8.8 Use the Complex Number System

Complex Numbers

Complex numbers let you solve equations like $x^2 = -4$ that have no solution among real numbers. By introducing the imaginary unit $i$, defined so that $i^2 = -1$, you gain a whole new number system where every polynomial equation has a solution. This section covers how to simplify square roots of negative numbers, perform arithmetic with complex numbers, and work with powers of $i$.

Square Roots of Negative Numbers

An imaginary number is the square root of a negative number, written in the form $bi$, where $b$ is a real number. The key definition that makes this work: $i = \sqrt{-1}$, which means $i^2 = -1$.

To simplify the square root of a negative number:

1. Factor $-1$ out of the radicand.
2. Simplify the square root of the remaining positive number.
3. Write the result with $i$.

Example: Simplify $\sqrt{-16}$

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

Example: Simplify $\sqrt{-50}$

$\sqrt{-50} = \sqrt{-1 \cdot 25 \cdot 2} = i \cdot 5\sqrt{2} = 5i\sqrt{2}$

A common mistake: don't try to apply $\sqrt{a} \cdot \sqrt{b} = \sqrt{ab}$ when both $a$ and $b$ are negative. Always extract $i$ first, then simplify.

Operations with Complex Numbers

A complex number has the form $a + bi$, where $a$ is the real part and $b$ is the imaginary part. Both $a$ and $b$ are real numbers. For example, in $3 + 2i$, the real part is 3 and the imaginary part is 2.

Adding and subtracting: Combine real parts with real parts and imaginary parts with imaginary parts, just like combining like terms.

• $(a + bi) + (c + di) = (a + c) + (b + d)i$
• $(a + bi) - (c + di) = (a - c) + (b - d)i$
• Example: $(3 + 2i) + (4 - 5i) = (3 + 4) + (2 + (-5))i = 7 - 3i$
• Example: $(6 - i) - (2 + 3i) = (6 - 2) + (-1 - 3)i = 4 - 4i$

Multiplying: Use the distributive property (FOIL works for two binomials), then replace $i^2$ with $-1$.

• $(a + bi)(c + di) = ac + adi + bci + bdi^2 = (ac - bd) + (ad + bc)i$

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

1. FOIL: $2 \cdot 4 + 2 \cdot (-i) + 3i \cdot 4 + 3i \cdot (-i) = 8 - 2i + 12i - 3i^2$

2. Replace $i^2$ with $-1$: $8 - 2i + 12i - 3(-1) = 8 + 3 + 10i$

3. Result: $11 + 10i$

Division of Complex Numbers

You can't leave $i$ in a denominator, so you multiply top and bottom by the complex conjugate of the denominator. The complex conjugate of $a + bi$ is $a - bi$. This works because multiplying a complex number by its conjugate always produces a real number:

$(c + di)(c - di) = c^2 + d^2$

Steps to divide complex numbers:

1. Identify the complex conjugate of the denominator.
2. Multiply both numerator and denominator by that conjugate.
3. FOIL the numerator and simplify using $i^2 = -1$.
4. Simplify the denominator (it'll be a real number: $c^2 + d^2$).
5. Write the result in $a + bi$ form.

Example: Simplify $\frac{2 + 3i}{4 - 5i}$

1. Conjugate of $4 - 5i$ is $4 + 5i$.

2. $\frac{(2 + 3i)(4 + 5i)}{(4 - 5i)(4 + 5i)}$

3. Numerator: $8 + 10i + 12i + 15i^2 = 8 + 22i + 15(-1) = -7 + 22i$

4. Denominator: $16 + 25 = 41$

5. Result: $-\frac{7}{41} + \frac{22}{41}i$

Powers of i

Powers of $i$ cycle through four values and then repeat:

After $i^4 = 1$, the pattern starts over: $i^5 = i$, $i^6 = -1$, and so on.

To simplify any power of $i$: Divide the exponent by 4 and use the remainder.

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

Example: Simplify $i^{13}$

Divide 13 by 4: $13 = 4 \cdot 3 + 1$, so the remainder is 1. That means $i^{13} = i$.

Example: Simplify $i^{46}$

Divide 46 by 4: $46 = 4 \cdot 11 + 2$, so the remainder is 2. That means $i^{46} = -1$.