Classify numbers within the real number system, work with absolute value, and add, subtract, multiply and simplify complex numbers using $i^{2} = -1$ (Number and Quantity).

What this topic is asking

The ACT expects you to classify numbers within the real system and to do basic arithmetic with complex numbers, the numbers that appear when a square root of a negative is involved. Most ACT complex-number questions are straightforward arithmetic with $i$, so a few rules cover almost everything you will see.

Classifying real numbers

The real numbers nest inside one another.

A quick test: a square root of a non-perfect-square integer (like $\sqrt{7}$) is irrational, while a square root of a perfect square (like $\sqrt{9} = 3$) is rational.

Absolute value

The absolute value $|x|$ is the distance of $x$ from $0$ on the number line, so it is never negative: $|5| = 5$ and $|-5| = 5$. On the ACT it appears in expressions ($|3 - 8| = |-5| = 5$) and occasionally in simple equations, where $|x| = 7$ has two solutions, $x = 7$ and $x = -7$, because two points are distance 7 from zero.

Complex number arithmetic

A complex number is $a + bi$, with real part $a$ and imaginary part $b$. The one fact that drives everything is $i^{2} = -1$.

The cycle of powers of $i$

The powers of $i$ repeat with period four:

• $i^{1} = i$
• $i^{2} = -1$
• $i^{3} = i^{2} \cdot i = -i$
• $i^{4} = (i^{2})^{2} = 1$

After $i^{4} = 1$ the pattern restarts, so a high power like $i^{10}$ is found by dividing the exponent by 4 and using the remainder: $10 = 4(2) + 2$, so $i^{10} = i^{2} = -1$. This remainder trick handles any power of $i$ without long multiplication.

Simplifying square roots of negatives

Before any arithmetic, rewrite a square root of a negative using $i$. The rule is $\sqrt{-n} = i\sqrt{n}$ for positive $n$, so $\sqrt{-16} = i\sqrt{16} = 4i$ and $\sqrt{-18} = i\sqrt{18} = 3i\sqrt{2}$. Pulling the $i$ out first, then simplifying the remaining radical, avoids the common slip of treating $\sqrt{-16}$ as a real number. Once every radical of a negative is written with $i$, the rest is ordinary complex arithmetic.

The complex conjugate

The conjugate of $a + bi$ is $a - bi$: the same number with the sign of the imaginary part flipped. Conjugates matter because multiplying a complex number by its conjugate gives a real result: $(a + bi)(a - bi) = a^{2} - (bi)^{2} = a^{2} + b^{2}$, since $-(bi)^{2} = -b^{2}i^{2} = b^{2}$. This is exactly the tool for dividing complex numbers: to simplify $\frac{1}{2 + i}$, multiply top and bottom by the conjugate $2 - i$ to clear the imaginary part from the denominator, giving $\frac{2 - i}{(2)^{2} + (1)^{2}} = \frac{2 - i}{5}$. The conjugate is also why complex solutions of a real quadratic always come in pairs $a \pm bi$.

Why complex numbers appear at all

Complex numbers arise when a quadratic has a negative discriminant: the formula produces a square root of a negative number, which has no real value but a perfectly good complex one. For example $x^{2} + 4 = 0$ gives $x = \pm\sqrt{-4} = \pm 2i$, a conjugate pair. The ACT mostly tests the arithmetic of $a + bi$ rather than deep theory, so being fluent with adding, subtracting and multiplying (always converting $i^{2}$ to $-1$), rewriting $\sqrt{-n}$ as $i\sqrt{n}$, and knowing the power cycle will cover the great majority of complex-number questions you meet.

Try this

Q1. Simplify $(5 - 2i) - (3 + i)$. [1 point]

• Cue. Real: $5 - 3 = 2$. Imaginary: $-2i - i = -3i$. Answer $2 - 3i$.

Q2. What is $i^{7}$? [1 point]

• Cue. $7 = 4(1) + 3$, so $i^{7} = i^{3} = -i$.