ACT Math – Complex Numbers

ACT Math – Complex Numbers

I. The Basics

You’ve probably learned in algebra class that you can’t take the square root of a negative number because the result is not a real number. Instead, the result is what’s called an imaginary number.

The square root of negative 1 (\(\sqrt{-1}\)) is represented by the imaginary number i.

Here are some multiples of i that are commonly seen in problems involving complex numbers:

\(i=\sqrt{-1}\) \(\newline\) \(i^2=i\cdot{i} = \sqrt{-1}\cdot\sqrt{-1}=(\sqrt{-1})^2 = -1\) \(\newline\) \(i^3=i\cdot{i}\cdot{i} = \sqrt{-1}\cdot\sqrt{-1}\cdot\sqrt{-1}=-1\cdot\sqrt{-1} = -i\) \(\newline\) \(i^4=i\cdot{i}\cdot{i}\cdot{i} = \sqrt{-1}\cdot\sqrt{-1}\cdot\sqrt{-1}\cdot\sqrt{-1}=-1\cdot{-1} = 1\)

…and so on.

A complex number is a number that has both a real part and an imaginary part, and is often written in the form a+bi, where a and b are real numbers (for example, 3+4i and 5-6i are complex numbers).

II. Applying Your Knowledge

Just like real numbers, complex numbers can be added, subtracted, multiplied, and divided. Here is an example of each.

Example 1 (Addition): \((4+10i)+(2-3i)\)

Combine like terms: \(4+2+10i-3i = \fbox{6+7i}\)

Example 2 (Subtraction): \((4+10i)-(2-3i)\)

Distribute the minus sign through the second term: \(4+10i-2+3i\)

Combine like terms: \(4-2+10i+3i=\fbox{2+13i}\)

Example 3 (Multiplication): \((4+10i)(2-3i)\)

Multiply the terms: \(8-12i+20i-30i^2\)

Remember \(i^2=-1: 8-12i+20i-30(-1)=8-12i+20i+30\)

Combine like terms: \(8+30-12i+20i=\fbox{38+8i}\)

Example 4 (Division): \(\mbox{$\frac {4+10i}{2-3i}$}\)

Multiply the numerator and denominator by the complex conjugate of the denominator: 2+3i. The complex conjugate is the same complex number but with the opposite sign in front of the imaginary number (Ex: 2-3i and 2+3i, 5+6i and 5-6i, etc.).

= \(\mbox{$\frac {4+10i}{2-3i}$}\cdot\mbox{$\frac{2+3i}{2+3i}$}\)

Multiply the terms in the numerator and the terms in the denominator:

= \(\mbox{$\frac {8+12i+20i+30i^2}{4+6i-6i-9i^2}$}\)

Simplify and remember \(i^2=-1\):

= \(\mbox{$\frac {8+32i+30(-1)}{4-9(-1)}$}=\mbox{$\frac {8+32i-30}{4+9}$}=\fbox{$\frac {-22+32i}{13}$}\)