Introduction to Complex Numbers - Complete Interactive Lesson
Part 1: The Imaginary Unit i
🌀 Introduction to Complex Numbers
Part 1 of 5 — The Imaginary Unit i
Topics in This Part
Section
Why We Need a New Number
Defining i and i2=−1
Simplifying −n
🔑 Key Concept: There is no real number whose square is negative. To solve equations like x2=−1, mathematicians invented a brand-new number, the imaginary uniti, defined so that i2.
Why We Need a New Number
Take any real number and square it:
32=9,(−3)2=9,0
Defining the Imaginary Unit
The imaginary uniti is defined by:
i=−1
Concept Check 🎯
Simplify the Square Roots 🧮
Write each as a real number times i. Enter just the coefficient in front of i (a whole number).
1)−49
Match Each Radical 🔽
Some of these don't simplify to a whole-number coefficient — leave them under the radical.
What You Built
You now have one new tool: the imaginary unit i with i2=−1, which lets you take the square root of any negative number.
🔑 Carry this forward: for . In Part 2 we attach a real number to an imaginary one and get a full .
Part 2: Complex Numbers in Standard Form
🌀 Introduction to Complex Numbers
Part 2 of 5 — Complex Numbers in Standard Form
🔑 The Idea: A complex number combines a real part and an imaginary part into one object written a+bi, where a and b are real numbers.
Standard Form a
Part 3: Adding, Subtracting & Multiplying
🌀 Introduction to Complex Numbers
Part 3 of 5 — Adding, Subtracting & Multiplying
🔑 The Idea: Treat i like a variable when you add, subtract, and multiply — then, at the very end, replace every i2 with −1.
Adding and Subtracting
Combine real parts with real parts and imaginary parts with imaginary parts — exactly like collecting like terms.
Part 4: Powers of i & Complex Conjugates
🌀 Introduction to Complex Numbers
Part 4 of 5 — Powers of i & Complex Conjugates
🔑 Two power tools: the powers of i repeat in a cycle of four, and the complex conjugate lets you turn a product into a real number — the key to dividing complex numbers.
The Cycle of Powers
Start multiplying by i and watch the pattern repeat every four steps:
Part 5: Dividing, Solving & Mastery Check
🌀 Introduction to Complex Numbers
Part 5 of 5 — Dividing, Solving & Mastery Check
You can now define i, write numbers in a+bi form, add, subtract, multiply, take powers of i, and find conjugates. Let's finish with division and solving equations, then prove your mastery.
Dividing Complex Numbers
To divide, multiply the top and bottom by the conjugate of the denominator. This makes the denominator real.
Example:
=
−1
2
=
0
A real number squared is never negative. So the equation
x2=−1
has no real solution — there is simply no real number to plug in.
Rather than give up, mathematicians defined a new number to fill the gap.
💡 This is not a "fake" idea. Complex numbers are used every day in electrical engineering, signal processing, and quantum physics. The name imaginary is just a historical accident.
and equivalently
i2
=
−1
That one rule, i2=−1, is the engine behind everything in this lesson.
We can now take the square root of any negative number. For a positive number n:
−n=−1⋅n=in
Examples
Expression
Rewrite
Simplified
−25
−1⋅25
5i
−9
−1
−7
−1
⚠️ Pull out the i first. Always factor −1 out before simplifying the rest, so you never accidentally multiply two negative radicands together.
=
?i
2)
−100=?i
3)
−1=?i
−
n
=
in
n>0
complex number
+
bi
Every complex number can be written as
a+bi
a is the real part, written Re(z)=a
b is the imaginary part, written Im(z)=b(it's the coefficient of i, not bi itself)
Reading Off the Parts
Complex number
Real part a
Imaginary part b
3+5i
3
5
−2+7i
−2
7
4−6i
4
−6
9
9
0
−8i
0
−8
💡 Every real number is also complex. A real number like 9 is 9+0i — it just has a zero imaginary part. A number like −8i is 0−8i and is called purely imaginary.
Concept Check 🎯
Classify Each Number 🔽
Decide whether each is real, purely imaginary, or a non-real complex number (has both a nonzero real and a nonzero imaginary part).
Rewriting in Standard Form
Sometimes a number arrives "messy" and you must rewrite it as a+bi.
Example: 5+−16
5+−16=5+
So a=5 and b=4.
Example: 2−6+−4
2−6+−4
So a=−3 and b=1.
⚠️ Convert every −n into in trying to identify and .
Find a and b 🧮
Rewrite each in standard form a+bi, then enter the two values.
1)8+−9: enter a then b.
2)2−10+−36: enter a then b.
(a+bi)+(c+di)=(a+c)+(b+d)i
(a+bi)−(c+di)=(a−c)+(b−d)i
Example: Add
(3+5i)+(4+2i)=(3+4)+(5+2)i=7+7i
Example: Subtract
(6+3i)−(2+8i)=(6−2)+(3−8)i=4−5i
⚠️ When subtracting, distribute the minus sign to both parts of the second number — including the imaginary part.
Combine Like Parts 🧮
Write each answer as a+bi and enter a then b.
1)(2+7i)+(5+i): enter a then b.
2)(9+4i)−(3+10i): enter a then b.
Multiplying
Multiply just like binomials (FOIL), then replace i2 with −1.
Example: (2+3i)(4+5i)
(2+
Example: A Single Imaginary Times Itself
(4i)(4i)=16i2=16(−1)=−16
🔑 The whole trick: the i2 term collapses to a real number (−1), which then merges with the real part. That is why a product of two non-real numbers can come out real.
Concept Check 🎯
Multiply It Out 🧮
Write each product as a+bi and enter a then b. (If a part is 0, enter 0.)
1)(2+i)(3+4i): enter a then b.
2)(6i)(2i): enter a then b.
i1=i,i2=−1,i3=i2⋅i=−i,i4=(i2)2=1
Then it cycles: i5=i, i6=−1, and so on.
Power
Value
i1
i
i2
−1
i3
−i
i4
1
The Shortcut
To find in, divide n by 4 and use the remainder:
in=i(nmod4)
Example:i23. Since 23=4⋅5+3, the remainder is 3, so i23=i3=−i.
The conjugate of a+bi is a−bi — flip the sign of the imaginary part. We write it z.
Number z
Conjugate z
3+5i
3−5i
2−7i
Why Conjugates Matter
Multiply a number by its conjugate and the imaginary part vanishes:
(a+bi)(a−bi)=a2−
The result a2+b2 is always a real, non-negative number.
Example: (3+2i)(3−2i)
(3+2i)(3−2i)=32+2
🔑 This "real-izing" property is exactly what makes division of complex numbers possible (Part 5).
Concept Check 🎯
Conjugate Products 🧮
Each product of conjugates equals a real number a2+b2. Enter that number.
1)(4+3i)(4−3i)=?2)(2+6i)(2−6i)=?3)(7i)(−7i)=?
3−i2+i
Multiply by 3+i3+i:
3−i2+i⋅3+i3+i=(3−i)(3+i)(2+i)(3+i)
Numerator:(2+i)(3+i)=6+2i+3i+i2=6+5i−1=5+5i
Denominator:(3−i)(3+i)=32+12=10
=105+5i=21+21i
🔑 The strategy is always the same: the conjugate turns the denominator into a real number, and then you split the fraction into real and imaginary parts.
Divide and Simplify 🧮
Simplify each quotient to a+bi form (here both parts are whole numbers). Enter a then b.
1)1+3i10: enter a then b.
2)2+i5i: enter a then b.
Solving Equations with Complex Roots
When the square root of a negative number appears, write the answer with i.
Example: x2+25=0
x2=−25⇒x=±−25=±5i
Example: x2+9=0
x2=−9⇒x=±−9
💡 Big picture: with complex numbers, every quadratic equation has a solution. A quadratic with a negative discriminant has two complex conjugate roots instead of two real ones.