Introduction to Complex Numbers

🌀 Introduction to Complex Numbers

Part 1 of 5 — The Imaginary Unit $i$

Topics in This Part

🔑 Key Concept: There is no real number whose square is negative. To solve equations like $x^2 = -1$, mathematicians invented a brand-new number, the imaginary unit $i$, defined so that $i^2 = -1$.

Why We Need a New Number

Take any real number and square it:

$3^2 = 9, \qquad (-3)^2 = 9, \qquad 0^2 = 0$

A real number squared is never negative. So the equation

$x^2 = -1$

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.

Defining the Imaginary Unit

The imaginary unit $i$ is defined by:

$i = \sqrt{-1} \qquad\text{and equivalently}\qquad i^2 = -1$

That one rule, $i^2 = -1$, is the engine behind everything in this lesson.

We can now take the square root of any negative number. For a positive number $n$:

$\sqrt{-n} = \sqrt{-1}\cdot\sqrt{n} = i\sqrt{n}$

Examples

⚠️ Pull out the $i$ first. Always factor $\sqrt{-1}$ out before simplifying the rest, so you never accidentally multiply two negative radicands together.

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) $\sqrt{-49} = \,?\,i$ 2) $\sqrt{-100} = \,?\,i$ 3) $\sqrt{-1} = \,?\,i$

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 $i^2 = -1$, which lets you take the square root of any negative number.

🔑 Carry this forward: $\sqrt{-n} = i\sqrt{n}$ for $n > 0$. In Part 2 we attach a real number to an imaginary one and get a full complex number.

🌀 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 $a + bi$

Every complex number can be written as

$a + bi$

• $a$ is the real part, written $\text{Re}(z) = a$
• $b$ is the imaginary part, written $\text{Im}(z) = b$ (it's the coefficient of $i$, not $bi$ itself)

Reading Off the Parts

💡 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: $5 + \sqrt{-16}$

$5 + \sqrt{-16} = 5 + \sqrt{-1}\cdot\sqrt{16} = 5 + 4i$

So $a = 5$ and $b = 4$.

Example: $\dfrac{-6 + \sqrt{-4}}{2}$

$\frac{-6 + \sqrt{-4}}{2} = \frac{-6 + 2i}{2} = \frac{-6}{2} + \frac{2i}{2} = -3 + i$

So $a = -3$ and $b = 1$.

⚠️ Convert every $\sqrt{-n}$ into $i\sqrt{n}$ before trying to identify $a$ and $b$.

Find $a$ and $b$ 🧮

Rewrite each in standard form $a + bi$, then enter the two values.

1) $8 + \sqrt{-9}$: enter $a$ then $b$. 2) $\dfrac{-10 + \sqrt{-36}}{2}$: enter $a$ then $b$.

🌀 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 $i^2$ with $-1$.

Adding and Subtracting

Combine real parts with real parts and imaginary parts with imaginary parts — exactly like collecting like terms.

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

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

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 $i^2$ with $-1$.

Example: $(2 + 3i)(4 + 5i)$

Example: A Single Imaginary Times Itself

$(4i)(4i) = 16i^2 = 16(-1) = -16$

🔑 The whole trick: the $i^2$ 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 $b$.

🌀 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:

$i^1 = i, \quad i^2 = -1, \quad i^3 = i^2\cdot i = -i, \quad i^4 = (i^2)^2 = 1$

Then it cycles: $i^5 = i$, $i^6 = -1$, and so on.

The Shortcut

To find $i^n$, divide $n$ by $4$ and use the remainder:

$i^n = i^{(n \bmod 4)}$

Example: $i^{23}$. Since $23 = 4\cdot 5 + 3$, the remainder is $3$, so $i^{23} = i^3 = -i$.

💡 Remainder $0 \to 1$, remainder $1 \to i$, remainder $2 \to -1$, remainder $3 \to -i$.

Evaluate the Powers 🔽

Use the remainder-on-division-by-$4$ shortcut.

Complex Conjugates

The conjugate of $a + bi$ is $a - bi$ — flip the sign of the imaginary part. We write it $\overline{z}$.

Why Conjugates Matter

Multiply a number by its conjugate and the imaginary part vanishes:

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

The result $a^2 + b^2$ is always a real, non-negative number.

Example: $(3 + 2i)(3 - 2i)$

$(3+2i)(3-2i) = 3^2 + 2^2 = 9 + 4 = 13$

🔑 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 $a^2 + b^2$. Enter that number.

1) $(4 + 3i)(4 - 3i) = \,?$ 2) $(2 + 6i)(2 - 6i) = \,?$ 3) $(7i)(-7i) = \,?$

🌀 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: $\dfrac{2 + i}{3 - i}$

Multiply by $\dfrac{3 + i}{3 + i}$:

$\frac{2 + i}{3 - i}\cdot\frac{3 + i}{3 + i} = \frac{(2+i)(3+i)}{(3-i)(3+i)}$

Numerator: $(2+i)(3+i) = 6 + 2i + 3i + i^2 = 6 + 5i - 1 = 5 + 5i$

Denominator: $(3-i)(3+i) = 3^2 + 1^2 = 10$

$= \frac{5 + 5i}{10} = \frac{1}{2} + \frac{1}{2}i$

🔑 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) $\dfrac{10}{1 + 3i}$: enter $a$ then $b$. 2) $\dfrac{5i}{2 + i}$: enter $a$ then $b$.

Solving Equations with Complex Roots

When the square root of a negative number appears, write the answer with $i$.

Example: $x^2 + 25 = 0$

$x^2 = -25 \;\Rightarrow\; x = \pm\sqrt{-25} = \pm 5i$

Example: $x^2 + 9 = 0$

$x^2 = -9 \;\Rightarrow\; x = \pm\sqrt{-9} = \pm 3i$

💡 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.

Concept Check 🎯

Mixed Review 🔽

One quick check from each part of the lesson.

Exit Quiz ✅

Answer all three to finish the lesson.