Introduction to complex numbers AI HL

IB Mathematics: Applications and Interpretation · AI HL

Lesson 1.12 · Introduction to Complex Numbers

Extend the real numbers, calculate with $a+bi$ , and represent complex numbers on the complex plane.

Faith connection · God’s love

When mathematicians encountered equations without real solutions, they developed a larger number system. That discovery reminds us to remain open to understanding more than we first expect. Scripture says, “Herein is love, not that we loved God, but that he loved us, and sent his Son” (1 John 4:10, KJV). God has shown His love through Jesus Christ. Come to Jesus, trust Him as your Savior, and follow Him as Lord.

Learning objectives

By the end of this lesson, you should be able to:

use $i^2=-1$ and simplify powers of $i$ ;
identify the real and imaginary parts of a complex number;
add, subtract, multiply and divide complex numbers in Cartesian form;
find the conjugate, modulus and argument of a complex number;
represent complex numbers on an Argand diagram;
solve quadratic equations with complex roots.

1 · Why do we need complex numbers?

The equation $x^2=-1$ has no real solution: the square of any real number is nonnegative. Complex numbers extend the number system by introducing $i$ , defined by:

$\boxed{i^2=-1}$

Consequently, the solutions of $x^2=-1$ are $x=i$ and $x=-i$ .

Worked example · Powers of $i$

Successive powers of $i$ follow a repeating cycle.

$i^1=i,\qquad i^2=-1,\qquad i^3=-i,\qquad i^4=1.$

Since the cycle repeats every four powers, $i^{27}=i^{24}i^3=(i^4)^6i^3=\boxed{-i}$ .

Calculator note: On a TI-84 Plus CE, the imaginary unit appears as i in the complex-number menu. Check that the calculator is set to a complex-number mode when verifying answers.

2 · Cartesian form: real and imaginary parts

A complex number in Cartesian form is written as $z=a+bi$ , where $a$ and $b$ are real numbers.

$\boxed{z=a+bi,\qquad \operatorname{Re}(z)=a,\quad\operatorname{Im}(z)=b}$

The imaginary part is the real number $b$ , rather than the term $bi$ . A real number is a complex number with $b=0$ .

Worked example · Reading the parts

For $z=-4+7i$ :

$\boxed{\operatorname{Re}(z)=-4,\qquad \operatorname{Im}(z)=7}.$

For two complex numbers to be equal, their real parts must be equal and their imaginary parts must be equal.

3 · Adding, subtracting and multiplying

To add or subtract complex numbers, collect the real terms and the imaginary terms separately.

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

Worked example · Addition and subtraction

Let $z_1=3+5i$ and $z_2=-2+4i$ . Then:

$z_1+z_2=(3-2)+(5+4)i=\boxed{1+9i}.$

$z_1-z_2=(3+2)+(5-4)i=\boxed{5+i}.$

To multiply, expand the brackets and replace $i^2$ with $-1$ .

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

Worked example · Multiplication

Calculate $(2+3i)(4-i)$ .

$(2+3i)(4-i) =8-2i+12i-3i^2 =\boxed{11+10i}.$

The real part is 11 because $-3i^2=-3(-1)=3$ .

4 · The complex conjugate and division

The conjugate of $z=a+bi$ has the same real part and the opposite imaginary part.

$\boxed{\overline z=a-bi}$

Multiplying a complex number by its conjugate gives a nonnegative real number.

$(a+bi)(a-bi)=a^2+b^2.$

This makes conjugates useful for division: multiply both the numerator and denominator by the conjugate of the denominator.

Worked example · Division

Write $\dfrac{5+i}{2-i}$ in Cartesian form.

$\frac{5+i}{2-i} =\frac{(5+i)(2+i)}{(2-i)(2+i)} =\frac{9+7i}{5} =\boxed{\frac95+\frac75i}.$

Check: The denominator after multiplication by its conjugate must be real. For $2-i$ , it is $2^2+1^2=5$ .

5 · The Argand diagram and modulus

An Argand diagram represents $z=a+bi$ as the point $(a,b)$ . The horizontal axis shows the real part and the vertical axis shows the imaginary part.

The modulus $|z|$ is the distance from the origin to $(a,b)$ . By Pythagoras:

$\boxed{|z|=\sqrt{a^2+b^2}}$

Worked example · Plotting and finding a modulus

The number $z=3+4i$ is represented by the point $(3,4)$ , shown above. Its modulus is:

$|z|=\sqrt{3^2+4^2}=\boxed{5}.$

Its conjugate $3-4i$ is represented by $(3,-4)$ , the reflection of $(3,4)$ in the real axis.

6 · The argument of a complex number

The argument $\arg(z)$ is the angle measured from the positive real axis to the line joining the origin to $z$ . Positive angles are measured anticlockwise. The argument of $0$ is undefined.

For $z=a+bi$ in the first quadrant, a right triangle gives $\tan\theta=b/a$ . In other quadrants, use the signs of $a$ and $b$ to choose the correct angle.

Worked example · Choosing the correct quadrant

Find the principal argument of $z=-1+\sqrt3\,i$ in degrees.

The real part is negative and the imaginary part is positive, so $z$ lies in quadrant II. Its reference angle is $60^\circ$ because $\tan60^\circ=\sqrt3$ .

$\arg(z)=180^\circ-60^\circ=\boxed{120^\circ}.$

Calculator check: An inverse tangent calculation alone may give a reference angle in the wrong quadrant. Sketch or identify the signs of the real and imaginary parts before stating the argument.

7 · Quadratic equations with complex solutions

The quadratic formula still works when the discriminant is negative. Use $i^2=-1$ to simplify the square root.

Worked example · Complex roots

Solve $x^2-4x+13=0$ .

$x=\frac{4\pm\sqrt{(-4)^2-4(1)(13)}}{2} =\frac{4\pm\sqrt{-36}}{2}.$

$x=\frac{4\pm6i}{2} =\boxed{2\pm3i}.$

These two roots are complex conjugates. The graph of $y=x^2-4x+13$ has no real $x$ -intercepts.

IB-style Paper 1 · Skill practice

Give exact answers in Cartesian form where appropriate.

Question 1 · [3 marks]

Simplify $i^{38}$ and $i^{43}$ .

Show solution

Powers of $i$ repeat every four.

$i^{38}=i^{36}i^2=(i^4)^9i^2=\boxed{-1}.$

$i^{43}=i^{40}i^3=(i^4)^{10}i^3=\boxed{-i}.$

Question 2 · [4 marks]

Let $z=5-2i$ and $w=-3+4i$ . Find $z+w$ and $z-w$ .

Show solution

$z+w=(5-3)+(-2+4)i=\boxed{2+2i}.$

$z-w=(5+3)+(-2-4)i=\boxed{8-6i}.$

Question 3 · [4 marks]

Expand $(3-2i)(1+4i)$ and write the result in Cartesian form.

Show solution

$(3-2i)(1+4i) =3+12i-2i-8i^2 =\boxed{11+10i}.$

Question 4 · [5 marks]

Write $\dfrac{4+3i}{1+2i}$ in Cartesian form.

Show solution

$\frac{4+3i}{1+2i} =\frac{(4+3i)(1-2i)}{(1+2i)(1-2i)} =\frac{10-5i}{5} =\boxed{2-i}.$

Question 5 · [5 marks]

Let $z=-6+8i$ . Find $\overline z$ , $|z|$ , and $z\overline z$ .

Show solution

$\overline z=\boxed{-6-8i}.$

$|z|=\sqrt{(-6)^2+8^2}=\boxed{10}.$

$z\overline z=|z|^2=\boxed{100}.$

Question 6 · [5 marks]

Find the principal argument of $z=-1-i$ in degrees and state the quadrant in which it lies.

Show solution

Both parts are negative, so $z$ lies in quadrant III. Its reference angle is $45^\circ$ . Using an argument between $-180^\circ$ and $180^\circ$ :

$\boxed{\arg(z)=-135^\circ}.$

The coterminal positive angle is $225^\circ$ .

Question 7 · [6 marks]

Solve $2x^2+4x+10=0$ . Give both roots in Cartesian form.

Show solution

First divide the equation by 2.

$x^2+2x+5=0.$

Apply the quadratic formula.

$x=\frac{-2\pm\sqrt{2^2-4(1)(5)}}2 =\frac{-2\pm\sqrt{-16}}2 =\boxed{-1\pm2i}.$

Applications · Advanced IB-style questions

Application 1 · Combining signals [7 marks]

Two signals are represented by the complex numbers $S_1=3+4i$ and $S_2=-1+2i$ . Their combined signal is $S=S_1+S_2$ .

(a) Find $S$ . (b) Find its magnitude $|S|$ . (c) Find its argument in degrees, to three significant figures.

Show solution

(a)

$S=(3+4i)+(-1+2i)=\boxed{2+6i}.$

(b)

$|S|=\sqrt{2^2+6^2} =\sqrt{40} =\boxed{2\sqrt{10}}.$

(c) The number lies in quadrant I, so:

$\arg(S)=\tan^{-1}\left(\frac62\right) =\tan^{-1}(3) \approx\boxed{71.6^\circ}.$

Application 2 · Electrical impedance [7 marks]

In a simplified alternating-current model, a component has impedance $Z=4+3i$ ohms. The complex voltage is $V=10+5i$ volts, and the complex current is modeled by $I=V/Z$ .