IB Analysis and Approaches HL · Complex numbers · Section 1 of 4

Cartesian form and operations

3 worked examples · By Álvaro López

Key idea

Complex numbers extend the reals so that every equation such as $x^{2} + 1 = 0$ has a solution. They are written $z = a + bi$ and manipulated like algebraic expressions, remembering that $i^{2} = - 1$ . In the IB exam they appear in Paper 1 (adding, multiplying, dividing, solving for $z$ ) and are the basis for the modulus–argument and Euler forms that come after.

The imaginary unit

In the reals, $x^{2} = - 1$ has no solution, because a square is never negative. So a new number is invented whose square is $- 1$ .

If $b = 0$ , $z = a$ is a real number, so $R \subset C$ . If $a = 0$ and $b \neq = 0$ , $z = bi$ is a pure imaginary number.

With $i$ , quadratic equations with negative discriminant can now be solved: if $Δ < 0$ , then $Δ = - Δ i$ . For example, $x^{2} - 6 x + 10 = 0$ gives $x = \frac{6 \pm - 4}{2} = 3 \pm i$ .

Powers of $i$

Multiplying by $i$ repeatedly, the results repeat every four steps:

$i^{1} = i, i^{2} = - 1, i^{3} = - i, i^{4} = 1, i^{5} = i, \dots$

This is true because $i^{4} = 1$ : $i^{4 k + r} = (i^{4})^{k} i^{r} = i^{r}$ .

Equality and the Argand diagram

This property is the key tool for solving equations: an equality between complex numbers is equivalent to two equalities between real numbers.

Since a complex number is a pair of numbers $(a, b)$ , it is drawn as a point in the plane, called the Argand diagram (or complex plane): the horizontal axis is the real axis and the vertical axis is the imaginary axis. The complex number $z = a + bi$ is the point $(a, b)$ , and it can also be seen as the vector from the origin to that point.

Argand diagram with the complex number z equal to a plus bi as the point with coordinates a and b, and its conjugate z star equal to a minus bi as the point symmetric about the real axis

Real numbers lie on the horizontal axis and pure imaginary numbers on the vertical axis. This picture helps interpret the conjugate (below) and, in the next section, the modulus and argument.

Addition, subtraction and multiplication

Let $z = a + bi$ and $w = c + d i$ .

There is no need to memorise the formula for the product: expand it as a product of binomials and substitute $i^{2} = - 1$ : $(a + bi) (c + d i) = a c + a d i + b c i + b d i^{2} = (a c - b d) + (a d + b c) i$

Adding is adding vectors in the Argand diagram; the product, on the other hand, has a subtler geometric meaning that you will see with the modulus–argument form.

The conjugate and division

The key is that, when a complex number is multiplied by its conjugate, the $i$ disappears: $z z^{*} = (a + bi) (a - bi) = a^{2} + b^{2} = ∣ z ∣^{2}$ where $∣ z ∣ = a^{2} + b^{2}$ is the modulus of $z$ (the distance from $z$ to the origin). It is the same idea as rationalising a denominator with roots.

$\frac{a + bi}{c + d i} = \frac{( a + bi ) ( c - d i )}{c ^{2} + d ^{2}} = \frac{a c + b d}{c ^{2} + d ^{2}} + \frac{b c - a d}{c ^{2} + d ^{2}} i$

There is no need to memorise this result: apply the method each time.

Properties of the conjugate

In words: the conjugate of an operation is the operation of the conjugates. Each property is checked by writing $z = a + bi$ and $w = c + d i$ and expanding both sides. For example, for the product: $z w = (a c - b d) + (a d + b c) i$ and $z^{*} w^{*} = (a - bi) (c - d i) = (a c - b d) - (a d + b c) i$ , which is the conjugate of $z w$ .

Equations with complex numbers

Square root of a complex number

Every complex number has two square roots, one the opposite of the other. To find them in Cartesian form, set up $(x + y i)^{2} = a + bi$ .

With simple numbers it is usually enough to try integers that satisfy $2 x y = b$ and $x^{2} - y^{2} = a$ . In a quotient or a quadratic equation with complex coefficients, this step is what lets you apply the formula $x = \frac{- B \pm Δ}{2 A}$ .

Worked examples

Example 1 Basic

Powers, product and quotient

a) Calculate $i^{103}$ .

b) Calculate $(4 - 3 i) (2 + i)$ and give its real and imaginary parts in the form $a + bi$ .

c) Write $\frac{1 + 2 i}{3 - i}$ in the form $a + bi$ .

Solution

a) Divide the exponent by $4$ : $103 = 4 \cdot 25 + 3$ , so the remainder is $3$ and $i^{103} = i^{3} = - i$ .

b) Expand as a product of binomials: $(4 - 3 i) (2 + i) = 8 + 4 i - 6 i - 3 i^{2} = 8 - 2 i + 3 = 11 - 2 i$

c) Multiply by the conjugate of the denominator, $3 + i$ : $\frac{1 + 2 i}{3 - i} = \frac{( 1 + 2 i ) ( 3 + i )}{( 3 - i ) ( 3 + i )} = \frac{3 + i + 6 i + 2 i ^{2}}{9 + 1} = \frac{1 + 7 i}{10} = \frac{1}{10} + \frac{7}{10} i$

Example 2 Medium

Equations with z, with conjugate and square root

a) Find the complex number $z$ that satisfies $(1 + 2 i) z = 7 - i$ .

b) Find the complex number $z$ that satisfies $2 z + z^{*} = 9 + 2 i$ .

c) Find the two square roots of $- 5 + 12 i$ .

Solution

a) Solve for $z$ and divide by the conjugate of $1 + 2 i$ : $z = \frac{7 - i}{1 + 2 i} = \frac{( 7 - i ) ( 1 - 2 i )}{( 1 + 2 i ) ( 1 - 2 i )} = \frac{7 - 14 i - i + 2 i ^{2}}{5} = \frac{5 - 15 i}{5} = 1 - 3 i$

b) Since $z^{*}$ appears, write $z = x + y i$ and $z^{*} = x - y i$ : $2 (x + y i) + (x - y i) = 3 x + y i = 9 + 2 i$ Equating real and imaginary parts: $3 x = 9$ and $y = 2$ . Hence $z = 3 + 2 i$ .

c) Set up $(x + y i)^{2} = - 5 + 12 i$ , that is, $x^{2} - y^{2} + 2 x y i = - 5 + 12 i$ : $x^{2} - y^{2} 2 x y = - 5 = 12$ From the second, $y = \frac{6}{x}$ . Substituting, $x^{2} - \frac{36}{x ^{2}} = - 5$ , that is, $x^{4} + 5 x^{2} - 36 = 0$ . With $u = x^{2}$ : $u^{2} + 5 u - 36 = 0$ , hence $u = 4$ or $u = - 9$ . As $x$ is real, $u = x^{2} = 4$ , so $x = \pm 2$ and $y = \frac{6}{x} = \pm 3$ . The roots are $2 + 3 i and - 2 - 3 i$ Check: $(2 + 3 i)^{2} = 4 + 12 i - 9 = - 5 + 12 i$ .

Example 3 Hard

A parameter and a system with conjugates

a) Find the real value of $p$ so that $\frac{p + 3 i}{2 + i}$ is a pure imaginary number, and then write that number down.

b) The complex numbers $z$ and $w$ satisfy $z + i w z^{*} + 2 w = 2 + 5 i = 7 - 4 i$ Find $z$ and $w$ in the form $a + bi$ .

Solution

a) Divide using the conjugate of $2 + i$ : $\frac{p + 3 i}{2 + i} = \frac{( p + 3 i ) ( 2 - i )}{5} = \frac{2 p - p i + 6 i - 3 i ^{2}}{5} = \frac{( 2 p + 3 ) + ( 6 - p ) i}{5}$ It is a pure imaginary number when the real part is zero: $2 p + 3 = 0$ , that is, $p = - \frac{3}{2}$ . The number is then $\frac{6 - ( - 3/2 )}{5} i = \frac{15/2}{5} i = \frac{3}{2} i$ .

b) Write $z = x + y i$ and $w = p + q i$ with $x, y, p, q \in R$ . Then $z^{*} = x - y i$ and $i w = p i + q i^{2} = - q + p i$ .

First equation: $(x - q) + (y + p) i = 2 + 5 i$ , so $x - q = 2, y + p = 5$ Second equation: $(x - y i) + 2 (p + q i) = (x + 2 p) + (2 q - y) i = 7 - 4 i$ , so $x + 2 p = 7, 2 q - y = - 4$ Solving the system of four equations. From the first, $x = 2 + q$ . From the third, $2 + q + 2 p = 7$ , so $q = 5 - 2 p$ . From the second, $y = 5 - p$ . In the fourth: $2 (5 - 2 p) - (5 - p) = - 4$ , that is, $5 - 3 p = - 4$ and $p = 3$ . Then $q = - 1$ , $x = 2 + q = 1$ and $y = 5 - p = 2$ .

Therefore $z = 1 + 2 i$ and $w = 3 - i$ . Check in the first equation: $1 + 2 i + i (3 - i) = 1 + 2 i + 3 i + 1 = 2 + 5 i$ .

In the exam: when dealing with $z^{*}$ , always convert it to $x + y i$ and separate the real and imaginary parts; each equality of parts earns marks. Check the solution in the original equations, because one wrong sign in $z^{*}$ changes the whole system.

Common mistakes

Summary

$i^{2} = - 1$ . The powers of $i$ repeat every 4: $i, - 1, - i, 1$ ; use the remainder of the exponent divided by 4.
$z = a + bi$ : $Re (z) = a$ , $Im (z) = b$ (real numbers). On the Argand diagram, $z$ is the point $(a, b)$ .
Equality: $a + bi = c + d i \Leftrightarrow a = c$ and $b = d$ . A complex equation is equivalent to a system of two real equations.
Addition and subtraction: part by part. Product: expand and substitute $i^{2} = - 1$ .
Conjugate $z^{*} = a - bi$ : $z z^{*} = a^{2} + b^{2} = ∣ z ∣^{2}$ . To divide, multiply numerator and denominator by the conjugate of the denominator.
The conjugate respects operations: $(z \pm w)^{*} = z^{*} \pm w^{*}$ , $(z w)^{*} = z^{*} w^{*}$ , $(z / w)^{*} = z^{*} / w^{*}$ , $(z^{n})^{*} = (z^{*})^{n}$ .
With $z^{*}$ in an equation, write $z = x + y i$ and equate parts. The square root of $a + bi$ comes from $(x + y i)^{2} = a + bi$ : $x^{2} - y^{2} = a$ , $2 x y = b$ .

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Cartesian form and operations

The imaginary unit

Powers of i

Equality and the Argand diagram

Addition, subtraction and multiplication

The conjugate and division

Properties of the conjugate

Equations with complex numbers

Square root of a complex number

Worked examples

Common mistakes

Summary

Want to work on it together?

Powers of $i$