Complex Numbers

Discover a number system where negative square roots make sense

You have probably been told at some point that you cannot take the square root of a negative number. And with the numbers you knew at the time, that was true. After all, what number times itself gives you $-9$? If you try $3$, you get $3 \times 3 = 9$ (positive). If you try $-3$, you get $(-3) \times (-3) = 9$ (still positive, because a negative times a negative is positive). No matter what real number you try, squaring it always gives you something positive or zero - never negative.

But here is the thing: mathematicians are stubborn. When they encounter an equation like $x^2 = -1$ and find that no existing number works, they do not give up. They ask, “What if we invented a number that does work?” This is exactly what happened with complex numbers. Someone said, “Let’s imagine a number that, when squared, gives $-1$.” And from that single idea came an entirely new number system - one that turned out to be not just a clever trick, but essential for describing how the universe actually works.

If this sounds like mathematical fantasy, consider that people once said the same thing about negative numbers (“How can you have less than nothing?”) and about zero itself (“How can nothing be a number?”). Each time we expanded our number system, we gained new tools to solve problems that were previously impossible. Complex numbers are the next step in that journey.

Core Concepts

Why We Need Complex Numbers

Think back to the quadratic formula. For the equation $ax^2 + bx + c = 0$, you have:

$$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$

That expression under the square root - the discriminant $b^2 - 4ac$ - determines what kind of solutions you get. When it is positive, you get two real solutions. When it is zero, you get one real solution. But what about when it is negative?

Consider the equation $x^2 + 1 = 0$. Using the quadratic formula with $a = 1$, $b = 0$, $c = 1$:

$$x = \frac{0 \pm \sqrt{0 - 4}}{2} = \frac{\pm \sqrt{-4}}{2}$$

With only real numbers, you are stuck. The square root of a negative number does not exist in the real number system. But if we extend our number system to include these square roots, suddenly the equation has solutions. The quadratic formula never fails - we just need the right kind of numbers.

The Imaginary Unit $i$

The foundation of complex numbers is a single definition:

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

This means that $i^2 = -1$.

The letter $i$ stands for “imaginary,” a name that stuck from the 1600s when mathematicians were still suspicious of these new numbers. The name is unfortunate because it makes complex numbers sound fake or less valid than “real” numbers. In truth, they are just as legitimate as any other numbers - perhaps more so, since they are essential to physics and engineering.

With $i$ defined, you can now take the square root of any negative number:

$$\sqrt{-9} = \sqrt{9 \cdot (-1)} = \sqrt{9} \cdot \sqrt{-1} = 3i$$

$$\sqrt{-49} = \sqrt{49 \cdot (-1)} = \sqrt{49} \cdot \sqrt{-1} = 7i$$

The pattern is straightforward: $\sqrt{-n} = i\sqrt{n}$ for any positive number $n$.

Powers of $i$: A Cycle

What happens when you raise $i$ to various powers? Let’s find out:

  • $i^1 = i$
  • $i^2 = -1$ (by definition)
  • $i^3 = i^2 \cdot i = (-1) \cdot i = -i$
  • $i^4 = i^2 \cdot i^2 = (-1)(-1) = 1$
  • $i^5 = i^4 \cdot i = 1 \cdot i = i$
  • $i^6 = i^4 \cdot i^2 = 1 \cdot (-1) = -1$

Do you see the pattern? The powers of $i$ cycle through four values: $i, -1, -i, 1$, and then repeat. This cycle continues forever:

Power Value
$i^1$ $i$
$i^2$ $-1$
$i^3$ $-i$
$i^4$ $1$
$i^5$ $i$
$i^6$ $-1$

To find $i^n$ for any positive integer $n$, divide $n$ by 4 and look at the remainder:

  • Remainder 0: $i^n = 1$
  • Remainder 1: $i^n = i$
  • Remainder 2: $i^n = -1$
  • Remainder 3: $i^n = -i$

Standard Form: $a + bi$

A complex number is any number of the form:

$$a + bi$$

where $a$ and $b$ are real numbers. This is called the standard form of a complex number.

The number $a$ is called the real part, and the number $b$ is called the imaginary part. Note that the imaginary part is just $b$, not $bi$ - it is the coefficient of $i$, not the whole term.

For example, in the complex number $3 + 2i$:

  • The real part is $3$
  • The imaginary part is $2$

Every real number is also a complex number - just with an imaginary part of zero. So $5$ is the same as $5 + 0i$. And pure imaginary numbers like $7i$ are complex numbers with a real part of zero: $0 + 7i$.

Adding and Subtracting Complex Numbers

Adding and subtracting complex numbers works exactly like combining like terms in algebra. You add the real parts together and the imaginary parts together:

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

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

For example: $$(3 + 2i) + (5 - 4i) = (3 + 5) + (2 + (-4))i = 8 + (-2)i = 8 - 2i$$

Think of it like adding vectors, or combining apples and oranges - you keep the different types separate.

Multiplying Complex Numbers

Multiplying complex numbers uses the distributive property (sometimes called FOIL for binomials). Multiply each term in the first number by each term in the second, then combine like terms.

$$(a + bi)(c + di) = ac + adi + bci + bdi^2$$

Since $i^2 = -1$, that last term becomes $bd(-1) = -bd$. Combining:

$$(a + bi)(c + di) = (ac - bd) + (ad + bc)i$$

You do not need to memorize this formula. Just multiply out the terms and remember that $i^2 = -1$.

For example: $$(2 + 3i)(4 - i) = 2(4) + 2(-i) + 3i(4) + 3i(-i)$$ $$= 8 - 2i + 12i - 3i^2$$ $$= 8 + 10i - 3(-1)$$ $$= 8 + 10i + 3$$ $$= 11 + 10i$$

Complex Conjugates

The complex conjugate of a complex number $a + bi$ is $a - bi$. You simply flip the sign of the imaginary part. The conjugate of $3 + 2i$ is $3 - 2i$. The conjugate of $5 - 7i$ is $5 + 7i$.

Complex conjugates have a remarkable property. When you multiply a complex number by its conjugate, the imaginary parts cancel out and you get a real number:

$$(a + bi)(a - bi) = a^2 - abi + abi - b^2i^2$$ $$= a^2 - b^2(-1)$$ $$= a^2 + b^2$$

This result is always a positive real number (unless both $a$ and $b$ are zero). This property is the key to dividing complex numbers.

Dividing Complex Numbers

You cannot simply divide complex numbers by splitting up the parts. Instead, you use a clever trick: multiply both the numerator and denominator by the conjugate of the denominator.

To divide $\frac{a + bi}{c + di}$:

  1. Find the conjugate of the denominator: $c - di$
  2. Multiply both numerator and denominator by this conjugate: $$\frac{a + bi}{c + di} \cdot \frac{c - di}{c - di} = \frac{(a + bi)(c - di)}{(c + di)(c - di)}$$
  3. The denominator becomes real: $(c + di)(c - di) = c^2 + d^2$
  4. Expand the numerator and simplify

This process eliminates the imaginary number from the denominator, leaving you with a complex number in standard form.

Solving Quadratics with Complex Solutions

Now you can solve any quadratic equation - even those with negative discriminants. When $b^2 - 4ac < 0$, the solutions involve $i$.

For example, to solve $x^2 + 4x + 13 = 0$:

Using the quadratic formula with $a = 1$, $b = 4$, $c = 13$:

$$x = \frac{-4 \pm \sqrt{16 - 52}}{2} = \frac{-4 \pm \sqrt{-36}}{2}$$

Since $\sqrt{-36} = \sqrt{36} \cdot \sqrt{-1} = 6i$:

$$x = \frac{-4 \pm 6i}{2} = -2 \pm 3i$$

The solutions are $x = -2 + 3i$ and $x = -2 - 3i$.

Notice that the solutions are complex conjugates of each other. This is always true for quadratics with real coefficients: complex solutions come in conjugate pairs.

The Complex Plane

You can visualize complex numbers by graphing them on the complex plane (also called the Argand diagram). It looks like a regular coordinate plane, but:

  • The horizontal axis represents the real part
  • The vertical axis represents the imaginary part

The complex number $a + bi$ is plotted at the point $(a, b)$.

For example:

  • $3 + 2i$ is plotted at $(3, 2)$
  • $-1 + 4i$ is plotted at $(-1, 4)$
  • $-2 - 3i$ is plotted at $(-2, -3)$
  • $5$ (which is $5 + 0i$) is plotted at $(5, 0)$ on the real axis
  • $3i$ (which is $0 + 3i$) is plotted at $(0, 3)$ on the imaginary axis

The complex plane gives you a geometric way to understand complex numbers. Adding two complex numbers is like adding vectors - you can use the parallelogram rule. The conjugate of a number is its reflection across the real axis.

Notation and Terminology

Term Meaning Example
$i$ Imaginary unit: $i = \sqrt{-1}$ $i^2 = -1$
Complex number $a + bi$ where $a, b$ are real $3 + 2i$
Real part The $a$ in $a + bi$ Real part of $3 + 2i$ is $3$
Imaginary part The $b$ in $a + bi$ Imaginary part of $3 + 2i$ is $2$
Complex conjugate $a - bi$ (flip sign of imaginary) Conjugate of $3 + 2i$ is $3 - 2i$
Pure imaginary Complex number with $a = 0$ $5i$
Complex plane Coordinate system for complex numbers Real axis horizontal, imaginary axis vertical
Standard form Writing a complex number as $a + bi$ $3 + 2i$ is in standard form

Examples

Example 1: Simplifying Square Roots of Negative Numbers

Simplify $\sqrt{-49}$.

Solution:

Separate the negative from the positive part: $$\sqrt{-49} = \sqrt{49 \cdot (-1)}$$

Use the property that $\sqrt{ab} = \sqrt{a} \cdot \sqrt{b}$: $$= \sqrt{49} \cdot \sqrt{-1}$$

Simplify each part: $$= 7 \cdot i = 7i$$

Check: Does $(7i)^2 = -49$?

$(7i)^2 = 49i^2 = 49(-1) = -49$ ✓

Note: Always write the $i$ after the number, not before. Write $7i$, not $i7$.

Example 2: Adding Complex Numbers

Add $(3 + 2i) + (5 - 4i)$.

Solution:

Group the real parts and the imaginary parts: $$= (3 + 5) + (2i + (-4i))$$

Combine like terms: $$= 8 + (2 - 4)i$$ $$= 8 + (-2)i$$ $$= 8 - 2i$$

Why this works: Adding complex numbers is just like combining like terms. The real numbers combine with real numbers, and the imaginary numbers combine with imaginary numbers.

Example 3: Multiplying Complex Numbers

Multiply $(2 + 3i)(4 - i)$.

Solution:

Use the distributive property (FOIL):

First: $2 \cdot 4 = 8$

Outer: $2 \cdot (-i) = -2i$

Inner: $3i \cdot 4 = 12i$

Last: $3i \cdot (-i) = -3i^2$

Combine all terms: $$= 8 - 2i + 12i - 3i^2$$

Combine the imaginary terms: $$= 8 + 10i - 3i^2$$

Remember that $i^2 = -1$: $$= 8 + 10i - 3(-1)$$ $$= 8 + 10i + 3$$ $$= 11 + 10i$$

Key insight: The crucial step is replacing $i^2$ with $-1$. This is what turns the product of two complex numbers back into a complex number in standard form.

Example 4: Simplifying Powers of $i$

Simplify $i^{47}$.

Solution:

Powers of $i$ cycle every 4: $i, -1, -i, 1, i, -1, -i, 1, …$

To find where $i^{47}$ falls in this cycle, divide 47 by 4: $$47 \div 4 = 11 \text{ remainder } 3$$

The remainder of 3 tells us that $i^{47}$ is the same as $i^3$.

From the cycle:

  • Remainder 0 gives $1$
  • Remainder 1 gives $i$
  • Remainder 2 gives $-1$
  • Remainder 3 gives $-i$

Therefore: $i^{47} = -i$

Check: $i^{47} = i^{44} \cdot i^3 = (i^4)^{11} \cdot i^3 = 1^{11} \cdot (-i) = -i$ ✓

The pattern to remember: Divide the exponent by 4 and use the remainder to find the answer.

Example 5: Dividing Complex Numbers

Divide $\frac{3 + 2i}{1 - 4i}$.

Solution:

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

The conjugate of $1 - 4i$ is $1 + 4i$.

$$\frac{3 + 2i}{1 - 4i} \cdot \frac{1 + 4i}{1 + 4i}$$

Multiply the denominators: $$(1 - 4i)(1 + 4i) = 1 + 4i - 4i - 16i^2$$ $$= 1 - 16(-1) = 1 + 16 = 17$$

Multiply the numerators: $$(3 + 2i)(1 + 4i) = 3 + 12i + 2i + 8i^2$$ $$= 3 + 14i + 8(-1)$$ $$= 3 + 14i - 8$$ $$= -5 + 14i$$

Put it together: $$\frac{-5 + 14i}{17} = \frac{-5}{17} + \frac{14}{17}i = -\frac{5}{17} + \frac{14}{17}i$$

Why conjugates work: When you multiply $(a - bi)(a + bi)$, the imaginary terms cancel, leaving $a^2 + b^2$. This always gives a real number in the denominator, which is what we need.

Example 6: Solving a Quadratic with Complex Solutions

Solve $x^2 + 4x + 13 = 0$.

Solution:

First, check the discriminant to see what type of solutions we will get: $$b^2 - 4ac = (4)^2 - 4(1)(13) = 16 - 52 = -36$$

Since the discriminant is negative, we will have complex solutions.

Apply the quadratic formula: $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-4 \pm \sqrt{-36}}{2(1)}$$

Simplify the square root of the negative number: $$\sqrt{-36} = \sqrt{36} \cdot \sqrt{-1} = 6i$$

Substitute back: $$x = \frac{-4 \pm 6i}{2}$$

Simplify by dividing each term by 2: $$x = \frac{-4}{2} \pm \frac{6i}{2} = -2 \pm 3i$$

The solutions are $x = -2 + 3i$ and $x = -2 - 3i$.

Check for $x = -2 + 3i$: $$(-2 + 3i)^2 + 4(-2 + 3i) + 13$$ $$= (4 - 12i + 9i^2) + (-8 + 12i) + 13$$ $$= (4 - 12i - 9) + (-8 + 12i) + 13$$ $$= -5 - 12i - 8 + 12i + 13$$ $$= 0$$ ✓

Notice: The two solutions are complex conjugates. This always happens when a quadratic with real coefficients has complex solutions.

Key Properties and Rules

Fundamental Identity

$$i^2 = -1$$

Everything about complex numbers flows from this one definition.

Powers of $i$ (Cycle of 4)

Remainder when dividing exponent by 4 Value of $i^n$
0 $1$
1 $i$
2 $-1$
3 $-i$

Operations Summary

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

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

Multiplication: $(a + bi)(c + di) = (ac - bd) + (ad + bc)i$

Division: Multiply numerator and denominator by the conjugate of the denominator.

Conjugate Properties

For a complex number $z = a + bi$ and its conjugate $\bar{z} = a - bi$:

  • $z \cdot \bar{z} = a^2 + b^2$ (always a non-negative real number)
  • $\overline{z_1 + z_2} = \bar{z_1} + \bar{z_2}$
  • $\overline{z_1 \cdot z_2} = \bar{z_1} \cdot \bar{z_2}$

Complex Solutions of Quadratics

For a quadratic $ax^2 + bx + c = 0$ with real coefficients:

  • If $b^2 - 4ac < 0$, the solutions are complex conjugates of each other
  • The solutions have the form $p + qi$ and $p - qi$

Common Mistakes to Avoid

  • Writing $\sqrt{-9} = -3$ - This is incorrect. $\sqrt{-9} = 3i$, not a negative real number.
  • Forgetting that $i^2 = -1$ - When you see $i^2$ in a calculation, replace it with $-1$.
  • Writing $5i$ as $i5$ - Always write the number before $i$.
  • Confusing the imaginary part - The imaginary part of $3 + 2i$ is $2$, not $2i$.
  • Multiplying $\sqrt{-4} \cdot \sqrt{-9}$ directly - You cannot use $\sqrt{a} \cdot \sqrt{b} = \sqrt{ab}$ when both are negative. Instead: $\sqrt{-4} \cdot \sqrt{-9} = 2i \cdot 3i = 6i^2 = -6$.

Real-World Applications

Electrical Engineering

Complex numbers are essential for analyzing alternating current (AC) circuits. In AC, voltage and current vary sinusoidally over time, and they can be out of phase with each other. Electrical engineers use complex numbers (which they call “phasors”) to represent these quantities, making calculations much simpler than working with trigonometric functions directly.

The impedance of a circuit - its opposition to AC current - is naturally expressed as a complex number. The real part represents resistance (energy lost as heat), and the imaginary part represents reactance (energy stored and released by capacitors and inductors).

Signal Processing

When you stream music or make a video call, complex numbers are working behind the scenes. The Fourier transform, which converts signals between time and frequency domains, relies heavily on complex exponentials. This mathematics enables MP3 compression, noise cancellation, and image processing.

Quantum Mechanics

In quantum physics, the state of a particle is described by a wave function - and that wave function is inherently complex-valued. The imaginary unit $i$ appears in the fundamental equation of quantum mechanics (the Schrodinger equation). Complex numbers are not just a convenience here; they are essential to how quantum theory works.

Fractals and Computer Graphics

The famous Mandelbrot set - that infinitely intricate, self-similar shape you may have seen - is defined entirely in terms of complex numbers. For each point $c$ in the complex plane, you iterate the formula $z_{n+1} = z_n^2 + c$ and see whether the sequence stays bounded. This simple operation on complex numbers creates stunning visual complexity.

Control Systems

Engineers who design autopilots, cruise control, and industrial robots use complex numbers to analyze system stability. The poles and zeros of a transfer function - crucial for understanding how a system responds - are complex numbers. Whether a pole is in the left or right half of the complex plane determines whether the system is stable or unstable.

Self-Test Problems

Problem 1: Simplify $\sqrt{-81}$.

Show Answer

$$\sqrt{-81} = \sqrt{81 \cdot (-1)} = \sqrt{81} \cdot \sqrt{-1} = 9i$$

Problem 2: Add $(7 - 3i) + (-2 + 5i)$.

Show Answer

Combine real parts and imaginary parts separately: $$(7 + (-2)) + ((-3) + 5)i = 5 + 2i$$

Problem 3: Multiply $(1 + 2i)(3 - 4i)$.

Show Answer

Using FOIL: $$= 1(3) + 1(-4i) + 2i(3) + 2i(-4i)$$ $$= 3 - 4i + 6i - 8i^2$$ $$= 3 + 2i - 8(-1)$$ $$= 3 + 2i + 8$$ $$= 11 + 2i$$

Problem 4: Simplify $i^{102}$.

Show Answer

Divide 102 by 4: $102 = 25 \times 4 + 2$

The remainder is 2, so $i^{102} = i^2 = -1$.

Problem 5: Find the complex conjugate of $-4 + 7i$ and verify that their product is real.

Show Answer

The conjugate of $-4 + 7i$ is $-4 - 7i$ (flip the sign of the imaginary part).

Their product: $$(-4 + 7i)(-4 - 7i) = (-4)^2 - (7i)^2$$ $$= 16 - 49i^2$$ $$= 16 - 49(-1)$$ $$= 16 + 49 = 65$$

The product is $65$, which is indeed a real number.

Problem 6: Divide $\frac{2 + i}{3 - 2i}$.

Show Answer

Multiply by the conjugate of the denominator: $$\frac{2 + i}{3 - 2i} \cdot \frac{3 + 2i}{3 + 2i}$$

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

Numerator: $(2 + i)(3 + 2i) = 6 + 4i + 3i + 2i^2 = 6 + 7i - 2 = 4 + 7i$

Result: $\frac{4 + 7i}{13} = \frac{4}{13} + \frac{7}{13}i$

Problem 7: Solve $x^2 - 2x + 5 = 0$.

Show Answer

Using the quadratic formula with $a = 1$, $b = -2$, $c = 5$:

Discriminant: $(-2)^2 - 4(1)(5) = 4 - 20 = -16$

$$x = \frac{2 \pm \sqrt{-16}}{2} = \frac{2 \pm 4i}{2} = 1 \pm 2i$$

The solutions are $x = 1 + 2i$ and $x = 1 - 2i$.

Summary

  • The imaginary unit $i$ is defined by $i = \sqrt{-1}$, which means $i^2 = -1$. This single definition opens up a whole new number system.

  • A complex number has the form $a + bi$, where $a$ is the real part and $b$ is the imaginary part. Every real number is a complex number with imaginary part zero.

  • Powers of $i$ cycle through four values: $i, -1, -i, 1$. To find $i^n$, divide $n$ by 4 and use the remainder.

  • Adding and subtracting complex numbers works like combining like terms: add real parts together and imaginary parts together.

  • Multiplying complex numbers uses the distributive property, with the key step being to replace $i^2$ with $-1$.

  • The complex conjugate of $a + bi$ is $a - bi$. Multiplying a number by its conjugate gives a real number: $(a + bi)(a - bi) = a^2 + b^2$.

  • To divide complex numbers, multiply the numerator and denominator by the conjugate of the denominator. This makes the denominator real.

  • Quadratics with negative discriminants have complex solutions that come in conjugate pairs. The quadratic formula works for all quadratics - you just need to know how to handle $\sqrt{\text{negative}}$.

  • The complex plane lets you visualize complex numbers as points, with the real part on the horizontal axis and the imaginary part on the vertical axis.

  • Complex numbers are not just mathematical curiosities - they are essential tools in electrical engineering, signal processing, quantum physics, and many other fields.

You have now expanded your number system beyond the real numbers. Where once $x^2 = -1$ had no solution, now it has two: $x = i$ and $x = -i$. This might feel like a mathematical trick, but these “imaginary” numbers turn out to be remarkably real in their applications. Every quadratic equation now has solutions, and you have the tools to find them.