Introduction

When we first learn about square roots, a natural question comes up: what is −1\sqrt{-1}? If your teacher told you it doesn't exist, they were completely wrong both right and wrong at the same time. It doesn't exist as a real number. But mathematicians decided that was a perfectly good reason to invent a new kind of number.

We define the imaginary unit ii by the equation

i2=−1,equivalentlyi=−1i^2 = -1, \quad \text{equivalently} \quad i = \sqrt{-1}

Numbers formed by multiplying ii by a real number — like 2i2i, −5i-5i, or πi\pi i — are called pure imaginary numbers. They cannot be simplified any further; 2i2i is already in its simplest form.

Warning: The rule ab=ab\sqrt{a}\sqrt{b} = \sqrt{ab} breaks down for negative numbers. For example,

−1−1≠(−1)2=1\sqrt{-1}\sqrt{-1} \neq \sqrt{(-1)^2} = 1

The correct answer is

i⋅i=i2=−1i \cdot i = i^2 = -1

Never combine square roots of negative numbers under one radical.

A complex number is any number of the form

z=a+bi,a,b∈R,i2=−1z = a + bi, \qquad a,b \in \mathbb{R}, \qquad i^2 = -1

The set of all complex numbers is denoted C\mathbb{C}. This is a strict superset: every real number is a complex number (take b=0b=0), and every pure imaginary number is a complex number (take a=0a=0).

Complex numbers ℂ (a + bi)

Real ℝ

(b = 0)

3, −7, π, √2 …

Pure imaginary

(a = 0)

2i, −5i …

For a complex number z=a+biz = a + bi:

  • Re⁡(z)=a\operatorname{Re}(z)=a is the real part
  • Im⁡(z)=b\operatorname{Im}(z)=b is the imaginary part

Note that bb itself is real — it is the coefficient of ii.

So,

Re⁡(3−7i)=3andIm⁡(3−7i)=−7\operatorname{Re}(3-7i)=3 \qquad \text{and} \qquad \operatorname{Im}(3-7i)=-7

not −7i-7i.

Powers of ii

If we keep multiplying ii by itself, we get a cycle that repeats every 44 steps:

i1=i,i2=−1,i3=−i,i4=1,i5=i,…i^1=i,\qquad i^2=-1,\qquad i^3=-i,\qquad i^4=1,\qquad i^5=i,\ldots

i¹ = i

+i

i² = −1

−1

i³ = −i

−i

i⁴ = 1

+1

cycle repeats every 4

×i each step

The key rule is:

in=i(nmod4)i^n = i^(n mod 4)

where the remainder is taken in {0,1,2,3}\{0,1,2,3\} and i0=1i^0=1.

2019 AMC 10A Problem 4

What is i2019i^{2019}?

Since

2019=4⋅504+32019 = 4 \cdot 504 + 3

we get

i2019=i3=−ii^{2019}=i^3=\boxed{-i}

The mod-44 trick is mechanical once you see it — and it appears on AMC/AIME more often than you'd expect.

Algebra of Complex Numbers

Addition and Subtraction

Add or subtract real and imaginary parts separately:

(a+bi)±(c+di)=(a±c)+(b±d)i(a+bi)\pm(c+di) = (a\pm c)+(b\pm d)i

This is exactly like vector addition componentwise.

Multiplication

Expand using FOIL and replace i2i^2 with −1-1:

(a+bi)(c+di)=ac+adi+bci+bdi2=(ac−bd)+(ad+bc)i(a+bi)(c+di) = ac+adi+bci+bdi^2 = (ac-bd)+(ad+bc)i

You do not need to memorize this formula.

Example: Compute (2+3i)(1−4i)(2+3i)(1-4i).

(2+3i)(1−4i)=2(1)+2(−4i)+3i(1)+3i(−4i)=2−8i+3i−12i2=2−5i+12=14−5i\begin{aligned} (2+3i)(1-4i) &=2(1)+2(-4i)+3i(1)+3i(-4i) \\ &=2-8i+3i-12i^2 \\ &=2-5i+12 \\ &=14-5i \end{aligned}

Equating Real and Imaginary Parts

Two complex numbers are equal iff their real parts and imaginary parts are equal:

a+bi=c+di  ⟺  a=c and b=da+bi=c+di \iff a=c \text{ and } b=d

This means one complex equation is secretly two real equations.

Example: Find real x,yx,y such that

(x+yi)(2+3i)=1+13i(x+yi)(2+3i)=1+13i

Expanding:

(2x−3y)+(3x+2y)i(2x-3y)+(3x+2y)i

Equating parts:

2x−3y=1and3x+2y=132x-3y=1 \qquad\text{and}\qquad 3x+2y=13

Multiply the first equation by 22 and the second by 33:

4x−6y=24x-6y=2
9x+6y=399x+6y=39

Adding:

13x=41  ⟹  x=411313x=41 \implies x=\frac{41}{13}

Substituting back:

y=−2313y=-\frac{23}{13}

Conjugates

The conjugate of

z=a+biz=a+bi

is

z‾=a−bi\overline{z}=a-bi

Geometrically, this reflects the point across the real axis.

The key identity is:

zz‾=(a+bi)(a−bi)=a2+b2z\overline{z} = (a+bi)(a-bi) = a^2+b^2

The product is always a nonnegative real number.

Two useful identities:

z+z‾=2Re⁡(z)z+\overline{z}=2\operatorname{Re}(z)
z−z‾=2iIm⁡(z)z-\overline{z}=2i\operatorname{Im}(z)

Properties of conjugation:

z+w‾=z‾+w‾\overline{z+w}=\overline{z}+\overline{w}
zw‾=z‾ w‾\overline{zw}=\overline{z}\,\overline{w}

Division

To divide complex numbers, multiply numerator and denominator by the conjugate of the denominator.

1a+bi=a−bia2+b2\frac{1}{a+bi} = \frac{a-bi}{a^2+b^2}

More generally,

a+bic+di=(a+bi)(c−di)c2+d2\frac{a+bi}{c+di} = \frac{(a+bi)(c-di)}{c^2+d^2}

start

a+bi

c+di

× c−di

× c−di

multiply out

(a+bi)(c−di)

c² + d²

simplify

result

p + qi

real denominator ✓

Example: Compute

3+2∗i1−4∗i\frac{3+2*i}{1-4*i}

Multiply top and bottom by 1+4i1+4i:

3+2∗i1−4∗i⋅1+4∗i1+4∗i=(3+2∗i)(1+4∗i)12+42\frac{3+2*i}{1-4*i} \cdot \frac{1+4*i}{1+4*i} = \frac{(3+2*i)(1+4*i)}{1^2+4^2}

Expand the numerator:

(3+2i)(1+4i)=3+12i+2i+8i2=3+14i−8=−5+14i\begin{aligned} (3+2i)(1+4i) &=3+12i+2i+8i^2 \\ &=3+14i-8 \\ &=-5+14i \end{aligned}

Denominator:

1+16=171+16=17

Therefore,

3+2∗i1−4∗i=−5+14∗i17=−517+1417i\frac{3+2*i}{1-4*i} = \frac{-5+14*i}{17} = -\frac{5}{17}+\frac{14}{17}i

Worked Examples

Example 1: Compute i2023i^{2023}.

Since

2023=4⋅505+32023=4\cdot505+3

we get

i2023=i3=−ii^{2023}=i^3=-i

Example 2: If z+6i=izz+6i=iz, find zz.

Rearranging:

z−iz=−6iz-iz=-6i
z(1−i)=−6iz(1-i)=-6i

Thus,

z=−6∗i1−i⋅1+i1+i=−6∗i(1+i)2=−6∗i+62=3−3iz = \frac{-6*i}{1-i} \cdot \frac{1+i}{1+i} = \frac{-6*i(1+i)}{2} = \frac{-6*i+6}{2} = 3-3i

Example 3: Let

z+z‾=6z+\overline{z}=6

and

zz‾=13z\overline{z}=13

Find zz.

Write

z=a+biz=a+bi

Then

2a=6  ⟹  a=32a=6 \implies a=3

Also,

a2+b2=13a^2+b^2=13

so

9+b2=13  ⟹  b2=4  ⟹  b=±29+b^2=13 \implies b^2=4 \implies b=\pm2

Therefore,

z=3+2iorz=3−2iz=3+2i \quad\text{or}\quad z=3-2i

Example 4 (AIME flavor): Find all complex zz such that

z2=2iz^2=2i

Write

z=a+biz=a+bi

Then

z2=(a2−b2)+2abiz^2=(a^2-b^2)+2abi

Equating parts:

a2−b2=0a^2-b^2=0
2ab=2  ⟹  ab=12ab=2 \implies ab=1

Since ab>0ab>0, we must have a=ba=b.

Then

a2=1a^2=1

giving

z=1+iorz=−1−iz=1+i \quad\text{or}\quad z=-1-i

Strategy Checklist

TaskTechnique
Simplify ini^nCompute n mod 4n \bmod 4
Solve for complex variablesExpand and equate real/imaginary parts
Eliminate ii from denominatorMultiply by conjugate
Extract Re⁡(z)\operatorname{Re}(z) or Im⁡(z)\operatorname{Im}(z)Use conjugate identities
Relate zz and z‾\overline{z}Use zz‾=a2+b2z\overline{z}=a^2+b^2
Solve equations like z2=wz^2=wWrite z=a+biz=a+bi and equate parts

Common Pitfalls

  • Treating Im⁡(z)\operatorname{Im}(z) as bibi instead of bb
  • Forgetting that i2=−1i^2=-1
  • Incorrectly combining square roots of negative numbers
  • Confusing z+z‾z+\overline{z} with 2∣z∣2|z|
  • Forgetting to multiply numerator and denominator by the conjugate
  • Changing the sign of the real part when conjugating

Remarks

Everything in this module has been purely algebraic. But there is a beautiful geometric picture underneath all of it. In the next module, we interpret

z=a+biz=a+bi

as the point (a,b)(a,b) in the plane and discover that conjugation is reflection, addition is vector addition, and multiplication becomes rotation and scaling.

Seeing the geometry makes many of the identities here feel obvious almost inevitable.

Practice Problems

StatusSourceProblem NameDifficultyTags
MathNetHard
Show TagsComplex numbers, Factorization techniques
Berkeley Math Circle Take-Home Contest #1Hard
Show TagsComplex numbers, Roots of unity, Vieta's formulas
Berkeley Math Circle Monthly Contest 2Hard
Show TagsComplex numbers, Recurrence relations
Berkeley Math Circle Take-Home Contest #2Hard
Show TagsComplex numbers, Roots of unity
Berkeley Math Circle Monthly Contest 3Hard
Show TagsComplex numbers, Polynomial operations
Berkeley Math CircleHard
Show TagsComplex numbers, Irreducibility: Rational Root Theorem, Gauss's Lemma, Eisenstein, Polynomials mod p
Berkeley Math Circle Monthly Contest 2Hard
Show TagsComplex numbers, Infinite descent / root flipping, Techniques: modulo, size analysis, order analysis, inequalities
Berkeley Math Circle Monthly Contest 1Hard
Show TagsComplex numbers, Recurrence relations, Symmetric functions, Vieta's formulas
Harvard-MIT Math TournamentHard
Show TagsComplex numbers, Polynomial operations, Roots of unity
Harvard-MIT Math TournamentHard
Show TagsComplex numbers
Harvard-MIT Math TournamentHard
Show TagsComplex numbers, Sums and products
Harvard-MIT Math TournamentHard
Show TagsComplex numbers, Polynomial operations
Harvard-MIT Mathematics TournamentHard
Show TagsComplex numbers, Sums and products
Harvard-MIT Mathematics TournamentHard
Show TagsComplex numbers, Polynomial operations
USA IMOHard
Show TagsComplex numbers, Enumeration with symmetry, Generating functions, Inclusion-exclusion, Roots of unity
Harvard-MIT Mathematics TournamentHard
Show TagsComplex numbers, Roots of unity, Vieta's formulas
Harvard-MIT Mathematics TournamentHard
Show TagsComplex numbers, Ring Theory, Roots of unity
Harvard-MIT Mathematics TournamentHard
Show TagsComplex numbers, Complex numbers in geometry, Vectors
Harvard-MIT Mathematics TournamentHard
Show TagsComplex numbers, Polynomial operations, Roots of unity
Harvard-MIT Mathematics TournamentHard
Show TagsComplex numbers, Roots of unity
Harvard-MIT Mathematics TournamentHard
Show TagsComplex numbers, Complex numbers in geometry, Roots of unity
Harvard-MIT Mathematics TournamentHard
Show TagsComplex numbers, Polynomial operations, Roots of unity
Harvard-MIT Mathematics TournamentHard
Show TagsComplex numbers, Polynomials
Harvard-MIT Mathematics TournamentHard
Show TagsComplex numbers, Vieta's formulas
Harvard-MIT Mathematics TournamentHard
Show TagsComplex numbers, Polynomial operations

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