negative one.
\[ \text{both equal } 1 \]
Note they are the same
Why: The relation cannot distinguish them.
Locate each
Why: Quadrant one and quadrant three.
Assign the angles
Why: Reference angle pi over 4 in each.
\[ \frac{\pi}{4}\text{ and } -3 \pi / 4 \]
Figure (svg): The solution to Worked example why the tangent is not enough shown as a ladder of expressions, one row per legal move
\[ \text{Arg}(1+i) = \tfrac{\pi}{4}, \quad \text{Arg}(-1-i) = -\tfrac{3\pi}{4} \]
Verify: check the second is in range
Why: Three quarters of pi taken negative is about negative 2.36, which lies in the interval from negative pi to pi. The alternative name of 5 pi over 4 is a genuine argument but is outside the interval, so it is not the principal one.
Error analysis
A student finds the principal argument of negative 1 minus i.
Annotate
On: \( \tan\theta = \frac{-1}{-1} = 1 \;\Longrightarrow\; \text{Arg}(z) = \arctan(1) = \tfrac{\pi}{4} \)
This is precisely the error from Lesson 11.4b, in new notation. The arctangent finds the reference angle and nothing more, and the fix has not changed: plot the number, then place the angle.
Faded example
Find the principal argument of negative 2 plus 2 root 3 i.
Fill in the blanks
\tan\theta = \frac2}3 = -\sqrt___ \;\Longrightarrow\; \text___ \tfrac______, \; \text___ \;\Longrightarrow\; \text___ = \tfrac___\pi}___}
Why: The reference angle is pi over 3, and quadrant two means subtracting it from pi, giving two thirds of pi. That value is under pi, so it lies in the principal interval and needs no further adjustment.
Prediction
A complex number is a negative real number.
Predict first
What is its principal argument?
Correct: pi.
Why: A negative real number lies on the negative real axis, reached by a half turn. Both pi and negative pi are arguments, since they are coterminal, but only one can be principal — and the convention keeps the interval's right endpoint, so pi is the principal argument. This is the single case in which the choice of endpoint matters.
Two truths and a lie
Three of these are true and one is false.
Eliminate the wrong options
One of these statements about arguments is wrong.
Survives elimination: B
Why: Statement B is false. The tangent relation gives only the reference angle, since a number and its negative share a tangent — so it cannot distinguish a quadrant one number from a quadrant three one. The quadrant has to come from the signs of the two parts, which is why the theorem is stated as a necessary condition rather than as a formula for the argument.
Section
Section 5
Concept
Substitute the polar conversion formulas into the rectangular form and factor out the distance. What remains in the bracket gets its own abbreviation.
Definition 11.3 — For a complex number z and any argument theta of it, the expression given by the modulus times cosine theta plus i sine theta is a polar form of z. The word a is deliberate: there are infinitely many, one for each argument.
\[ z = |z|\,\text{cis}(\theta) = |z|\left[\cos(\theta) + i\sin(\theta)\right] \]
Converting to polar form needs two pieces of information: the modulus and any one argument. The principal argument is convenient but not required.
Figure (svg): The polar form of a complex number, built by substituting the polar conversion formulas into the rectangular form and factoring out the modulus
Stitz & Zeager, College Trigonometry, Ch. 11 Applications of Trigonometry — §11.7 Polar Form of Complex Numbers §11.7, pp. 995-996
Picture it
Two substitutions and a factorisation.
Figure (svg): The polar form of a complex number, built by substituting the polar conversion formulas into the rectangular form and factoring out the modulus
Nothing about cis is deep. Its value is that it makes the next lesson's product rule expressible in one line rather than five.
Worked example
Example 11.7.2, part 1. Expand the cis and evaluate.
\[ \text{Find the rectangular form of } 4\,\text{cis}\left(\tfrac{2\pi}{3}\right). \]
Expand the cis
Why: Cosine plus i times sine.
\[ 4 [\cos(2 \pi / 3) + i \sin(2 \pi / 3)] \]
Evaluate the cosine
Why: Reference angle pi over 3 in quadrant two.
\[ \cos = -\frac{1}{2} \]
Evaluate the sine
Why: Positive in quadrant two.
\[ \sin = \sqrt{3} / 2 \]
Multiply through
Why: By the modulus 4.
\[ -2 + 2 i \sqrt{3} \]
Figure (svg): The solution to Worked example polar to rectangular shown as a ladder of expressions, one row per legal move
\[ 4\,\text{cis}\left(\tfrac{2\pi}{3}\right) = -2 + 2i\sqrt{3} \]
Verify: check the modulus
Why: The modulus of the answer is the root of 4 plus 12, which is 4 — matching the modulus in the polar form, as it must. That check catches most arithmetic slips in this direction.
Stitz & Zeager, College Trigonometry, Ch. 11 Applications of Trigonometry — §11.7 Polar Form of Complex Numbers §11.7, pp. 996-996
Faded example
Convert 2 cis of negative 3 pi over 4 into rectangular form.
Fill in the blanks
2\left[\cos\left(-\tfrac-sqrt2/2-sqrt2 - i sqrt2\right) + i\sin\left(-\tfrac______\right)\right] = 2\left[___ + i___\right] = ___
Why: Negative three quarters of pi lies in quadrant three, where both the cosine and the sine are negative root two over two. Multiplying by 2 gives negative root two for each part, so the real and imaginary parts are equal here.
Worked example
Example 11.7.2, part 2. Reusing the earlier calculations.
\[ \text{Find a polar form of } \sqrt{3} - i \text{ and of } -2 + 4i. \]
Recall the first modulus and argument
Why: From the earlier example.
\[ 2\text{ and } -\frac{\pi}{6} \]
Assemble the first polar form
Why: Modulus times cis of the argument.
\[ 2 \operatorname{cis}(-\frac{\pi}{6}) \]
Recall the second
Why: Modulus 2 root 5, argument pi minus arctan 2.
\[ 2 \sqrt{5}, \pi - \arctan 2 \]
Assemble the second
Why: Same pattern.
\[ 2 \sqrt{5} \operatorname{cis}(\pi - \arctan 2) \]
Figure (svg): The solution to Worked example rectangular to polar shown as a ladder of expressions, one row per legal move
\[ 2\,\text{cis}\left(-\tfrac{\pi}{6}\right), \qquad 2\sqrt{5}\,\text{cis}\left(\pi - \arctan(2)\right) \]
Verify: convert the first back
Why: Two times the cosine of negative pi over 6 is two times root three over two, which is root three; and two times the sine is two times negative one half, which is negative 1. So the rectangular form returns as root three minus i.
Stitz & Zeager, College Trigonometry, Ch. 11 Applications of Trigonometry — §11.7 Polar Form of Complex Numbers §11.7, pp. 996-996
Trap
\[ z = 2\,\text{cis}\left(-\tfrac{\pi}{6}\right) \quad \text{the polar form of } z \]
Give the polar form using the principal argument
Why: That produces a unique answer, so it looks like the polar form.
But any argument may be used, and there are infinitely many. The expression with 11 pi over 6, or with negative 13 pi over 6, is equally a polar form of the same number.
\[ z = 2\,\text{cis}\left(-\tfrac{\pi}{6} + 2\pi k\right) \text{ for any integer } k \]
Say a polar form, not the polar form
Why: The definition uses the indefinite article deliberately.
This matters more than it looks. The next lesson's root-finding produces different arguments for different roots, and treating polar forms as unique would make that impossible to state. The plurality is the useful part.
Prediction
A complex number has modulus 1.
Predict first
Where does it lie in the plane?
Correct: On the unit circle.
Why: A modulus of 1 means a distance of 1 from the origin, and the set of all such points is the unit circle. Its polar form is simply cis of theta, with no leading factor, so the numbers of modulus 1 are exactly the values of cis. This is why cis of pi over 2 turns out to be i, which sits on the circle a quarter turn round.
Sorting
Match each rectangular number to its modulus.
Sort into buckets
Sort each number by its modulus.
Real world
An electrical engineer analysing an alternating-current circuit represents each voltage and current as a phasor: a complex number whose modulus is the amplitude and whose argument is the phase.
Discussion prompt
Explain why the polar form is the natural representation here, and what the two parameters mean physically.
Hint: What two things describe a sinusoidal signal of known frequency?
Answer:
A sinusoid of fixed frequency is completely described by two numbers: how big it is and when it peaks. Those are exactly a modulus and an argument, so a phasor is a complex number in polar form and nothing else.
The modulus is the amplitude — the voltage or current in volts or amps — and the argument is the phase, how far the signal leads or lags a reference. Both are read directly off the polar form and neither is visible in the rectangular one.
What makes this more than notation is the next lesson: multiplying phasors multiplies the amplitudes and adds the phases, which is exactly what passing a signal through a component does. Circuit analysis becomes complex multiplication, and the reason it becomes easy is the reason this lesson exists.
Engineers even write the phase in degrees while the mathematics uses radians, and the notation with a leading modulus and an angle symbol is the same object as cis. The polar form of a complex number is standard engineering vocabulary, learned in one course as mathematics and in another as electricity.
Comparison
Fill the blanks from memory. The last row is why the next lesson exists.
Comparison matrix
| Rectangular form | Polar form | |
|---|---|---|
| written as | a + bi | the modulus times cis of an argument |
| the two parameters | real and imaginary parts | modulus and argument |
| uniqueness | exactly one form | infinitely many, one per argument |
| addition | easy: add the parts | awkward |
| multiplication | awkward: four products to expand | easy, as the next lesson shows |
The two forms are good at opposite operations, which is the whole reason for keeping both. Addition wants rectangular; multiplication, powers and roots want polar.
Pattern
Five moves, covering both directions of conversion.
Say a polar form rather than the polar form. Any argument works, and the plurality is used deliberately in the next lesson.
OpenStax Algebra and Trigonometry 2e, §10.5 Polar Form of Complex Numbers §10.5
Check
The modulus.
Check your understanding
What is the modulus of negative 5 plus 12i?
Answer: B
Why: The modulus is the root of 25 plus 144, which is the root of 169, namely 13. Both parts are squared, so the negative sign on the real part makes no difference.
Check
The principal argument.
Check your understanding
What is the principal argument of negative 1 plus i?
Answer: B
Why: The number lies in quadrant two, with a reference angle of pi over 4 since the two parts have equal magnitude. Quadrant two means pi minus the reference angle, giving three quarters of pi, which lies in the principal interval.
Check
The polar form.
Check your understanding
What is the rectangular form of 6 cis of pi?
Answer: C
Why: Expanding gives 6 times the cosine of pi plus i times the sine of pi, which is 6 times negative 1 plus i times 0, namely negative 6. Geometrically, going 6 units out and turning a half turn lands on the negative real axis.
Real world
A signal-processing engineer applies a Fourier transform to an audio recording. The output is a complex number for each frequency, and the engineer plots two graphs from it: one of moduli against frequency and one of arguments against frequency.
Discussion prompt
Explain what each graph shows physically, and why the polar description rather than the rectangular one is the useful one.
Hint: What do the two parameters mean for a sinusoid?
Answer:
Each complex output describes a sinusoid at one frequency, and the polar parameters are exactly its two physical properties: the modulus is how loud that frequency is and the argument is its phase, where in its cycle it starts.
The first graph is the magnitude spectrum, which is what a spectrum analyser draws and what corresponds to what you hear. The second is the phase spectrum, which governs how the frequencies line up in time and therefore the shape of transients.
The rectangular parts have no direct physical meaning at all. There is no property of the sound corresponding to the real part of a Fourier coefficient; it is an artefact of the coordinate system, and no engineer plots it.
This is worth generalising. The useful coordinates are the ones whose entries mean something in the problem, and for anything oscillating that is nearly always amplitude and phase — which is to say, modulus and argument. The mathematics of this lesson is what lets those two be extracted from a single complex number.
Commit first
Answer, then rate your confidence honestly. Confident and wrong is the combination worth finding.
Predict first
How many polar forms does a non-zero complex number have?
Correct: Infinitely many.
\[ 2\,\text{cis}\left(-\tfrac{\pi}{6}\right) = 2\,\text{cis}\left(\tfrac{11\pi}{6}\right) = 2\,\text{cis}\left(-\tfrac{13\pi}{6}\right) = \cdots \]
Why: A polar form uses any argument, and there are infinitely many arguments, differing by whole turns. So there are infinitely many polar forms of the same number, which is why the definition says a polar form rather than the polar form. Exactly one of them uses the principal argument, and that one is unique — but it is a choice made afterwards, not a property of the number.
Explain it
They are comfortable with complex arithmetic but see no reason for the modulus and argument, which look like extra vocabulary for something already understood.
Discussion prompt
In no more than five sentences, give them a reason to care.
Hint: What operation is painful in rectangular form?
Answer:
Adding complex numbers in rectangular form is easy: add the parts. Multiplying is not — four products, a sign flip from i squared, and terms to collect, and raising to the tenth power is out of the question.
In polar form, multiplication is multiply the moduli and add the arguments, and a tenth power is a tenth power of one real number and a multiplication of one angle by ten. That is the whole payoff, and the next lesson establishes it.
So the modulus and argument are not extra vocabulary; they are the coordinates in which multiplication is simple, in exactly the way the real and imaginary parts are the coordinates in which addition is simple. Having both is the point.
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: The modulus is fixed by one formula and never has a sign to worry about. The principal argument is fixed by plotting the number before reaching for the arctangent. The two notations are fixed by remembering that the lower-case one is a set and the capital one is a single element of it. Conversion is fixed by expanding the cis in one direction and assembling modulus and argument in the other. 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 draw the complex plane with both axes labelled, plot the four numbers root three minus i, negative 2 plus 4i, 3i and negative 117, and beside each write its modulus and its principal argument. Underneath, write the definitions of modulus, argument and principal argument in your own words, and note in one line why the modulus is a number and the argument is a set. In the middle of the page list the six properties of the modulus, marking which three follow from the product rule. In the bottom left, write the three-line derivation of the polar form from the rectangular form. In the bottom right, convert 4 cis of two thirds of pi into rectangular form and convert root three minus i into polar form. Finally, circle the one complex number on the page for which the principal argument is undefined, or write none if there is not one.
There is none on the page, because zero is not among the four numbers. That is worth noticing rather than glossing: zero is the only complex number with no principal argument, and it is the same exceptional point as the pole in the polar lessons.
Recap
Five things, and the last one is the setup for everything in the next lesson.
| If the question says | Your first move is |
|---|---|
| Find the modulus | Square both parts, add, take the root |
| Find the argument | Plot it, then use the tangent for the reference angle |
| The real part is zero | Read the angle off the axis directly |
| Find the principal argument | Reduce into the interval from -pi to pi |
| Convert to rectangular form | Expand the cis and evaluate |
The polar form is now assembled but has not yet been used for anything. The next lesson supplies the payoff: multiplication multiplies moduli and adds arguments, powers become DeMoivre's Theorem, and the nth roots of a complex number turn out to be evenly spaced points on a circle.
Stitz & Zeager, College Trigonometry, Ch. 11 Applications of Trigonometry — §11.7 Polar Form of Complex Numbers §11.7, pp. 991-997 — everything on these slides traces back here
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