Algebra 2 · N.CN.A.2

Adding, subtracting and multiplying complex numbers

The square root of negative one is now a number. Can you still add and multiply the way you always have?

You need

Imaginary and complex numbersAlgebra 2 Adding and subtracting polynomialsAlgebra 1

Unlocks

The fundamental theorem of algebraAlgebra 2 Conjugates and dividing complex numbersAlgebra 2 The complex plane, modulus and argumentPrecalculus

Where you are 5 in Complex numbers

Number9

Complex numbers5

Adding, subtracting and multiplying complex numbers

Conjugates and dividing complex numbers Imaginary and complex numbers Complex roots come in pairs The complex plane, modulus and argument

When you reach for it

Once i is a number with i^2 = -1, the complex numbers a + bi need arithmetic. You reach for it whenever two of them meet: a sum, a difference or a product to put back into the form a + bi, or a complex solution from the quadratic formula that you want to check by substituting it into the equation.

Almost nothing new has to be learned. A complex number a + bi behaves like a binomial whose “variable” is i, and you already know how to add and subtract polynomials and how to multiply two binomials. The one new fact is what happens when i meets i: i^2 is not a new kind of term but the real number -1.

The procedure

Why the old rules still apply. The complex numbers are built so that the commutative, associative and distributive properties hold for them exactly as they do for real numbers. So every regrouping and every expansion you would make with x in place of i is allowed here too. The only difference comes at the end: where a polynomial keeps its x^2, here i^2 is replaced by -1.

Adding and subtracting. The real parts are like terms of each other, and so are the imaginary parts. Reorder (commutative property), regroup (associative property) and collect:

(a + bi) + (c + di) = (a + c) + (b + d)i

To subtract, take away both parts. The minus sign distributes over the whole bracket, just as -(c + dx) = -c - dx:

(a + bi) - (c + di) = (a - c) + (b - d)i

Multiplying. Distribute each part of the first number over each part of the second. That gives four products, the same four as in (a + bx)(c + dx):

(a + bi)(c + di) = ac + adi + bci + bdi^2

Now the step polynomials never had: bdi^2 = bd(-1) = -bd, which is real. It joins ac, and the two i terms join each other:

(a + bi)(c + di) = (ac - bd) + (ad + bc)i

There is no need to memorize this line, since expanding and replacing i^2 reaches it every time. What it shows is that the answer always lands back in the form a + bi: sums, differences and products of complex numbers are complex numbers. The minus sign in the real part is the i^2 at work, and it is the only place this differs from multiplying binomials.

A square is the case where both factors are the same, so the square of a binomial carries over:

\begin{aligned} (a + bi)^2 &= a^2 + 2abi + b^2 i^2 \\ &= (a^2 - b^2) + 2abi \end{aligned}

Reading the answer. With a, b, c, d real, a + bi = c + di exactly when a = c and b = d. So a + bi is 0 only when both a = 0 and b = 0, and it is a real number exactly when b = 0. Example 6 uses this to find an unknown.

Checking a solution. To test whether a complex number solves an equation, substitute it, carry out the arithmetic above, and compare the two sides. When one side is 0, both parts of the other side must come out 0.

Worked examples

Example 1

Adding and subtracting

Simplify (a) (2 + 3i) + (4 - 7i) \qquad (b) (6 - i) - (-2 + 5i)

Solution

(a) Drop the brackets and gather the real parts together and the imaginary parts together, which the commutative and associative properties allow:

\begin{aligned} (2 + 3i) + (4 - 7i) &= (2 + 4) + (3 - 7)i \\ &= 6 - 4i \end{aligned}

(b) The minus sign in front of the second bracket changes the sign of both its parts, so -(-2 + 5i) = 2 - 5i:

\begin{aligned} (6 - i) - (-2 + 5i) &= 6 - i + 2 - 5i \\ &= (6 + 2) + (-1 - 5)i \\ &= 8 - 6i \end{aligned}

Check by adding back: (8 - 6i) + (-2 + 5i) = 6 - i.

Example 2

Multiplying by a pure imaginary number

Write 3i(7 - 2i) in standard form.

Solution

Distribute 3i over both parts, as when multiplying by a monomial:

3i(7 - 2i) = 21i - 6i^2

Replace i^2 with -1, so -6i^2 = -6(-1) = 6. Standard form puts the real part first:

3i(7 - 2i) = 6 + 21i

The 7, which was real, became part of an imaginary term, and the -2i, which was imaginary, became real. Multiplying by i always does this: i(a + bi) = -b + ai.

Example 3

Multiplying two complex numbers

Multiply (2 + 3i)(4 - i).

Solution

There are four products, one for each pairing of a part of the first number with a part of the second. The grid lays them out.

Figure 1Each part of 2 + 3i meets each part of 4 - i. The shaded cell holds i^2, so it is real, -3i^2 = 3.

\begin{aligned} (2 + 3i)(4 - i) &= 8 - 2i + 12i - 3i^2 \\ &= 8 - 2i + 12i + 3 \\ &= 11 + 10i \end{aligned}

The second line replaces -3i^2 with -3(-1) = 3. The third collects the real terms, 8 + 3 = 11, and the imaginary terms, -2i + 12i = 10i. The general line from the procedure agrees: ac - bd = 2 \cdot 4 - 3 \cdot (-1) = 11 and ad + bc = 2 \cdot (-1) + 3 \cdot 4 = 10.

Example 4

Squaring a complex number

Simplify (5 - 2i)^2.

Solution

Use the square of a binomial, (p - q)^2 = p^2 - 2pq + q^2, with p = 5 and q = 2i:

\begin{aligned} (5 - 2i)^2 &= 25 - 20i + (2i)^2 \\ &= 25 - 20i + 4i^2 \\ &= 25 - 20i - 4 \\ &= 21 - 20i \end{aligned}

With a real q, the last term q^2 would be positive. Here q is imaginary, so q^2 = (2i)^2 = -4 is negative, and the real part of the answer is 25 - 4.

Example 5

Checking a complex solution

Show that x = -2 + 3i is a solution of x^2 + 4x + 13 = 0, and that x = 2 + 3i is not.

Solution

Work out each term on its own, then add. For x = -2 + 3i:

\begin{aligned} (-2 + 3i)^2 &= 4 - 12i + 9i^2 \\ &= -5 - 12i \\ 4(-2 + 3i) &= -8 + 12i \end{aligned}

Adding these and the 13:

\begin{aligned} &(-5 - 12i) + (-8 + 12i) + 13 \\ &\quad = (-5 - 8 + 13) + (-12 + 12)i \\ &\quad = 0 \end{aligned}

Both parts are 0, so -2 + 3i is a solution.

For x = 2 + 3i: (2 + 3i)^2 = 4 + 12i - 9 = -5 + 12i and 4(2 + 3i) = 8 + 12i, so

(-5 + 12i) + (8 + 12i) + 13 = 16 + 24i

That is not 0, so 2 + 3i is not a solution.

This is how an answer from the quadratic formula gets checked. Here b^2 - 4ac = 16 - 52 = -36 is negative, so the discriminant says there are no real solutions, and the formula gives \displaystyle x = \frac{-4 \pm 6i}{2} = -2 \pm 3i. Substituting -2 - 3i gives 0 as well; why complex solutions come in pairs like this is the subject of conjugate pairs of roots.

Example 6

Choosing a number to make a product real

Find the real number k for which (2 + ki)(3 - i) is a real number, and find that number.

Solution

Multiply with k left as a letter, then collect the parts. Since -ki^2 = k,

\begin{aligned} (2 + ki)(3 - i) &= 6 - 2i + 3ki - ki^2 \\ &= (6 + k) + (3k - 2)i \end{aligned}

A complex number is real exactly when its imaginary part is 0, so

3k - 2 = 0 \qquad\Longrightarrow\qquad k = \frac{2}{3}

The product is then its real part, \displaystyle 6 + \frac{2}{3} = \frac{20}{3}. Check by multiplying:

\begin{aligned} \left(2 + \tfrac{2}{3}i\right)(3 - i) &= 6 - 2i + 2i - \tfrac{2}{3}i^2 \\ &= 6 + \tfrac{2}{3} \end{aligned}

Example 7

Voltage in an electrical circuit

In alternating-current circuits, engineers describe how strongly a part opposes the current by a complex number called its impedance, measured in ohms, and describe the current by a complex number too. Two facts from that subject are all this example needs: the impedances of parts connected one after another (in series) add, and the voltage is the current times the impedance.

Branwen’s circuit has two parts in series, with impedances 3 + 4i ohms and 2 - 6i ohms, and carries a current of 2 + i amps. Find the total impedance and the voltage.

Solution

Add the impedances, real parts and imaginary parts separately:

(3 + 4i) + (2 - 6i) = 5 - 2i \text{ ohms}

The voltage is the current times this total:

\begin{aligned} (2 + i)(5 - 2i) &= 10 - 4i + 5i - 2i^2 \\ &= 12 + i \text{ volts} \end{aligned}

since -2i^2 = 2 joins the 10. Engineers write j for the square root of -1, because the letter i already stands for current in their formulas; the arithmetic is the same.

Boundaries

i^2 = -1 is the one new rule. The product of two imaginary terms bi and di is the real number -bd, and a result that still contains i^2 is not yet in standard form.

In a sum, the two parts stay apart. 4 + 3i cannot be shortened, for the same reason 4 + 3x cannot: the terms are not like terms.

Square roots of negatives are written with i first. The rule \displaystyle \sqrt{a}\,\sqrt{b} = \sqrt{ab} gives the wrong sign when a and b are both negative, as imaginary and complex numbers shows. So rewrite each square root of a negative as a multiple of i before multiplying.

Dividing is a different move. Adding, subtracting and multiplying land in the form a + bi directly. Dividing by a complex number needs the conjugate to clear i out of the denominator.

Complex numbers have no order. They can be added and multiplied, but 3 + i < 2 + 5i means nothing, so a result is “positive” or “larger” only when it is real.

A check must zero both parts. A substituted number solves an equation with 0 on one side only when the other side comes out with real part 0 and imaginary part 0.

Where this goes next

The product of a + bi and a - bi comes out real, and that one fact is what dividing complex numbers is built on. The complex plane gives the arithmetic a picture: adding is adding arrows, and multiplying turns out to rotate and stretch. Substituting a complex number into a polynomial, as in Example 5, is the basic tool of conjugate pairs of roots and of the fundamental theorem of algebra, which says a polynomial of degree n has exactly n complex roots, counting repeats.

Exercises

Add and subtract

Write each answer in the form a + bi.

(5 + 2i) + (3 - 6i)
(-4 + 7i) + (9 - 2i)
(7 - 3i) - (2 + 4i)
(-1 + 8i) - (-6 - 5i)
(3 + i) - (4 - 2i) + (-2 + 5i)

Multiply by a real or pure imaginary number

4(2 - 7i)
-2i(6 + 5i)
5i(-3 + 4i)

Multiply two complex numbers

(3 + 2i)(1 + 4i)
(4 - 3i)(2 + 5i)
(-1 + 2i)(3 - 4i)
(6 - i)(2 - 3i)
(1 - 4i)(-2 - i)

Squares and longer products

(3 + 4i)^2
(1 - 3i)^2
(1 + i)^2 and (1 + i)^3
(2 + i)(1 - i)(1 + 2i)

Check a solution

Is x = 4 + i a solution of x^2 - 8x + 17 = 0?
Is x = 1 - i a solution of x^2 + 2x + 2 = 0?
Show that \displaystyle x = \frac{1}{2} + \frac{1}{2}i is a solution of 2x^2 - 2x + 1 = 0.

Unknowns and a circuit

Find the real number k for which (k + 3i)(2 - i) is real, and find that real number.
Find real numbers a and b with (a + bi)(2 + i) = 7 + i.
Eskander connects two parts in series, with impedances 4 - i ohms and 2 + 3i ohms, and the current is 1 - 2i amps. Find the total impedance and the voltage, using the two facts from Example 7.

Reasoning

Prove that the product of two complex numbers a + bi and c + di, with a, b, c, d real, is again a complex number in standard form, and say what its real and imaginary parts are.
Show that (a + bi)^2, with a and b real, is a real number exactly when a = 0 or b = 0, and say when that real number is negative.

Answers

8 - 4i. (5 + 3) + (2 - 6)i.
5 + 5i. (-4 + 9) + (7 - 2)i.
5 - 7i. 7 - 3i - 2 - 4i = (7 - 2) + (-3 - 4)i.
5 + 13i. -1 + 8i + 6 + 5i = (-1 + 6) + (8 + 5)i.
-3 + 8i. 3 + i - 4 + 2i - 2 + 5i; the real parts give 3 - 4 - 2 = -3 and the imaginary parts 1 + 2 + 5 = 8.
8 - 28i. Distribute the 4 over both parts.
10 - 12i. -2i(6 + 5i) = -12i - 10i^2 = -12i + 10.
-20 - 15i. 5i(-3 + 4i) = -15i + 20i^2 = -15i - 20.
-5 + 14i. 3 + 12i + 2i + 8i^2 = (3 - 8) + 14i.
23 + 14i. 8 + 20i - 6i - 15i^2 = (8 + 15) + 14i.
5 + 10i. -3 + 4i + 6i - 8i^2 = (-3 + 8) + 10i.
9 - 20i. 12 - 18i - 2i + 3i^2 = (12 - 3) - 20i.
-6 + 7i. -2 - i + 8i + 4i^2 = (-2 - 4) + 7i.
-7 + 24i. 9 + 24i + 16i^2 = (9 - 16) + 24i.
-8 - 6i. 1 - 6i + 9i^2 = (1 - 9) - 6i.
(1 + i)^2 = 2i and (1 + i)^3 = -2 + 2i. First 1 + 2i + i^2 = 2i. Then (1 + i)^3 = 2i(1 + i), which is 2i + 2i^2 = -2 + 2i.
5 + 5i. (2 + i)(1 - i) = 2 - 2i + i - i^2, which is 3 - i. Then (3 - i)(1 + 2i) = 3 + 6i - i - 2i^2, which is 5 + 5i.
Yes. (4 + i)^2 = 15 + 8i and -8(4 + i) = -32 - 8i. Adding these and 17 gives (15 - 32 + 17) + (8 - 8)i = 0.
No. (1 - i)^2 = -2i and 2(1 - i) = 2 - 2i, so the left side is -2i + 2 - 2i + 2 = 4 - 4i, not 0.
\displaystyle x^2 = \frac{1}{4} + \frac{1}{2}i + \frac{1}{4}i^2 = \frac{1}{2}i, so 2x^2 = i, and -2x = -1 - i. The left side is i - 1 - i + 1 = 0.
k = 6, and the product is 15. (k + 3i)(2 - i) = 2k - ki + 6i - 3i^2, which is (2k + 3) + (6 - k)i. Its imaginary part is 0 when k = 6, leaving 2 \cdot 6 + 3 = 15.
a = 3 and b = -1. The product (a + bi)(2 + i) is (2a - b) + (a + 2b)i, so matching parts, 2a - b = 7 and a + 2b = 1. The first gives b = 2a - 7; then a + 4a - 14 = 1, so a = 3. Check: (3 - i)(2 + i) = 6 + 3i - 2i - i^2, which is 7 + i.
Total impedance 6 + 2i ohms, voltage 10 - 10i volts. (4 - i) + (2 + 3i) = 6 + 2i, and (1 - 2i)(6 + 2i) = 6 + 2i - 12i - 4i^2, which is 10 - 10i.
Distributing gives ac + adi + bci + bdi^2, and i^2 = -1 turns the last term into -bd. Collecting, the product is (ac - bd) + (ad + bc)i. Sums and products of real numbers are real, so ac - bd and ad + bc are real: the real part is ac - bd and the imaginary part is ad + bc.
(a + bi)^2 = (a^2 - b^2) + 2abi, which is real exactly when 2ab = 0, that is when a = 0 or b = 0. If b = 0 the square is a^2 \ge 0. If a = 0 it is -b^2, negative unless b = 0. So the square is a negative real number exactly when a = 0 and b \ne 0, that is, when the number is a non-zero multiple of i.