Complex numbers for quantum computing

A number with a size and a direction

A complex number $z = a + bi$ has a real part $a$ , an imaginary part $b$ , and $i^2 = -1$ . Draw it as the point $(a, b)$ in the plane. Its magnitude (or modulus) is the distance from the origin, $|z| = \sqrt{a^2 + b^2}$ . Its phase (or argument) is the angle from the positive real axis, measured counter-clockwise. This page covers what the lessons need.

In quantum mechanics we almost always need $|z|^2$ , which you can compute without square roots: $|z|^2 = a^2 + b^2 = z\,z^*$ , where $z^* = a - bi$ is the complex conjugate (the reflection across the real axis).

Check yourself

What is $|z|^2$ for $z = \tfrac{1}{2} + \tfrac{i}{2}$ ?

$|z|^2 = \;?$

Hints

Which two real numbers are $a$ and $b$ here?
Add the squares of the real and imaginary parts: $a^2 + b^2$ .

Show the answer

Answer: $|z|^2 = \frac{1}{2}$

$(1/2)^2 + (1/2)^2 = 1/4 + 1/4 = 1/2$ .

Common slips

$1$ : That squares the sum, $(\tfrac12 + \tfrac12)^2$ . Square each part first, then add: $a^2 + b^2$ .

Check yourself

Compute $z\,z^*$ for $z = 3 - 4i$ .

Hints

What is $z^*$ ? Flip the sign of the imaginary part.
Multiply out $(3-4i)(3+4i)$ : the two cross terms cancel.
Use $i^2 = -1$ on the last term.

Show the answer

Answer: $25$

$(3-4i)(3+4i) = 9 + 12i - 12i - 16i^2 = 9 + 16 = 25$ , which is $|z|^2$ .

Common slips

$-7$ : That treats $i^2$ as $+1$ . With $i^2 = -1$ , the last term $-16i^2$ is $+16$ .
$-7-24i$ : That is $z^2 = z\,z$ . The conjugate flips the sign of the imaginary part: $z^* = 3 + 4i$ .

Polar form: $re^{i\varphi}$

Euler’s formula $e^{i\varphi} = \cos\varphi + i\sin\varphi$ puts a point on the unit circle at angle $\varphi$ . So any complex number can be written $z = r e^{i\varphi}$ with $r = |z|$ . Two facts carry most of the weight in these lessons:

$|e^{i\varphi}| = 1$ for every real $\varphi$ : multiplying by a pure phase never changes a magnitude.
Multiplying complex numbers multiplies magnitudes and adds phases: ${r_1 e^{i\varphi_1} \cdot r_2 e^{i\varphi_2} = r_1 r_2\, e^{i(\varphi_1 + \varphi_2)}}$ .

Check yourself

What is $\abs{e^{i\pi/3}\cdot\tfrac{1}{\sqrt{2}}}^2$ ?

A $\tfrac{1}{2}e^{2i\pi/3}$
B $\cos^2(\pi/3)/2$
C $\tfrac{1}{4}$
D $\tfrac{1}{2}$

Hints

What is the magnitude of $e^{i\pi/3}$ ?
Magnitudes multiply, so $|e^{i\pi/3}\cdot\tfrac{1}{\sqrt2}| = |e^{i\pi/3}|\cdot\tfrac{1}{\sqrt2}$ . Square that.

Show the answer

Answer: D. $\tfrac{1}{2}$

$|e^{i\pi/3}| = 1$ , so $|e^{i\pi/3}/\sqrt2|^2 = 1/2$ . A phase factor never changes a probability on its own.

Why not the others

A. That squares the number itself, $z^2$ , which still carries a phase. Probabilities use $|z|^2 = z z^*$ , which is always real and non-negative.
B. That keeps only the real part of the phase factor. The whole factor $e^{i\pi/3}$ has magnitude exactly 1.
C. That shrinks the number by the phase factor, as if $e^{i\pi/3}$ counted as its real part $\cos(\pi/3) = \tfrac12$ . But $|e^{i\pi/3}| = 1$ , so only $(1/\sqrt2)^2 = \tfrac12$ is left.

Check yourself

Write $e^{i\pi/4}\cdot e^{i\pi/4}$ as a single complex number (e.g. i, -1, 1/sqrt(2)).

Hints

Multiplying adds phases. What is $\pi/4 + \pi/4$ ?
Which point on the unit circle sits at angle $\pi/2$ ?

Show the answer

Answer: $i$

$e^{i\pi/4}e^{i\pi/4} = e^{i\pi/2} = i$ .

Common slips

$2\,i$ : That adds the two numbers’ sizes. Multiplying multiplies magnitudes, $1 \times 1 = 1$ , and adds the phases: $e^{i\pi/2}$ .

Practice

Check yourself

What is $|z|^2$ when $z = \tfrac{1}{\sqrt{2}}e^{i\pi/4}$ ?

Hints

What is $|e^{i\theta}|$ for any real $\theta$ ?
So $|z|$ is just the positive number in front of the phase factor. Square it.

Show the answer

Answer: $\frac{1}{2}$

The phase has magnitude 1; $(1/\sqrt{2})^2 = 1/2$ .

Check yourself

Compute $(1 + i)^2$ .

Hints

Expand $(1+i)(1+i)$ term by term.
Use $i^2 = -1$ on the last term.

Show the answer

Answer: $2\,i$

$1 + 2i + i^2 = 2i$ .

Common slips

$0$ : That is $1^2 + i^2$ , which drops the cross terms. $(1+i)(1+i) = 1 + i + i + i^2$ .
$2$ : That is $|1+i|^2 = (1+i)(1-i)$ , which uses the conjugate. Here $1+i$ multiplies itself: $1 + 2i + i^2$ .

Check yourself

Which number has the same magnitude as $\tfrac{1}{\sqrt2}$ but a phase of $\pi$ ?

A $\tfrac{1}{\sqrt2}e^{i}$
B $-\tfrac{1}{\sqrt2}$
C $\tfrac{i}{\sqrt2}$

Show the answer

Answer: B. $-\tfrac{1}{\sqrt2}$

$e^{i\pi} = -1$ , so a phase of $\pi$ just flips the sign.

Why not the others

A. That phase is 1 radian, not $\pi$ radians.
C. $i = e^{i\pi/2}$ : that is a phase of $\pi/2$ , a quarter turn.

Go further

Free resources for when you want more than this refresher covers.

Imaginary numbers are realWelch Labs · Video series
A video series on why mathematicians needed i, from Cardano’s cubic equations to the complex plane and Riemann surfaces.
Precalculus 2e: Polar form of complex numbersOpenStax · Free book
Worked examples and exercises on polar form, multiplying and dividing in it, and taking powers with De Moivre’s theorem.
Intuitive understanding of Euler’s formulaKalid Azad, BetterExplained · Article
Reads e^(iφ) as a rotation rather than a formula to memorize, with many pictures and analogies.

Lessons that use it

Lesson 0: Orientation and diagnosticHow the lessons work, five key words, how states are written, and a quick check of the math the lessons assume
Lesson 1: Qubits and measurementKets, amplitudes, the Born rule, and what a measurement leaves behind
Lesson 2: Phase and the ± basisA global phase changes nothing; a relative phase shows up in another basis

A number with a size and a direction

Polar form: reiφre^{i\varphi}reiφ

Practice

Go further

Lessons that use it

Polar form: $re^{i\varphi}$