Introduction to Complex Numbers

When the discriminant of a quadratic is negative, the quadratic formula asks for the square root of a negative number, and the real numbers have no answer. Complex numbers fix that by adding one new number, $i$, with $i^2 = -1$. With $i$ in hand, every quadratic has solutions, and you get a number system that engineers use for alternating current and that the rest of this unit builds on. This page covers what both IB HL courses need first (AA HL and AI HL, subtopic AHL 1.12).

Key ideas

The number i

The imaginary unit $i$ is defined by

$$i^2 = -1$$

This lets you take square roots of negative numbers: $\sqrt{-16} = \sqrt{16}\,\sqrt{-1} = 4i$, and $\sqrt{-12} = 2\sqrt{3}\,i$.

Powers of $i$ repeat in a cycle of four:

$i^1$	$i^2$	$i^3$	$i^4$	$i^5$	$i^6$
$i$	$-1$	$-i$	$1$	$i$	$-1$

To simplify $i^n$, divide $n$ by $4$ and use the remainder. For example, $27 = 4(6) + 3$, so $i^{27} = i^3 = -i$.

Cartesian form

A complex number in Cartesian form is

$$z = a + bi, \qquad a, b \in \mathbb{R}$$

• $a$ is the real part, written $\text{Re}(z)$.
• $b$ is the imaginary part, written $\text{Im}(z)$. Notice that the imaginary part is the real number $b$, not $bi$.

The set of all complex numbers is $\mathbb{C}$. Every real number is also complex (its imaginary part is $0$), and a number like $5i$ with real part $0$ is called purely imaginary.

Equality. Two complex numbers are equal only when their real parts are equal and their imaginary parts are equal:

$$a + bi = c + di \quad\Longleftrightarrow\quad a = c \text{ and } b = d$$

So one complex equation gives you two real equations. That’s a powerful trick for finding unknowns (Practice 7 and 8).

Arithmetic

Treat $i$ like a variable, then replace $i^2$ with $-1$.

• Add and subtract real parts and imaginary parts separately: $(3 + 2i) + (1 - 5i) = 4 - 3i$.
• Multiply by expanding brackets: $(2 + i)(3 + 4i) = 6 + 8i + 3i + 4i^2 = 6 + 11i - 4 = 2 + 11i$.
• Divide by multiplying the top and bottom by the conjugate of the denominator (below). This makes the denominator real, just like rationalizing a denominator with a square root.

The conjugate

The complex conjugate of $z = a + bi$ is

$$z^* = a - bi$$

Only the sign of the imaginary part changes. (Some books write $\bar{z}$; the IB uses $z^*$.) A number times its conjugate is always a non-negative real number:

$$z z^* = (a + bi)(a - bi) = a^2 - b^2 i^2 = a^2 + b^2$$

That’s why the conjugate is the tool for division.

The complex plane (Argand diagram)

You can picture $z = a + bi$ as the point $(a, b)$, or as the vector from the origin to that point. The horizontal axis is the real axis and the vertical axis is the imaginary axis. This picture is called the complex plane or Argand diagram.

Because complex numbers add part by part, adding them on an Argand diagram is the same as adding vectors (tip to tail). Taking the conjugate reflects the point in the real axis.

$z = 3 + 2i$ and its conjugate $z^* = 3 - 2i$ are reflections of each other in the real axis. The length of the arrow is $|z|$ and the angle is $\arg z$.

Modulus and argument

The modulus $|z|$ is the distance from the origin to $z$:

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

The argument $\arg z$ is the angle $\theta$ from the positive real axis to the arrow, measured anticlockwise (clockwise angles are negative). The IB normally uses the principal argument, in radians, with $-\pi \lt \theta \le \pi$. To find it:

1. Sketch $z$ on an Argand diagram so you know which quadrant it’s in.
2. Find the reference angle $\alpha = \arctan\left|\dfrac{b}{a}\right|$.
3. Adjust for the quadrant: $\theta = \alpha$ (first), $\pi - \alpha$ (second), $-(\pi - \alpha)$ (third), $-\alpha$ (fourth).

Numbers on the axes are quick: $\arg 5 = 0$, $\arg(-5) = \pi$, $\arg(3i) = \dfrac{\pi}{2}$, $\arg(-3i) = -\dfrac{\pi}{2}$. The argument of $0$ is undefined. You’ll use modulus and argument much more in polar and Euler form.

Quadratics with a negative discriminant

For $ax^2 + bx + c = 0$ with real coefficients, the quadratic formula still works when $b^2 - 4ac \lt 0$. The square root of the negative discriminant is imaginary, so you get two non-real roots that are conjugates of each other:

$$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-b}{2a} \pm \frac{\sqrt{4ac - b^2}}{2a}\,i$$

On the graph, this is exactly the case where the parabola $y = ax^2 + bx + c$ doesn’t cross the x-axis. The roots still exist; they just aren’t on the real number line. (For more on why the roots come in conjugate pairs, see complex conjugate roots.)

Using technology

Your GDC has a complex mode (often written $a + bi$). You can type $i$ directly and let it calculate sums, differences, products, quotients and powers, for example $(2 - i)^5 = -38 - 41i$. (AI HL) The AI course expects you to do arithmetic by hand and with technology, and to calculate powers of complex numbers in Cartesian form with technology. (AA HL) Paper 1 is non-calculator, so make sure you can do all the arithmetic on this page by hand.

Worked examples

Example 1: Adding, subtracting and multiplying

Let $z = 3 - 2i$ and $w = 1 + 4i$. Find $z + w$, $z - w$, $zw$ and $z^*$.

Solution. Add and subtract the parts separately:

$$z + w = (3 + 1) + (-2 + 4)i = 4 + 2i$$ $$z - w = (3 - 1) + (-2 - 4)i = 2 - 6i$$

Multiply by expanding, then use $i^2 = -1$:

$$\begin{aligned} zw &= (3 - 2i)(1 + 4i) \\ &= 3 + 12i - 2i - 8i^2 \\ &= 3 + 10i + 8 && i^2 = -1 \\ &= 11 + 10i \end{aligned}$$

The conjugate changes the sign of the imaginary part only: $z^* = 3 + 2i$.

Example 2: Dividing

Write $\dfrac{5 + i}{2 - 3i}$ in the form $a + bi$.

Solution. The conjugate of the denominator is $2 + 3i$. Multiply the top and bottom by it:

$$\begin{aligned} \frac{5 + i}{2 - 3i} &= \frac{(5 + i)(2 + 3i)}{(2 - 3i)(2 + 3i)} \\ &= \frac{10 + 15i + 2i + 3i^2}{4 + 9} && \text{bottom: } a^2 + b^2 \\ &= \frac{7 + 17i}{13} \\ &= \frac{7}{13} + \frac{17}{13}i \end{aligned}$$

Check: multiply back. $\left(\dfrac{7}{13} + \dfrac{17}{13}i\right)(2 - 3i) = \dfrac{(7 + 17i)(2 - 3i)}{13} = \dfrac{14 - 21i + 34i + 51}{13} = \dfrac{65 + 13i}{13} = 5 + i$. ✓

Example 3: Modulus and argument

Plot $z_1 = -1 + \sqrt{3}\,i$ and $z_2 = 2 - 2i$ on an Argand diagram, and find the modulus and principal argument of each.

Solution. $z_1$ is in the second quadrant and $z_2$ is in the fourth.

$\arg z_1 = \dfrac{2\pi}{3}$ is measured anticlockwise; $\arg z_2 = -\dfrac{\pi}{4}$ is measured clockwise.

For $z_1$:

$$|z_1| = \sqrt{(-1)^2 + (\sqrt{3})^2} = \sqrt{1 + 3} = 2$$

Reference angle: $\alpha = \arctan\dfrac{\sqrt{3}}{1} = \dfrac{\pi}{3}$. In the second quadrant, $\arg z_1 = \pi - \dfrac{\pi}{3} = \dfrac{2\pi}{3}$.

For $z_2$:

$$|z_2| = \sqrt{2^2 + (-2)^2} = \sqrt{8} = 2\sqrt{2}$$

Reference angle: $\alpha = \arctan\dfrac{2}{2} = \dfrac{\pi}{4}$. In the fourth quadrant, $\arg z_2 = -\dfrac{\pi}{4}$.

Notice that a calculator’s $\arctan\left(\dfrac{\sqrt{3}}{-1}\right)$ gives $-\dfrac{\pi}{3}$, which points into the fourth quadrant: wrong for $z_1$. The sketch is what tells you to adjust.

Example 4: A quadratic with no real roots

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

Solution. Here $a = 1$, $b = -4$, $c = 13$. The discriminant is $(-4)^2 - 4(1)(13) = 16 - 52 = -36$, so the roots are not real. Use the quadratic formula with $\sqrt{-36} = 6i$:

$$z = \frac{4 \pm \sqrt{-36}}{2} = \frac{4 \pm 6i}{2} = 2 \pm 3i$$

Check $z = 2 + 3i$: $(2 + 3i)^2 = 4 + 12i + 9i^2 = -5 + 12i$, so

$$(-5 + 12i) - 4(2 + 3i) + 13 = -5 + 12i - 8 - 12i + 13 = 0 \quad ✓$$

The graph of $y = x^2 - 4x + 13 = (x - 2)^2 + 9$ has its vertex at $(2, 9)$, above the x-axis, so it has no x-intercepts. That matches the negative discriminant. Notice the real part of the roots, $2$, is the x-coordinate of the vertex.

Common mistakes

Writing $i^2 = 1$. The whole point of $i$ is that $i^2 = -1$. A sign slip here ruins every product, so after expanding, circle each $i^2$ and replace it with $-1$ before collecting terms.

Multiplying square roots of negatives as if they were positive. $\sqrt{-4}\,\sqrt{-9}$ is not $\sqrt{36} = 6$. Convert to $i$ first: $\sqrt{-4}\,\sqrt{-9} = (2i)(3i) = 6i^2 = -6$. The rule $\sqrt{a}\sqrt{b} = \sqrt{ab}$ only works when $a$ and $b$ are not both negative.

Including the i in the imaginary part. For $z = 3 - 2i$, $\text{Im}(z) = -2$, not $-2i$. The imaginary part is a real number.

Dividing the parts separately. $\dfrac{5 + i}{2 - 3i}$ is not $\dfrac{5}{2} + \dfrac{i}{-3i}$. You must multiply the top and bottom by the conjugate of the denominator.

Trusting arctan for the argument. $\arctan\dfrac{b}{a}$ only gives angles between $-\dfrac{\pi}{2}$ and $\dfrac{\pi}{2}$, so it’s wrong for any number in the second or third quadrant. Always sketch the point first, and use the reference angle.

Changing both signs for the conjugate. The conjugate of $-2 + i$ is $-2 - i$, not $2 - i$. Only the imaginary part changes sign (a reflection in the real axis).

Practice

1. (Warm-up) Simplify.

• (a) $\sqrt{-49}$
• (b) $\sqrt{-12}$
• (c) $i^6$
• (d) $i^{27}$

Solution

(a) $\sqrt{-49} = \sqrt{49}\,\sqrt{-1} = 7i$.

(b) $\sqrt{-12} = \sqrt{4 \cdot 3}\,\sqrt{-1} = 2\sqrt{3}\,i$.

(c) $i^6 = i^4 \cdot i^2 = 1 \cdot (-1) = -1$.

(d) $27 = 4(6) + 3$, so $i^{27} = (i^4)^6 \cdot i^3 = i^3 = -i$.

2. (Warm-up) Let $z = 4 - 7i$. Write down $\text{Re}(z)$, $\text{Im}(z)$ and $z^*$, and find $|z|$.

Solution

$\text{Re}(z) = 4$, $\text{Im}(z) = -7$ and $z^* = 4 + 7i$.

$$|z| = \sqrt{4^2 + (-7)^2} = \sqrt{16 + 49} = \sqrt{65}$$

3. (Core) Let $z = 2 + 5i$ and $w = -3 + i$. Find, in the form $a + bi$:

• (a) $2z - 3w$
• (b) $zw$
• (c) $z z^*$

Solution

(a) $2z - 3w = (4 + 10i) - (-9 + 3i) = 13 + 7i$.

(b)

$$zw = (2 + 5i)(-3 + i) = -6 + 2i - 15i + 5i^2 = -6 - 13i - 5 = -11 - 13i$$

(c) $z z^* = 2^2 + 5^2 = 29$ (a real number, as always).

4. (Core) Write $\dfrac{3 + 4i}{1 - 2i}$ in the form $a + bi$, without a calculator.

Solution

Multiply the top and bottom by $1 + 2i$:

$$\begin{aligned} \frac{3 + 4i}{1 - 2i} &= \frac{(3 + 4i)(1 + 2i)}{1^2 + 2^2} \\ &= \frac{3 + 6i + 4i + 8i^2}{5} \\ &= \frac{-5 + 10i}{5} \\ &= -1 + 2i \end{aligned}$$

Check: $(-1 + 2i)(1 - 2i) = -1 + 2i + 2i - 4i^2 = -1 + 4i + 4 = 3 + 4i$. ✓

5. (Core) Solve $2z^2 + 2z + 5 = 0$.

Solution

The discriminant is $2^2 - 4(2)(5) = 4 - 40 = -36$, and $\sqrt{-36} = 6i$.

$$z = \frac{-2 \pm 6i}{2(2)} = \frac{-2 \pm 6i}{4} = -\frac{1}{2} \pm \frac{3}{2}i$$

The two roots are conjugates, as expected for a quadratic with real coefficients.

6. (Core) Find the modulus and principal argument of each number. Give the argument exactly where possible, otherwise in radians to 3 s.f.

• (a) $-5$
• (b) $-2i$
• (c) $-\sqrt{3} - i$
• (d) $-3 + 4i$

Solution

(a) $-5$ is on the negative real axis: $|-5| = 5$ and $\arg(-5) = \pi$.

(b) $-2i$ is on the negative imaginary axis: $|-2i| = 2$ and $\arg(-2i) = -\dfrac{\pi}{2}$.

(c) $|z| = \sqrt{3 + 1} = 2$. The point is in the third quadrant. Reference angle $\alpha = \arctan\dfrac{1}{\sqrt{3}} = \dfrac{\pi}{6}$, so

$$\arg z = -\left(\pi - \frac{\pi}{6}\right) = -\frac{5\pi}{6}$$

(d) $|z| = \sqrt{9 + 16} = 5$. The point is in the second quadrant. Reference angle $\alpha = \arctan\dfrac{4}{3} = 0.9273\ldots$, so

$$\arg z = \pi - 0.9273\ldots = 2.21 \text{ (3 s.f.)}$$

7. (Core) Find the real numbers $x$ and $y$ such that $(x + yi)(1 + 2i) = 7 + 4i$.

Solution

Expand the left side:

$$(x + yi)(1 + 2i) = x + 2xi + yi + 2yi^2 = (x - 2y) + (2x + y)i$$

Equate real parts and imaginary parts:

$$x - 2y = 7 \qquad\text{and}\qquad 2x + y = 4$$

From the second equation, $y = 4 - 2x$. Substitute: $x - 2(4 - 2x) = 7$, so $5x = 15$ and $x = 3$. Then $y = 4 - 6 = -2$.

Check: $(3 - 2i)(1 + 2i) = 3 + 6i - 2i - 4i^2 = 7 + 4i$. ✓

8. (Challenge) Find the two square roots of $-5 + 12i$, in the form $a + bi$ with $a, b \in \mathbb{R}$.

Solution

Suppose $(a + bi)^2 = -5 + 12i$. Expanding, $a^2 - b^2 + 2abi = -5 + 12i$. Equate parts:

$$a^2 - b^2 = -5 \qquad\text{and}\qquad 2ab = 12$$

From the second equation, $b = \dfrac{6}{a}$. Substitute into the first:

$$\begin{aligned} a^2 - \frac{36}{a^2} &= -5 \\ a^4 + 5a^2 - 36 &= 0 && \times a^2 \\ (a^2 + 9)(a^2 - 4) &= 0 \end{aligned}$$

Since $a$ is real, $a^2 = 4$, so $a = 2$ (then $b = 3$) or $a = -2$ (then $b = -3$). The square roots are $2 + 3i$ and $-2 - 3i$, that is, $\pm(2 + 3i)$.

Check: $(2 + 3i)^2 = 4 + 12i + 9i^2 = -5 + 12i$. ✓

9. (Challenge) In AC circuits, the impedance of a component is a complex number in ohms. Engineers often write $j$ for $\sqrt{-1}$, but here we’ll use $i$. Two components have impedances $Z_1 = 3 + 4i$ and $Z_2 = 2 - 6i$.

• (a) In series, the total impedance is $Z = Z_1 + Z_2$. Find it.
• (b) In parallel, the total impedance satisfies $\dfrac{1}{Z} = \dfrac{1}{Z_1} + \dfrac{1}{Z_2}$, which simplifies to $Z = \dfrac{Z_1 Z_2}{Z_1 + Z_2}$. Find $Z$ in the form $a + bi$, exactly and to 3 s.f.

Solution

(a) $Z = (3 + 4i) + (2 - 6i) = 5 - 2i$ ohms.

(b) First the product: $Z_1 Z_2 = (3 + 4i)(2 - 6i) = 6 - 18i + 8i - 24i^2 = 30 - 10i$. Then divide by $Z_1 + Z_2 = 5 - 2i$, multiplying by the conjugate $5 + 2i$:

$$\begin{aligned} Z &= \frac{(30 - 10i)(5 + 2i)}{5^2 + 2^2} \\ &= \frac{150 + 60i - 50i - 20i^2}{29} \\ &= \frac{170 + 10i}{29} \\ &= \frac{170}{29} + \frac{10}{29}i \approx 5.86 + 0.345i \text{ ohms (3 s.f.)} \end{aligned}$$

With a GDC in $a + bi$ mode you can type $\dfrac{(3 + 4i)(2 - 6i)}{(3 + 4i) + (2 - 6i)}$ directly and get the same result.