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Add, multiply and conjugate complex numbers, find a modulus, and solve quadratics whose discriminant is negative.
Paper packet. Every task here also exists on screen, where it is checked automatically; answers written on paper are not assessed by Nydus. When you are back at a device, enter your answers there.
In this lesson you meet the number $i$, defined by $i^2 = -1$, and the complex numbers $a + bi$ built from it. You add and multiply them with the algebra you already have, plus one substitution; you use conjugates to turn a complex product or a complex denominator back into a real number; and you measure a complex number's size with its modulus. Then you go back to the quadratics that had "no solution" because their discriminant was negative, and solve them properly — the roots always arriving as a conjugate pair, which also lets you build an equation from roots you were handed.
You already collect like terms, so $5x + 2x = 7x$ and $3x \times 4x = 12x^2$. You already expand two brackets into four products. You already know $(a + b)(a - b) = a^2 - b^2$, and you already know Pythagoras. And you know one thing that is about to become useful rather than annoying: no real number squares to a negative, because a positive times a positive and a negative times a negative are both positive.
Imaginary unit $i$: the number defined by $i^2 = -1$. It is not an approximation and not a mistake; it is a new number, invented for exactly this job.
Complex number: anything of the form $a + bi$ with $a$ and $b$ real.
Real part $a$ and imaginary part $b$ — note that the imaginary part is $b$, a real number, not $bi$.
Conjugate: $a - bi$ is the conjugate of $a + bi$; only the sign in front of $i$ changes.
Modulus, written $|a + bi|$: the distance from $0$, equal to $\sqrt{a^2 + b^2}$.
Define one new number, $i$, by $i^2 = -1$. A complex number is then $a + bi$ with $a$ and $b$ real, and every ordinary rule of algebra still applies — with $i^2$ replaced by $-1$ wherever it appears.
Adding and subtracting is collecting like terms: $(3 + 2i) + (5 - 7i) = 8 - 5i$. Real with real, imaginary with imaginary; nothing crosses over.
Multiplying is four products, then one substitution: $(2 + 3i)(4 + 5i) = 8 + 10i + 12i + 15i^2 = 8 + 22i - 15 = -7 + 22i$. The $i^2$ product always changes sign and moves into the real part.
Powers of $i$ cycle with period four: $i^1 = i$, $i^2 = -1$, $i^3 = -i$, $i^4 = 1$, and then round again, so $i^{27} = i^{24} \times i^3 = -i$.
The conjugate $a - bi$ is the partner that clears $i$ away: $(a + bi)(a - bi) = a^2 - b^2i^2 = a^2 + b^2$, always real and never negative. That product is the square of the modulus $|a + bi| = \sqrt{a^2 + b^2}$, and it is what lets you divide: multiply top and bottom by the conjugate of the bottom, and the denominator becomes a plain real number.
Another way: picture
A plane with the real numbers along the horizontal axis and the multiples of $i$ up the vertical one. The number $3 + 4i$ is the point $3$ right and $4$ up; its modulus $5$ is the length of the arrow to it, and its conjugate $3 - 4i$ is its mirror image below the horizontal axis.
Another way: story
Treat $i$ as a letter you may not simplify away, with one extra rule: whenever $i^2$ appears, cross it out and write $-1$. Every technique in this lesson is ordinary algebra plus that one substitution.
"$i^2 = 1$, because a square is positive." That is the rule for real numbers, and $i$ is not one. $i^2 = -1$ is the whole definition; every other fact here follows from it.
Losing the $i^2$ term. In $(2 + 3i)(4 + 5i)$ the last product is $15i^2$, which is $-15$ — a real number. It leaves the imaginary part and joins the real one. Answers that keep it as $15$ or as $15i$ are the commonest error in the topic.
"$|3 + 4i| = 7$." The modulus is a hypotenuse, not a sum: $\sqrt{9 + 16} = 5$. Square, add, then root.
Four products: $12$, $20i$, $-6i$ and $-10i^2$.
Exactly as with $(4 - 2x)(3 + 5x)$.
$-10i^2 = -10 \times (-1) = +10$, so it joins the real part: $12 + 10 = 22$.
This is the step that makes complex arithmetic different.
The $i$ terms give $20i - 6i = 14i$, so the answer is $22 + 14i$.
Multiply top and bottom by the conjugate $2 + i$.
Multiplying by $\frac{2 + i}{2 + i}$ is multiplying by $1$, so the value is unchanged.
The bottom becomes $(2 - i)(2 + i) = 4 + 1 = 5$, a real number.
$a^2 + b^2$, with no $i$ left.
The top is $5(2 + i) = 10 + 5i$, so the quotient is $\frac{10 + 5i}{5} = 2 + i$.
Conjugates multiply to $a^2 + b^2$: here $1 + 36 = 37$.
$|1 + 6i| = \sqrt{1 + 36} = \sqrt{37}$, so its square is $37$ — the same number, which is exactly what the conjugate rule says.
Compute $(-6 - 7i) + (8 - 9i)$.
(-6 - 7i) + (8 - 9i) = re + (im)i
Expand $(-2 + 3i)(-6 + 4i)$ and write it as $a + bi$.
(-2 + 3i)(-6 + 4i) = re + (im)i
Multiply $-6 + 2i$ by its conjugate.
The product is answer.
What is $|3 + 4i|$?
The modulus is answer.
You already solve $ax^2 + bx + c = 0$ with the quadratic formula, and you already know that the discriminant $b^2 - 4ac$ decides how many real roots there are: positive gives two, zero gives one, negative gives none. You have just learned what to do with the square root of a negative number, so "none" is about to become "two, but not on the real number line".
Discriminant: $b^2 - 4ac$, the part under the root in the quadratic formula.
Complex conjugate pair: two roots $a + bi$ and $a - bi$, differing only in the sign of the imaginary part.
Pure imaginary: a complex number whose real part is $0$, like $7i$.
Monic: a polynomial whose leading coefficient is $1$, like $x^2 + bx + c$.
Sum and product of roots: for a monic quadratic, the roots add to $-b$ and multiply to $c$.
Nothing about solving changes; only the reading of the answer does. For $ax^2 + bx + c = 0$ the formula still gives $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$. When the discriminant is negative, write $\sqrt{-k} = \sqrt{k}\,i$ and carry on.
The simplest case has no linear term at all: $x^2 + 25 = 0$ gives $x^2 = -25$ and $x = \pm 5i$, a pure imaginary pair. With a linear term, $x^2 - 6x + 13 = 0$ has discriminant $36 - 52 = -16$, so $x = \frac{6 \pm 4i}{2} = 3 \pm 2i$ — and both halves of the numerator were divided by $2$.
The two roots are always a conjugate pair when the coefficients are real, and the reason is worth knowing rather than memorising: the $\pm$ in the formula puts the same real part in front of both and opposite signs on the same imaginary part. Read backwards, this builds equations. A monic quadratic with roots $r$ and $s$ is $x^2 - (r + s)x + rs$; for the pair $p \pm qi$ the sum is $2p$ and the product is $p^2 + q^2$, both real, so the equation is $x^2 - 2px + (p^2 + q^2) = 0$. Every real quadratic that fails to cross the $x$-axis is one of these.
Another way: picture
The parabola $y = x^2 - 6x + 13$ floating entirely above the $x$-axis, with its vertex at $(3, 4)$. It never crosses, which is what "no real roots" looks like; the roots $3 \pm 2i$ are read off the vertex, $3$ across and $2$ for the height $4 = 2^2$.
Another way: story
The discriminant is a verdict on which number system the answer lives in, not on whether an answer exists. Negative just means: step off the line and into the plane.
"A negative discriminant means no solutions." It means no real solutions. Over the complex numbers every quadratic has two roots, counted properly — that is the whole point of inventing $i$.
Halving only half of the answer. From $x = \frac{6 \pm 8i}{2}$ the roots are $3 \pm 4i$, not $3 \pm 8i$. The division by $2a$ applies to the real part and the imaginary part.
"The other root is $-a - bi$." Only the sign in front of $i$ flips. The real part of a conjugate is unchanged, because it is the imaginary parts that have to cancel for the coefficients to come out real.
Discriminant: $4^2 - 4 \times 1 \times 13 = 16 - 52 = -36$, negative, so the roots are complex.
Compute it first; it tells you what kind of answer to expect.
$\sqrt{-36} = 6i$, so $x = \frac{-4 \pm 6i}{2}$.
The root of a negative is the root of its size, times $i$.
Divide both parts by $2$: $x = -2 \pm 3i$.
Both parts, not just the first.
Real coefficients force the conjugate $5 + 2i$ to be a root as well.
Complex roots of real polynomials come in pairs.
Sum of roots: $10$. Product: $(5 - 2i)(5 + 2i) = 25 + 4 = 29$.
The conjugate product is $a^2 + b^2$.
So $x^2 - 10x + 29 = 0$. Check the discriminant: $100 - 116 = -16$, negative, as it must be.
Discriminant: $4 - 20 = -16$, so $\sqrt{-16} = 4i$.
$x = \frac{2 \pm 4i}{2} = 1 \pm 2i$, a conjugate pair.
Solve $x^2 + 16 = 0$. The roots are $x = \pm bi$; what is $b$?
x = \pm ri
Solve $x^2 - 6x + 18 = 0$. The roots are $x = a \pm bi$; give $a$ and $b$.
x = a \pm bi
A quadratic with real coefficients has $-6 + 4i$ as one root. What is the other?
Solve $x^2 + 4 = 0$. The roots are $x = \pm bi$; what is $b$?
x = \pm ri
A control engineer needs a quadratic $x^2 + bx + c$ with real coefficients whose roots are $4 \pm 5i$. What is $c$?
The constant c is answer.
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
Compute $(-8 - 9i) + (2 - 9i)$.
(-8 - 9i) + (2 - 9i) = re + (im)i
Solve $x^2 + 25 = 0$. The roots are $x = a \pm bi$; give $a$ and $b$.
x = a \pm bi
You can compute with complex numbers and solve a quadratic with a negative discriminant. Expand $(3 - 2i)(1 + 4i)$, find $|5 + 12i|$, and solve $x^2 - 4x + 20 = 0$.
8. Your turn: compute $(1 + 6i)(1 - 6i)$ and $|1 + 6i|^2$, step 2
19. Your turn: solve $x^2 - 2x + 5 = 0$, step 2