Enlarges the number system so that every quadratic has roots. Defines the imaginary unit, plots complex numbers on a plane, and works out the arithmetic — addition and multiplication behave like ordinary algebra plus the single rule that i squared is negative one, and division is handled by multiplying through by the conjugate.
Subject: Precalculus · 65 slides · symbolic lesson
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Title
Precalculus · Chapter 3 — Polynomial and Rational Functions
§3.1 Complex Numbers, pp. 274-285
Objectives
Five things, each one you can check yourself on paper.
OpenStax, Precalculus, §3.1 Complex Numbers §3.1, pp. 274-285 — the pages these objectives are drawn from
Warm-up
Every number system in your education was built to solve an equation the previous one could not.
Discussion prompt
Which equation forced the invention of negative numbers, which forced fractions, and which one is left over?
Hint: Try to solve each of: x plus 5 equals 2, then 3x equals 1, then x squared equals negative 1.
Answer:
The first has no solution among the counting numbers, so negatives were invented. The second has none among the integers, so fractions were.
The third has none among the reals: every real number squares to something at or above zero, so nothing squares to negative 1. The pattern is complete and one more extension is needed.
So complex numbers are the next step in a sequence you have already climbed three times. Each step is the same move: name the missing solution and work out the arithmetic that follows, and each time the result turned out to be useful far beyond the equation that prompted it.
Concept
Define i to be a number whose square is negative one. All the ordinary rules of algebra continue to hold, with one extra substitution available whenever i squared appears.
imaginary unit — The number i, defined by the property that i squared equals negative one. It is not a real number, and the numbers of the form a plus b i are called complex numbers.
\[ i^2 = -1, \qquad a+bi \text{ with } a,b \in \mathbb{R} \]
The word 'imaginary' is an accident of history and an unhelpful one. These numbers are no less real than negative numbers, which were themselves called absurd for centuries. What matters is that the arithmetic is consistent and that it answers questions the smaller system could not.
Figure (svg): A parabola sitting entirely above the horizontal axis with a horizontal line at zero that it never meets, labelled to show that the equation has no real solutions and therefore needs a larger number system
OpenStax, Precalculus, §3.1 Complex Numbers §3.1, pp. 274-277
Section
Section 1
Concept
The square root of a negative number is written by factoring out negative one, replacing its root by i, and simplifying what remains.
\[ \sqrt{-a} = i\sqrt{a} \quad \text{for } a>0 \]
That last convention matters more than it looks. Written with the i inside or immediately before a radical it is easy to misread which factor the root applies to, and the resulting expression is ambiguous. Putting i in front, with the radical after it, removes the doubt.
Figure (svg): A parabola sitting entirely above the horizontal axis with a horizontal line at zero that it never meets, labelled to show that the equation has no real solutions and therefore needs a larger number system
OpenStax, Precalculus, §3.1 Complex Numbers §3.1, pp. 274-278
Picture it
The parabola clears the axis, so no real input makes the output zero.
Figure (svg): A parabola sitting entirely above the horizontal axis with a horizontal line at zero that it never meets, labelled to show that the equation has no real solutions and therefore needs a larger number system
Once i exists, this equation has the two solutions i and negative i. The graph does not change; what changes is the set of numbers we are willing to look in.
Worked example
Factor out the negative one before anything else.
\[ \text{Simplify } \sqrt{-48}. \]
Factor out negative one
Why: Separating the sign from the size.
\[ \sqrt{-1} \sqrt{48} \]
Replace the root of negative one
Why: By definition.
\[ i \sqrt{48} \]
Simplify the remaining radical
Why: Forty eight is 16 times 3.
\[ i \sqrt{16} \sqrt{3} \]
Finish
Why: The root of 16 is 4.
\[ 4 i \sqrt{3} \]
Figure (svg): The solution to Worked example simplify a root of a negative shown as a ladder of expressions, one row per legal move
\[ \sqrt{-48} = 4i\sqrt{3} \]
Verify: square the answer
Why: Squaring gives 16 times i squared times 3, which is 16 times negative 1 times 3, giving negative 48 — the original radicand. Note that the i is written before the radical, so there is no doubt that the root applies only to the 3.
OpenStax, Precalculus, §3.1 Complex Numbers §3.1, pp. 275-276
Faded example
Write the square root of negative 75 in the standard form.
Fill in the blanks
\sqrt5 = i\sqrt3 = i\sqrt___\sqrt___ = ___i\sqrt___}
Why: Seventy five is 25 times 3, and the root of 25 is 5, leaving the root of 3 under the radical. The i goes in front of the radical rather than inside it, so the expression cannot be misread as the root of 3i.
Worked example
The algebra is unchanged; only the last step is new.
\[ \text{Solve } x^2+9=0. \]
Isolate the square
Why: Subtract 9 from both sides.
\[ x ^{2} = -9 \]
Take square roots of both sides
Why: Remembering the plus or minus.
\[ x = +- \sqrt{-9} \]
Simplify the root
Why: Factor out negative one.
\[ \sqrt{-9} = 3 i \]
Write both solutions
Why: The plus and the minus.
\[ x = 3 i, x = -3 i \]
Figure (svg): The solution to Worked example solve a quadratic with no real roots shown as a ladder of expressions, one row per legal move
\[ x = \pm 3i \]
Verify: substitute one back
Why: At 3i the square is 9 times i squared, which is negative 9, and adding 9 gives zero. The other root works identically. Note the two roots are a conjugate pair, which §3.6 will show is always the case for a polynomial with real coefficients.
OpenStax, Precalculus, §3.1 Complex Numbers §3.1, pp. 277-278
Trap
\[ \sqrt{-4}\cdot\sqrt{-9} = \sqrt{36} = 6 \]
Combine the two radicals into one
Why: The rule that the product of roots is the root of the product is applied as usual.
The answer is given as 6.
That rule requires non-negative radicands, and both of these are negative. Converting first gives 2i times 3i, which is 6 times i squared, and that is negative 6.
The correct answer is negative 6, not 6 — the sign is wrong, not merely the form.
Convert every root of a negative into i form before multiplying anything. The radical rules you learned were stated for non-negative radicands, and this is the one place where using them out of range silently gives a wrong sign.
Sorting
A complex number is purely real when its imaginary part is zero and purely imaginary when its real part is.
Sort into buckets
Sort each number.
Prediction
A quadratic equation has a negative discriminant.
Predict first
How many solutions does it have among the complex numbers?
Correct: Two, forming a conjugate pair.
Why: A negative discriminant means the quadratic formula asks for the square root of a negative number, which is now available. The plus and the minus give two distinct solutions differing only in the sign of their imaginary parts — a conjugate pair. Over the reals there are none, which is the same equation looked at in a smaller number system.
Socratic
These numbers have been called imaginary since Descartes, dismissively.
Discussion prompt
In what sense are complex numbers no less real than negative numbers, and why does the name persist?
Hint: Can you hold negative three apples?
Answer:
Negative numbers were called absurd and fictitious for centuries, on the grounds that you cannot have negative three of anything. They earned acceptance by being useful and consistent, not by becoming tangible.
Complex numbers are in the same position and have the same credentials: the arithmetic is consistent, and they describe real phenomena — alternating current, quantum states, wave behaviour — better than the reals alone can.
The name persists because terminology is sticky, not because anyone still means it. It is worth being explicitly aware of, because students who take the word literally tend to treat results involving i as provisional, when they are as final as any other.
Section
Section 2
Concept
A complex number needs two real numbers to specify it, so it is plotted as a point on a plane rather than on a line: the real part horizontally and the imaginary part vertically.
The plane is not decoration. It makes the conjugate a reflection, the modulus a distance, and multiplication a rotation combined with a scaling — though that last fact has to wait until §8.5, where polar form makes it visible.
Figure (svg): The complex plane with a point plotted, showing the real part measured along the horizontal axis and the imaginary part along the vertical, with the modulus drawn as the distance to the origin
OpenStax, Precalculus, §3.1 Complex Numbers §3.1, pp. 278-280
Picture it
The two parts are the two coordinates, and the modulus is the hypotenuse.
Figure (svg): The complex plane with a point plotted, showing the real part measured along the horizontal axis and the imaginary part along the vertical, with the modulus drawn as the distance to the origin
The number 3 plus 4i sits 3 across and 4 up, and its distance from the origin is 5 — the familiar right triangle, which is why moduli of small complex numbers are so often whole numbers.
Worked example
Two coordinates, then Pythagoras.
\[ \text{Plot } -2+3i \text{ and find its modulus.} \]
Read the real part
Why: It is the horizontal coordinate.
\[ \text{across } -2 \]
Read the imaginary part
Why: It is the vertical coordinate.
\[ \text{up } 3 \]
Apply the Pythagorean theorem
Why: Square both, add, take the root.
\[ \sqrt{4 + 9} \]
Simplify
Why: Thirteen has no square factors.
\[ \sqrt{13} \]
Figure (svg): The complex plane with a point plotted, showing the real part measured along the horizontal axis and the imaginary part along the vertical, with the modulus drawn as the distance to the origin
\[ |-2+3i| = \sqrt{13} \approx 3.61 \]
Verify: check the modulus is positive
Why: A modulus is a distance and must be positive regardless of the signs of the parts. The negative real part contributed 4 after squaring, exactly as a positive 2 would have, which is what makes the modulus insensitive to the signs — as a distance should be.
OpenStax, Precalculus, §3.1 Complex Numbers §3.1, pp. 279-280
Matching
Real part across, imaginary part up.
Match the pairs
Why: Numbers with zero imaginary part lie on the horizontal axis, which is where the reals live. Purely imaginary numbers lie on the vertical one. A number with both parts nonzero sits off both axes, and the signs of the two parts decide which quadrant.
Worked example
Adding complex numbers adds the coordinates.
\[ \text{Add } (3+2i)+(-1+4i). \]
Add the real parts
Why: The horizontal coordinates.
\[ 3 + (-1) = 2 \]
Add the imaginary parts
Why: The vertical coordinates.
\[ 2 + 4 = 6 \]
Assemble
Why: Real part plus imaginary part times i.
\[ 2 + 6 i \]
Note the geometry
Why: The point is the tip of the two arrows joined head to tail.
Figure (svg): The solution to Worked example add graphically and algebraically shown as a ladder of expressions, one row per legal move
\[ (3+2i)+(-1+4i) = 2+6i \]
Verify: check it componentwise
Why: The real parts 3 and negative 1 give 2, and the imaginary parts 2 and 4 give 6. Adding complex numbers never mixes the two parts, which is why it is exactly vector addition and why it is the easiest of the four operations.
OpenStax, Precalculus, §3.1 Complex Numbers §3.1, pp. 280-281
Error analysis
A student simplifies a sum.
Annotate
On: \( (3+2i)+(4+5i) = 14i \)
Treat a plus b i as two separate quantities that travel together, exactly like the two coordinates of a point. Nothing ever converts one into the other except multiplication by i.
Faded example
Find the distance from the origin to the point 5 minus 12i.
Fill in the blanks
|5-12i| = \sqrt144}} = \sqrt13 = ___
Why: Squaring negative 12 gives positive 144, and 25 plus 144 is 169, whose root is 13. The sign of the imaginary part vanishes when squared, which is correct: a distance does not depend on direction. This is the 5-12-13 right triangle.
Prediction
A complex number is replaced by its conjugate, which negates the imaginary part.
Predict first
What happens to its modulus?
Correct: It is unchanged.
Why: The modulus squares the imaginary part, and squaring destroys the sign, so negating that part leaves the distance untouched. Geometrically the conjugate reflects the point across the horizontal axis, and a reflection never changes distance from a point on the mirror line.
Explain it to yourself
Real numbers fit on a line; complex numbers need a plane.
Discussion prompt
Explain why one dimension is not enough for the complex numbers.
Hint: How many real numbers do you need to write down to specify one complex number?
Answer:
A complex number is specified by two independent real numbers, its real and imaginary parts, and neither is determined by the other. Two independent coordinates need two dimensions.
Trying to fit them on a line would mean ordering them, and complex numbers cannot be ordered in any way compatible with their arithmetic — asking whether i is greater than 1 has no sensible answer.
The plane also explains why so many complex operations have geometric descriptions: the conjugate is a reflection, the modulus is a distance, addition is vector addition, and — as §8.5 will show — multiplication is a rotation. None of that would be visible on a line.
Section
Section 3
Concept
Complex numbers add and multiply exactly as binomials in i do. The only extra step is replacing i squared by negative one wherever it appears.
The reason multiplication feels surprising the first few times is that a product of two purely imaginary numbers is real. That is not an anomaly — it is exactly the substitution doing its job, and it is what makes the conjugate trick work in the next section.
Figure (svg): A complex number and its conjugate plotted as mirror images across the horizontal axis, with their product shown to be a real number
OpenStax, Precalculus, §3.1 Complex Numbers §3.1, pp. 281-283
Picture it
The product of these two mirror-image points is a real number.
Figure (svg): A complex number and its conjugate plotted as mirror images across the horizontal axis, with their product shown to be a real number
The cross terms cancel because they have opposite signs, and the i squared term turns real. That is the whole mechanism behind division.
Worked example
Expand, then substitute.
\[ \text{Multiply } (2+3i)(4-5i). \]
Expand as binomials
Why: Every term against every term.
\[ 8 - 10 i + 12 i - 15 i ^{2} \]
Substitute for i squared
Why: It is negative one.
\[ 8 - 10 i + 12 i + 15 \]
Combine the real parts
Why: Eight plus fifteen.
\[ 23 \]
Combine the imaginary parts
Why: Negative ten plus twelve.
\[ +2 i \]
Figure (svg): The solution to Worked example multiply two complex numbers shown as a ladder of expressions, one row per legal move
\[ (2+3i)(4-5i) = 23+2i \]
Verify: check the substitution changed a sign
Why: The last term was negative 15 i squared, and substituting turned it into positive 15 — a sign change that is easy to miss. Without it the real part would have been negative 7 instead of 23, so this is the step that carries the whole difference between complex and ordinary binomial multiplication.
OpenStax, Precalculus, §3.1 Complex Numbers §3.1, pp. 282-283
Faded example
Multiply 1 plus 4i by 2 minus 3i.
Fill in the blanks
2 - 3i + 8i - 12i^2 = 2 + 5i + 12 = 14 + 5i
Why: The term negative 12 i squared becomes positive 12 after substitution, so the real part is 2 plus 12, which is 14. The imaginary terms negative 3i and 8i combine to 5i. The sign change on the i squared term is the step that turns this from binomial multiplication into complex multiplication.
Worked example
The minus sign applies to both parts.
\[ \text{Subtract } (7-2i)-(3+5i). \]
Distribute the minus sign
Why: Both terms of the second number change sign.
\[ 7 - 2 i - 3 - 5 i \]
Combine the real parts
Why: Seven minus three.
\[ 4 \]
Combine the imaginary parts
Why: Negative two minus five.
\[ -7 i \]
Assemble
Why: Standard form.
\[ 4 - 7 i \]
Figure (svg): The solution to Worked example subtract carefully shown as a ladder of expressions, one row per legal move
\[ (7-2i)-(3+5i) = 4-7i \]
Verify: check the second imaginary part changed sign
Why: The subtracted number had imaginary part positive 5, and after distributing it contributes negative 5. Failing to distribute across both terms is the standard error, and it would have given 4 plus 3i — wrong in the imaginary part only, which makes it easy to overlook.
OpenStax, Precalculus, §3.1 Complex Numbers §3.1, pp. 281-282
Trap
\[ (3+2i)(3+2i) = 9 + 12i + 4i^2 = 9 + 12i + 4 = 13 + 12i \]
Expand the square and simplify
Why: The expansion is correct and the i squared term is identified.
The four i squared is replaced by positive 4 rather than negative 4.
i squared is negative one, so 4 i squared is negative 4, not positive 4. The real part is 9 minus 4, which is 5.
The correct answer is 5 plus 12i. The error is a single sign and it changes the real part by 8.
Write the substitution out explicitly rather than doing it in your head: replace i squared by the bracketed negative one and then simplify. It is one extra line and it removes the commonest error in the section.
Prediction
A complex number is multiplied by its own conjugate.
Predict first
What kind of number results?
Correct: A real number, equal to the squared modulus.
Why: The cross terms have opposite signs and cancel, and the i squared term turns real, leaving the sum of the squares of the two parts — which is exactly the modulus squared. This is the reason conjugates are the right tool for clearing a complex denominator, and it never fails.
Discrimination
Some operations keep the real and imaginary parts separate and some do not.
Sort into buckets
Sort each operation.
Counterexample
A classmate claims that multiplying two numbers with nonzero imaginary parts always gives a number with a nonzero imaginary part.
Discussion prompt
Find a counterexample.
Hint: What multiplies to something real?
Answer:
Any conjugate pair. Multiply 3 plus 2i by 3 minus 2i and the result is 13, entirely real, though both factors had nonzero imaginary parts.
A simpler one: i times i gives negative one. Both factors are purely imaginary and the product is real, which is the defining property of i itself.
The general lesson is that the imaginary part of a product is not built from the imaginary parts of the factors alone. Multiplication genuinely mixes the two coordinates, and treating a plus b i as two independent quantities works for addition but not here.
Section
Section 4
Concept
A complex number is not in standard form while an i remains in a denominator. Multiplying numerator and denominator by the denominator's conjugate makes the denominator real.
\[ \frac{a+bi}{c+di} \cdot \frac{c-di}{c-di} \]
This is exactly the rationalising trick used to clear a radical from a denominator in Algebra 2, and for exactly the same reason: multiplying by a conjugate turns a difference of two things into a difference of their squares, which removes the awkward one.
Figure (svg): A complex number and its conjugate plotted as mirror images across the horizontal axis, with their product shown to be a real number
OpenStax, Precalculus, §3.1 Complex Numbers §3.1, pp. 283-285
Picture it
The product of the pair lands on the real axis, which is what a denominator needs.
Figure (svg): A complex number and its conjugate plotted as mirror images across the horizontal axis, with their product shown to be a real number
Nine plus four is 13, with no i term surviving. The cross terms cancelled because they differed only in sign, which is what conjugation guarantees.
Worked example
Multiply top and bottom by the bottom's conjugate.
\[ \text{Simplify } \frac{3+2i}{1-i}. \]
Write the conjugate of the denominator
Why: Change the sign of its imaginary part.
\[ \text{conjugate is } 1 + i \]
Multiply top and bottom by it
Why: This is multiplying by one.
\[ (3 + 2 i) (1 + i) / ((1 - i) (1 + i)) \]
Expand the denominator
Why: The cross terms cancel.
\[ 1 + 1 = 2 \]
Expand the numerator and substitute
Why: Three plus 3i plus 2i minus 2.
\[ 1 + 5 i \]
Figure (svg): The solution to Worked example divide complex numbers shown as a ladder of expressions, one row per legal move
\[ \frac{3+2i}{1-i} = \frac{1+5i}{2} = \tfrac{1}{2}+\tfrac{5}{2}i \]
Verify: multiply back
Why: Taking the answer times the original denominator: one half plus five halves i, times 1 minus i, gives one half minus one half i plus five halves i plus five halves, which is 3 plus 2i — the original numerator. Multiplying back is the reliable check on any division.
OpenStax, Precalculus, §3.1 Complex Numbers §3.1, pp. 284-285
Faded example
Simplify 2 over the quantity 1 plus 3i.
Fill in the blanks
\frac910\cdot\frac______ = \frac______}} = \frac______}
Why: The denominator becomes 1 squared plus 3 squared, which is 1 plus 9, giving 10. The numerator expands to 2 minus 6i. The answer in standard form is one fifth minus three fifths i, obtained by dividing both parts by 10.
Worked example
The same method, and a shortcut worth knowing.
\[ \text{Simplify } \frac{5}{2i}. \]
Identify the conjugate
Why: Of 2i, which is 0 plus 2i.
\[ \text{conjugate is } -2 i \]
Multiply top and bottom
Why: By negative 2i.
\[ -10 i / (-4 i ^{2}) \]
Simplify the denominator
Why: Negative 4 times negative 1.
\[ \text{denominator } 4 \]
Finish
Why: Divide through.
\[ -10 i / 4 = -5 i / 2 \]
Figure (svg): The solution to Worked example a purely imaginary denominator shown as a ladder of expressions, one row per legal move
\[ \frac{5}{2i} = -\tfrac{5}{2}i \]
Verify: check by multiplying back
Why: Negative five halves i times 2i gives negative 5 i squared, which is positive 5 — the original numerator. Note that dividing by i has the effect of multiplying by negative i, which is a shortcut worth remembering and which §8.5 will explain as a rotation.
OpenStax, Precalculus, §3.1 Complex Numbers §3.1, pp. 285-285
Error analysis
A student clears a complex denominator.
Annotate
On: \( \frac{4}{3-i} = \frac{4}{(3-i)(3+i)} = \frac{4}{10} \)
Multiplying top and bottom by the same thing is multiplying by one and preserves the value. Multiplying only one of them is a different operation entirely, and it is not allowed.
Prediction
A denominator of a plus b i is multiplied by its conjugate.
Predict first
What does it become?
Correct: a squared plus b squared, a positive real number.
Why: The cross terms cancel and the i squared term contributes a plus sign, so the result is a sum rather than a difference of squares — which is why it is always positive unless the number was zero. This is the squared modulus, and it is what guarantees the method never produces a zero denominator except when the original one was zero.
Matching
Change the sign of the imaginary part, and only that.
Match the pairs
Why: Only the imaginary part changes sign; the real part is left alone, including when it is negative. The last case is worth noticing: a real number is its own conjugate, since its imaginary part is zero and negating zero changes nothing. That is consistent with the geometry, since a point on the mirror line is its own reflection.
Analogy
Clearing a complex denominator is the rationalising trick you already know.
Match the pairs
Why: The two techniques are the same idea in two settings: multiply by a conjugate so that the cross terms cancel and the awkward part squares away. The one difference is the sign in the result — a plus for complex numbers because i squared is negative, a minus for radicals because a square root squares to a positive.
Section
Section 5
Concept
The powers of i repeat every four steps: i, negative one, negative i, one, and then round again. Any power is found by dividing the exponent by four and keeping the remainder.
\[ i^1=i, \; i^2=-1, \; i^3=-i, \; i^4=1 \]
The cycle is really a rotation. Multiplying by i turns a point a quarter turn about the origin, and four quarter turns is a full turn back to the start. That explanation belongs to §8.5, but it is worth mentioning now because it makes the period of four inevitable rather than a curiosity.
Figure (svg): The powers of i arranged in a cycle of four, showing i, then negative one, then negative i, then one, returning to the start, with the remainder rule written beneath
OpenStax, Precalculus, §3.1 Complex Numbers §3.1, pp. 278-280
Picture it
Each step multiplies by i, and the fourth step lands back at 1.
Figure (svg): The powers of i arranged in a cycle of four, showing i, then negative one, then negative i, then one, returning to the start, with the remainder rule written beneath
Because the cycle closes after four, only the remainder on division by four matters. The exponent 27 and the exponent 3 give the same answer.
Worked example
Divide by four and keep the remainder.
\[ \text{Simplify } i^{35}. \]
Divide the exponent by 4
Why: Find the quotient and the remainder.
\[ 35 = 4(8) + 3 \]
Discard the complete cycles
Why: Each group of four contributes a factor of 1.
\[ \text{remainder } 3 \]
Read the cycle at that remainder
Why: The third power.
\[ i ^{3} = -i \]
State the answer
Why: The complete cycles changed nothing.
\[ -i \]
Figure (svg): The powers of i arranged in a cycle of four, showing i, then negative one, then negative i, then one, returning to the start, with the remainder rule written beneath
\[ i^{35} = i^3 = -i \]
Verify: check a nearby power
Why: The exponent 36 is a multiple of 4 and gives 1, and multiplying by i once more gives i for the exponent 37. So 35 is two steps back from 37, which is negative i — consistent. Checking against a neighbouring multiple of four is a quick way to confirm the remainder was read correctly.
OpenStax, Precalculus, §3.1 Complex Numbers §3.1, pp. 279-279
Faded example
Simplify i to the fiftieth power.
Fill in the blanks
50 = 4\cdot 12 + 2 \;\Longrightarrow\; i^-1 = i^___} = ___
Why: Fifty divided by four is twelve remainder two, so the answer is the second power of i, which is negative one. The twelve complete cycles each contribute a factor of one and can be discarded entirely.
Worked example
Powers of i turn up whenever a complex number is raised to a power.
\[ \text{Expand } (1+i)^4. \]
Square it first
Why: One plus 2i plus i squared.
\[ (1 + i) ^{2} = 2 i \]
Square that result
Why: Squaring the simplified form.
\[ (2 i) ^{2} \]
Expand
Why: Four times i squared.
\[ 4(-1) \]
Simplify
Why: The answer is real.
\[ -4 \]
Figure (svg): The solution to Worked example use the cycle inside an expansion shown as a ladder of expressions, one row per legal move
\[ (1+i)^4 = -4 \]
Verify: notice how much the intermediate simplification saved
Why: Squaring twice, with the intermediate result simplified to 2i, took four short steps. Expanding the fourth power directly would have produced five terms with powers of i up to the fourth, all needing conversion. Simplifying at each stage rather than at the end is the practical lesson here.
OpenStax, Precalculus, §3.1 Complex Numbers §3.1, pp. 280-281
Trap
\[ i^{22}: \quad 22 \div 4 = 5 \text{ remainder } 2 \;\Longrightarrow\; i^{22} = i^5 \]
Divide the exponent by four
Why: The division is carried out correctly and both numbers are found.
The quotient 5 is used as the reduced exponent instead of the remainder.
The remainder is what survives, not the quotient. Each complete group of four contributes a factor of 1, so only the leftover matters.
The remainder is 2, so the answer is i squared, which is negative one.
Say what the two numbers mean: the quotient counts the complete cycles, which do nothing, and the remainder is the position within the last incomplete one. Naming them prevents the swap.
Sorting
Divide the exponent by four and read the remainder.
Sort into buckets
Sort each power by its value.
Prediction
The powers of i repeat with period four.
Predict first
Why four rather than two?
Correct: Because i squared is negative one, not one.
Why: If squaring returned to 1 the period would be two, as it is for negative one. Because squaring gives negative one, it takes another squaring to return to 1, making the period four. The connection with quadrants is real but is a consequence rather than the cause: multiplying by i is a quarter turn, and it takes four to come back.
Edge cases
The cycle was described for positive exponents.
Discussion prompt
What is i to the power negative one, and does the cycle still work?
Hint: What is one divided by i?
Answer:
It is negative i. Dividing 1 by i and multiplying top and bottom by negative i gives negative i over 1, which is negative i.
The cycle does still work: going backwards one step from i to the zeroth power, which is 1, lands on the previous entry, and that is i cubed, which is negative i. Consistent.
So the cycle extends in both directions, and negative exponents are handled by adding a multiple of four until the exponent is positive. Negative 1 plus 4 is 3, giving i cubed, which is negative i — the same answer by a route that avoids any division.
Comparison
Fill the blanks from memory. Only one of the four needs a technique the reals did not have.
Comparison matrix
| how it works | what to watch | |
|---|---|---|
| addition | combine like parts | never mix real with imaginary |
| subtraction | combine like parts | distribute the minus across BOTH parts |
| multiplication | expand as binomials | substitute -1 for i squared |
| division | multiply top and bottom by the conjugate | multiply BOTH, not just the denominator |
The first two are componentwise and need nothing new. The third needs one substitution, and the fourth is the third applied to a cleverly chosen factor of one.
Pattern
Standard form means a single real part plus a single imaginary part, with no i in any denominator.
Step 1 is first for a reason: the radical rules that let you combine roots require non-negative radicands, and applying them to negatives is the one error in this section that produces a wrong sign rather than a wrong form.
OpenStax Algebra and Trigonometry 2e, §2.4 Complex Numbers §2.4
Check
Convert before multiplying.
Check your understanding
What is the square root of negative 4, times the square root of negative 25?
Answer: A
Why: Converting first gives 2i times 5i, which is 10 times i squared, and that is negative 10. The two i factors combine to produce the sign change.
Check
Expand, then substitute.
Check your understanding
What is (2 - i)(3 + 4i)?
Answer: A
Why: Expanding gives 6 plus 8i minus 3i minus 4 i squared. Substituting turns the last term into positive 4, so the real part is 6 plus 4, which is 10, and the imaginary terms give 5i.
Check
Remainder, not quotient.
Check your understanding
What is i to the power 43?
Answer: A
Why: Forty three is four times ten remainder three, so the answer is the third power of i, which is negative i. The ten complete cycles each contribute a factor of one.
Real world
Complex numbers are the standard tool for anything that oscillates, which is a great deal of engineering.
Discussion prompt
Electrical engineers describe alternating current using complex numbers. What do the two parts represent, and why is a plane the right picture?
Hint: What two things describe a wave, besides its frequency?
Answer:
A wave is described by a size and a timing — its amplitude and its phase. Those are two independent quantities, exactly like the two parts of a complex number, and neither is derivable from the other.
The plane is right because the modulus carries the amplitude and the angle carries the phase. Adding two waves becomes adding two points, which is why complex arithmetic replaces a page of trigonometry with one addition.
This is the payoff of the section that is hardest to see from inside it. The extension was made to solve an equation, and it turned out to describe oscillation better than anything built for the purpose — which is roughly what happened with negative numbers and with fractions too.
Commit first
State your confidence along with your answer.
Predict first
What kind of number is the product of 4 plus 3i and its conjugate?
Correct: A positive real number, namely 25.
Why: The product of a conjugate pair is the sum of the squares of the two parts, here 16 plus 9, which is 25. It is always real and always positive unless the original number was zero, and it equals the square of the modulus — which is why it is exactly what a division needs in the denominator.
Explain it
The test of understanding is being able to say why, not just what.
Discussion prompt
Explain to a classmate why complex numbers were invented, and why the word 'imaginary' is misleading.
Hint: What sequence of number systems came before, and why was each one added?
Answer:
Each number system was built to solve an equation the previous one could not: negatives for subtraction, fractions for division, irrationals for roots, and complex numbers for the square root of a negative.
The word is misleading because it suggests these numbers are less legitimate, when they satisfy exactly the same standard as the others: the arithmetic is consistent and it is useful. Negative numbers were called absurd by mathematicians for centuries on identical grounds.
The clinching point is that they turned out to describe real physical things — currents, waves, quantum states — better than the reals do. A tool invented to close an algebraic gap became the natural language for oscillation, which is a strange thing to say about something imaginary.
Exit ticket
One honest answer, so the next lesson can start in the right place.
Predict first
Which idea from this lesson would you most want to see again?
Correct: Any of these is a legitimate answer; the useful one is the honest one.
Why: There is no correct choice here. The second causes more wrong answers than any other idea here, because the substitution is easy to perform mentally and easy to perform with the wrong sign. The third is the only genuinely new technique, and it is the one that recurs whenever a complex denominator appears.
Connect it up
One page, drawn from memory, is worth more than rereading the section.
Draw it
Draw the complex plane and plot a number with both parts nonzero, marking its real part, imaginary part and modulus. Plot its conjugate too, and write out the product of the pair, showing why it is real. Then draw the cycle of four powers of i around a circle, and beside it write how you would reduce i to the hundredth power.
If your conjugate is a reflection across the horizontal axis and your product came out real, you have the mechanism that makes division work.
Recap
Five things, and the whole chapter depends on the first.
| if you remember one thing | it should be this |
|---|---|
| about the definition | i squared is negative one, and that single fact drives everything |
| about roots of negatives | convert to i form BEFORE combining any radicals |
| about division | multiply top and bottom by the conjugate, never just the bottom |
| about powers | use the remainder on division by four, not the quotient |
Section 3.2 returns to quadratics with this larger number system available, so that every quadratic now has two roots and the question becomes what kind rather than whether.
OpenStax, Precalculus, §3.1 Complex Numbers §3.1, pp. 274-285 — everything on these slides traces back here
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