Complex Numbers and Operations

🌀 Complex Numbers and Operations

Part 1 of 7 — The Imaginary Unit & Square Roots of Negatives

Topics in This Part

🔑 Key Concept: There is no real number whose square is negative. Mathematicians fixed this by inventing one — the imaginary unit $i$, defined so that $i^2 = -1$. That single idea unlocks the complex numbers, and once you master it, equations that "had no solution" suddenly do.

Why We Need a New Number

Try to solve $x^2 = -1$. No real number works: a positive squared is positive, a negative squared is positive, and $0^2 = 0$. So we define a brand-new number to plug the hole:

$i = \sqrt{-1}, \qquad\text{which means}\qquad i^2 = -1$

With $i$ in hand, we can take the square root of any negative number:

$\sqrt{-9} = \sqrt{9}\cdot\sqrt{-1} = 3i, \qquad \sqrt{-7} = \sqrt{7}\,i$

The General Rule

$\sqrt{-a} = i\sqrt{a} \quad (a > 0)$

⚠️ Pull out the $i$ first. The rule $\sqrt{a}\cdot\sqrt{b} = \sqrt{ab}$ fails for negatives. For example $\sqrt{-4}\cdot\sqrt{-9} = (2i)(3i) = 6i^2 = -6$, not $\sqrt{36} = 6$. Always convert to $i$ before multiplying.

Concept Check 🎯

Small Powers of $i$ and Complex Zeros

Because $i^2 = -1$, the next powers follow by multiplying by $i$ once more:

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

These small powers are all you need to test whether a complex number is a zero of a polynomial.

Example 1: Is $3i$ a zero of $p(x) = x^2 + 9$?

$p(3i) = (3i)^2 + 9 = 9i^2 + 9 = -9 + 9 = 0$

Yes. Its conjugate $-3i$ is a zero too.

Example 2: Is $i$ a zero of $q(x) = x^3 + x$?

$q(i) = i^3 + i = -i + i = 0$

Yes. So $q$ has the zeros $0$, $i$ and $-i$: three complex zeros for a degree-3 polynomial.

Square Roots of Negatives Inside a Formula

The quadratic formula often produces expressions such as $\dfrac{6 \pm \sqrt{-16}}{2}$. Convert the root to $i$ form first, then divide both terms:

$\frac{6 \pm \sqrt{-16}}{2} = \frac{6 \pm 4i}{2} = 3 \pm 2i$

⚠️ Square the whole term. $(3i)^2 = 3^2\cdot i^2 = 9(-1) = -9$, not $9i$ and not $3i^2$.

Powers and Roots 🧮

1) $(3i)^2 = \,?$ (enter an integer) 2) Write $\dfrac{4 + \sqrt{-36}}{2}$ as $a + bi$. Enter $b$. 3) For $q(x) = x^3 + x$, what is $q(2i)$? (enter in the form -5i)

Match Each Result 🔽

What You Can Now Do

You can simplify the square root of any negative number, use $i^2 = -1$ and $i^3 = -i$, and check whether a complex number is a zero of a polynomial. These are the building blocks for complex zeros.

In Part 2, we combine real and imaginary pieces into a single object — a complex number $a + bi$ — and learn to add, subtract, and multiply them.

🌀 Complex Numbers and Operations

Part 2 of 7 — Adding, Subtracting & Multiplying

🔑 A complex number has the form $a + bi$, where $a$ is the real part and $b$ is the imaginary part. Think of it as two numbers riding together: the real $a$ and the imaginary $b$. Operations keep the two parts separate — until $i^2$ collapses an imaginary product back into a real number.

Addition & Subtraction

Combine like parts: real with real, imaginary with imaginary. It works exactly like combining like terms in algebra (treat $i$ as the "variable").

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

Worked Example

$(3 + 5i) + (4 - 2i) = (3 + 4) + (5 - 2)i = 7 + 3i$

For subtraction, distribute the minus sign to both parts:

$(3 + 5i) - (4 - 2i) = (3 - 4) + (5 - (-2))i = -1 + 7i$

⚠️ Watch the sign on the imaginary part. Subtracting $-2i$ adds $2i$, giving $5i + 2i = 7i$, not $3i$.

Add & Subtract 🧮

For each result $a + bi$, enter the real part $a$ in the first box and the imaginary part $b$ in the second.

1) $(6 + 2i) + (1 + 9i)$ → $a = \,?,\ b = \,?$ 2) $(5 - 3i) - (2 + 4i)$ → $a = \,?,\ b = \,?$

Multiplication: FOIL, Then Replace $i^2$

Multiply two complex numbers just like two binomials (FOIL), then use $i^2 = -1$ to simplify.

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

$\underbrace{2\cdot 4}_{8} + \underbrace{2\cdot 5i}_{10i} + \underbrace{3i\cdot 4}_{12i} + \underbrace{3i\cdot 5i}_{15i^2}$

Combine the imaginary terms and replace $i^2 = -1$:

$= 8 + 22i + 15i^2 = 8 + 22i + 15(-1) = 8 + 22i - 15 = -7 + 22i$

💡 The key step is $i^2 = -1$. The last product $15i^2$ becomes the real number $-15$, which merges with the $8$.

Concept Check 🎯

Multiply It Out 🧮

Write each product as $a + bi$. Enter $a$, then $b$.

1) $(1 + i)(1 - i)$ → $a = \,?,\ b = \,?$ 2) $(2 + 5i)(3 - i)$ → $a = \,?,\ b = \,?$

🌀 Complex Numbers and Operations

Part 3 of 7 — Conjugates & Conjugate Zeros

🔑 The conjugate of $a + bi$ is $a - bi$ — same real part, flipped sign on the imaginary part. Multiplying a complex number by its conjugate always produces a real number. For a polynomial with real coefficients, non-real zeros come in conjugate pairs, and this product is what lets each pair combine into a quadratic factor with real coefficients.

The Complex Conjugate

The conjugate of $z = a + bi$ is written $\bar{z} = a - bi$.

The Magic Property

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

The middle terms cancel and $i^2 = -1$ turns the subtraction into addition. The answer $a^2 + b^2$ is real and non-negative — no $i$ left.

Worked Example

$(3 + 4i)(3 - 4i) = 3^2 + 4^2 = 9 + 16 = 25$

💡 This is a difference of squares in disguise: $(a+bi)(a-bi) = a^2 + b^2$. Memorize it: it turns a conjugate pair of zeros into a quadratic factor with real coefficients.

Concept Check 🎯

Conjugate Pairs of Zeros

If a polynomial has real coefficients and $a + bi$ (with $b \neq 0$) is a zero, then its conjugate $a - bi$ is also a zero. Multiplying the two linear factors gives a quadratic with real coefficients:

$\big(x - (a + bi)\big)\big(x - (a - bi)\big) = x^2 - 2a\,x + (a^2 + b^2)$

Worked Example 1: Zeros $2 + i$ and $2 - i$

Sum $= 4$ and product $= 2^2 + 1^2 = 5$, so the factor is $x^2 - 4x + 5$.

Worked Example 2: Finishing a Cubic

$p(x) = x^3 - 5x^2 + 11x - 15$ has real coefficients and $1 + 2i$ as a zero. Then $1 - 2i$ is also a zero, with sum $2$ and product $1^2 + 2^2 = 5$, so $x^2 - 2x + 5$ is a factor. Long division gives

$p(x) = (x^2 - 2x + 5)(x - 3)$

so the zeros are $1 + 2i$, $1 - 2i$ and $3$.

🔑 A polynomial of degree $n$ has exactly $n$ complex zeros, counted with multiplicity. Non-real zeros pair off, so a cubic with real coefficients always has at least one real zero.

Build the Quadratic Factor 🔽

A polynomial with real coefficients has $3 - 2i$ as a zero. Choose what follows at each stage.

Conjugate Pairs 🧮

1) The zeros $-1 + 3i$ and $-1 - 3i$ give the factor $x^2 + bx + c$. Enter $b$, then $c$. 2) A cubic with real coefficients has zeros $4i$ and $5$. What is the product of its two non-real zeros?

🌀 Complex Numbers and Operations

Part 4 of 7 — The Complex Plane & Modulus

🔑 Every complex number is a point. Plot $a + bi$ at coordinates $(a, b)$ in the complex plane: the horizontal axis is real, the vertical axis is imaginary. Its distance from the origin is the modulus $\lvert z\rvert$.

Plotting in the Complex Plane

The complex plane (also called the Argand diagram) replaces the usual $x$- and $y$-axes:

• Real axis (horizontal) ← the real part $a$
• Imaginary axis (vertical) ← the imaginary part $b$

So $z = a + bi$ lives at the point $(a, b)$.

💡 A real number sits on the horizontal axis ($b = 0$); a pure imaginary number sits on the vertical axis ($a = 0$).

Locate Each Number 🔽

The Modulus (Absolute Value)

The modulus $\lvert z\rvert$ is the distance from the origin to the point $(a, b)$ — straight from the Pythagorean theorem:

$\lvert a + bi\rvert = \sqrt{a^2 + b^2}$

Worked Example: $z = 3 + 4i$

$\lvert z\rvert = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5$

Worked Example: $z = -5 + 12i$

$\lvert z\rvert = \sqrt{(-5)^2 + 12^2} = \sqrt{25 + 144} = \sqrt{169} = 13$

⚠️ Square first, so signs disappear. $(-5)^2 = 25$, not $-25$. The modulus is always $\ge 0$.

Concept Check 🎯

Compute the Modulus 🧮

Each of these gives a whole-number answer. Enter the modulus.

1) $\lvert 9 + 12i\rvert = \,?$ 2) $\lvert -7 + 24i\rvert = \,?$ 3) $\lvert 6 - 8i\rvert = \,?$

🌀 Complex Numbers and Operations

Part 5 of 7 — Polar (Trigonometric) Form

🔑 A second address for every complex number. Instead of "go right $a$, up $b$," polar form says "go out a distance $r$ at an angle $\theta$." This is the same polar-coordinate idea you use for points in the plane (CED Topic 3.13).

Rectangular vs. Polar Form

A complex number $z = a + bi$ can be described by its distance $r$ from the origin and the angle $\theta$ it makes with the positive real axis:

$r = \lvert z\rvert = \sqrt{a^2 + b^2}, \qquad a = r\cos\theta, \qquad b = r\sin\theta$

Substituting gives the trigonometric (polar) form:

$z = r(\cos\theta + i\sin\theta)$

Here $r$ is the modulus and $\theta$ is the argument (the angle).

Finding $\theta$

$\tan\theta = \frac{b}{a}$

…but you must check the quadrant of $(a, b)$ to pick the correct angle, because $\tan$ repeats every $\pi$ radians.

💡 The reference angle from $\tan^{-1}\left\lvert\frac{b}{a}\right\rvert$ tells you the size; the quadrant tells you the true direction.

Worked Example: Convert $z = 1 + i\sqrt{3}$ to Polar Form

Step 1 — modulus: $r = \sqrt{1^2 + (\sqrt{3})^2} = \sqrt{1 + 3} = \sqrt{4} = 2$

Step 2 — argument. The point $(1, \sqrt{3})$ is in Quadrant I: $\tan\theta = \frac{\sqrt{3}}{1} = \sqrt{3} \;\Rightarrow\; \theta = \frac{\pi}{3}$

Step 3 — write it: $z = 2\left(\cos \frac{\pi}{3} + i\sin \frac{\pi}{3}\right)$

✅ Check by going back: $2\cos \frac{\pi}{3} = 2\cdot\tfrac{1}{2} = 1$ and $2\sin \frac{\pi}{3} = 2\cdot\tfrac{\sqrt{3}}{2} = \sqrt{3}$ ✓

Concept Check 🎯

Reminder: $\cos \frac{\pi}{2} = 0,\ \sin \frac{\pi}{2} = 1,\ \cos \pi = -1,\ \sin \pi = 0$, and $\cos \frac{\pi}{4} = \sin \frac{\pi}{4} = \dfrac{\sqrt{2}}{2}$.

Polar Form in Radians (How the AP Exam Writes It)

AP Precalculus measures angles in radians. A complex number $a + bi$ sits at the point $(a, b)$ in the complex plane, and its polar coordinates $(r, \theta)$ give

$a + bi = (r\cos\theta) + i\,(r\sin\theta)$

To find $\theta$ from $(a, b)$, use the arctangent and fix the quadrant:

Example: $z = -1 + i$

• $r = \sqrt{(-1)^2 + 1^2} = \sqrt{2}$
• $a < 0$, so $\theta = \arctan\left(\dfrac{1}{-1}\right) + \pi = -\dfrac{\pi}{4} + \pi = \dfrac{3\pi}{4}$

$-1 + i = \sqrt{2}\cos\tfrac{3\pi}{4} + i\,\sqrt{2}\sin\tfrac{3\pi}{4}$

⚠️ Using $\arctan(b/a)$ alone gives $-\frac{\pi}{4}$, which points to $1 - i$ in Quadrant IV. Always check which quadrant the point is in. Multiplying or raising complex numbers to powers in polar form is not assessed on the AP Precalculus exam.

Polar Coordinates of Complex Numbers 🔽

Polar Form Practice 🧮

1) The modulus $r$ of $z = -4 + 3i$ is $\,?$ (whole number) 2) For $z = 3(\cos\pi + i\sin\pi)$, the real part $a$ is $\,?$ 3) The number $z = -5i$ has angle $\theta = \dfrac{k\pi}{2}$ in $[0, 2\pi)$. Enter $k$: $\,?$

🌀 Complex Numbers and Operations

Part 6 of 7 — Complex Roots of Quadratics

🔑 Where complex numbers earn their keep. When the discriminant $b^2 - 4ac$ is negative, a quadratic has no real solutions — but it has two complex solutions, and they always come as a conjugate pair.

The Discriminant Tells You What to Expect

For $ax^2 + bx + c = 0$, the quadratic formula is

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

The discriminant $\Delta = b^2 - 4ac$ decides the type of roots:

💡 When $\Delta < 0$, the $\sqrt{\;}$ contains a negative number — and that is precisely where $i$ steps in: $\sqrt{-k} = i\sqrt{k}$.

Concept Check 🎯

Worked Example: Solve $x^2 + 4x + 13 = 0$

Step 1 — identify $a = 1,\ b = 4,\ c = 13$.

Step 2 — discriminant: $\Delta = 4^2 - 4(1)(13) = 16 - 52 = -36$

Negative, so expect a complex conjugate pair.

Step 3 — quadratic formula: $x = \frac{-4 \pm \sqrt{-36}}{2(1)} = \frac{-4 \pm 6i}{2}$

Step 4 — simplify by dividing each term by $2$: $x = -2 \pm 3i$

So the roots are $-2 + 3i$ and $-2 - 3i$ — conjugates, as promised.

⚠️ Divide BOTH terms by $2a$. Writing $\frac{-4 \pm 6i}{2}$ as $-4 \pm 3i$ is wrong; the $-4$ must also be divided, giving $-2 \pm 3i$.

Solve the Quadratic 🧮

Solve $x^2 - 6x + 25 = 0$. The roots are $p \pm qi$.

1) Discriminant $b^2 - 4ac = \,?$ 2) Real part $p = \,?$ 3) Imaginary coefficient $q = \,?$ (the positive value)

Build the Equation 🎯

🌀 Complex Numbers and Operations

Part 7 of 7 — Mixed Mastery & Exit Quiz

You can now (1) simplify square roots of negatives and use $i^2 = -1$, (2) add, subtract, and multiply, (3) use conjugates to pair complex zeros, (4) find the modulus, plot in the complex plane and convert to polar form, and (5) find complex roots of quadratics. Let's tie it all together.

Quick Reference

⚠️ Top mistakes to avoid: dropping the $i$ from a negative square root; forgetting $i^2 = -1$ in a product; and not dividing both terms by $2a$ in the quadratic formula.

Mixed Practice 🔽

Mixed Computation 🧮

1) Write $(2 + i)(3 + 4i)$ as $a + bi$: $a = \,?,\ b = \,?$ 2) The zeros $3 + i$ and $3 - i$ give the factor $x^2 + bx + c$: $b = \,?,\ c = \,?$

Exit Quiz ✅

Answer all three to finish the lesson.