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\( i \)

Understanding
Complex Numbers

Discover the number system beyond the reals: learn what \( i \) means, how every complex number takes the form \( a + bi \), and how to visualize them on the complex plane.

Algebra 2

What Are Complex Numbers?

For centuries, mathematicians had no answer to the equation \( x^2 = -1 \). No real number, when squared, gives a negative result. The complex numbers were invented to fill this gap: they extend the real number line into a full two-dimensional number plane.

The Imaginary Unit \( i \)

The imaginary unit \( i \) is defined by a single, powerful property:

This is not a "made-up" trick: \( i \) is as rigorously defined as the number \( \pi \). It simply lives in a direction perpendicular to the real number line.

\( i^1 = i \)
The unit itself
\( i^2 = -1 \)
The defining property
\( i^3 = -i \)
Since \( i^2 \cdot i = -i \)
\( i^4 = 1 \)
The cycle repeats!

Anatomy of \( a + bi \)

Every complex number has exactly two parts. Understanding these parts is the key to working with the entire system.

Real Part: \( a \)
The component along the horizontal real axis. For \( 3 + 5i \), the real part is 3.
Imaginary Part: \( b \)
The coefficient of \( i \) along the vertical imaginary axis. For \( 3 + 5i \), the imaginary part is 5.

The Complex Plane

We visualize complex numbers on a two-dimensional plane (also called the Argand diagram). The horizontal axis is the real axis, and the vertical axis is the imaginary axis.

Re Im -2 -1 1 2 3 1 2 -1 -2
\( 3 + 2i \)
\( -1 - i \)
0

Core Definition

1
\( i \) is defined by \( i^2 = -1 \). This is not derived from other numbers: it is an axiom that extends the real number system.
2
Every complex number is \( a + bi \). The set of all such numbers is denoted \( \mathbb{C} \). The reals \( \mathbb{R} \) are a subset where \( b = 0 \).
3
Two complex numbers are equal if and only if their real parts are equal and their imaginary parts are equal: \( a + bi = c + di \iff a = c \) and \( b = d \).

The Cyclic Pattern of \( i \)

The powers of \( i \) repeat in a cycle of four. This pattern lets you simplify any power of \( i \) instantly.

\( i^1 = i \)
\( i^2 = -1 \)
\( i^3 = -i \)
\( i^4 = 1 \)
Cycle repeats every 4 powers

Real vs. Imaginary vs. Complex

These three terms describe overlapping categories, not separate worlds. Every real number and every imaginary number is also a complex number.

Real Pure Imaginary Complex
Form \( a + 0i \) \( 0 + bi \) \( a + bi \)
Example 5 \( 3i \) \( 2 + 3i \)
Position on plane On real axis On imaginary axis Anywhere
Is it complex? Yes Yes Yes

How to Identify the Parts

1
Write in standard form \( a + bi \). Rearrange the expression so the real term comes first and the imaginary term (with \( i \)) comes second.
2
Read off \( a \): the coefficient standing alone (no \( i \)). This is the real part.
3
Read off \( b \): the coefficient multiplying \( i \). This is the imaginary part. Note: the imaginary part is the number \( b \), not \( bi \).
4
Plot on the complex plane: go \( a \) units along the real axis and \( b \) units along the imaginary axis.

Example 1: Identifying Parts

Problem: Identify the real and imaginary parts of \( z = -4 + 7i \).

1
The expression is already in standard form: \( a + bi \).
2
Real part: \( a = -4 \).
3
Imaginary part: \( b = 7 \).
Re Im 0 -4 7
\( -4 + 7i \)

Example 2: Simplifying \( \sqrt{-25} \)

Problem: Simplify \( \sqrt{-25} \) and write it in the form \( a + bi \).

1
Factor out \( -1 \): \( \sqrt{-25} = \sqrt{25 \cdot (-1)} \).
2
Split the radical: \( \sqrt{25} \cdot \sqrt{-1} = 5i \).
3
Standard form: \( 0 + 5i \). The real part is 0, the imaginary part is 5.

Example 3: Powers of \( i \)

Problem: Simplify \( i^{23} \).

1
Divide the exponent by 4: \( 23 \div 4 = 5 \) remainder 3.
2
Use the remainder: remainder 3 corresponds to \( i^3 \).
3
Recall the cycle: \( i^3 = -i \).

Adding Complex Numbers

Addition is performed component-wise: add the real parts together and the imaginary parts together.

Example: \( (3 + 2i) + (-1 + 4i) \)

1
Real parts: \( 3 + (-1) = 2 \)
2
Imaginary parts: \( 2 + 4 = 6 \)
3
Result: \( 2 + 6i \)

Multiplying Complex Numbers

Use the distributive property (FOIL), then replace \( i^2 \) with \( -1 \).

Example: \( (2 + 3i)(1 - 4i) \)

1
FOIL: \( 2 \cdot 1 + 2 \cdot (-4i) + 3i \cdot 1 + 3i \cdot (-4i) \)
2
Expand: \( 2 - 8i + 3i - 12i^2 \)
3
Replace \( i^2 = -1 \): \( 2 - 5i - 12(-1) = 2 - 5i + 12 \)
4
Result: \( 14 - 5i \)

Common Mistakes to Avoid

✕ Forgetting \( i^2 = -1 \)
When multiplying, students leave \( i^2 \) in the answer instead of replacing it with \( -1 \). Always simplify!
✕ Confusing \( b \) with \( bi \)
The imaginary part of \( 3 + 5i \) is 5, not \( 5i \). The part is the coefficient, not the term.
✕ Wrong sign on \( \sqrt{-k} \)
Writing \( \sqrt{-9} = -3 \) instead of \( 3i \). Remember: \( \sqrt{-k} = i\sqrt{k} \), not a negative real number.

Summary Reference

Concept Formula / Rule Example
Imaginary unit \( i^2 = -1 \) \( i = \sqrt{-1} \)
Standard form \( a + bi \) \( 3 + 2i \)
Real part \( \text{Re}(a + bi) = a \) \( \text{Re}(3 + 2i) = 3 \)
Imaginary part \( \text{Im}(a + bi) = b \) \( \text{Im}(3 + 2i) = 2 \)
Addition \( (a+bi)+(c+di) = (a+c)+(b+d)i \) \( (3+2i)+(-1+4i) = 2+6i \)
Powers of \( i \) Cycle of 4 \( i, -1, -i, 1, \ldots \)

Question 1 of 5

What is the value of \( i^2 \)?

Question 2 of 5

What is the imaginary part of the complex number \( 6 - 9i \)?

Question 3 of 5

Simplify \( \sqrt{-49} \).

Question 4 of 5

What is \( i^{46} \)?

Question 5 of 5

Which of the following is a pure imaginary number?