Real and Complex Numbers

Real and Complex Numbers

This chapter covers two ideas that show up on the ACT as quick, high-accuracy questions: deciding whether a number is rational or irrational, and doing arithmetic with the imaginary unit $i$. Both reward a clean method far more than a clever trick.

Rational and irrational numbers

A rational number can be written as a ratio of two integers. That includes every integer, every terminating or repeating decimal, and every fraction, even a fraction like $\frac{22}{7}$ that is commonly used to approximate an irrational number. An irrational number cannot be written as a ratio of integers; its decimal expansion never terminates and never repeats.

The key habit: a number's rationality depends on what it equals, not on how familiar or unfamiliar it looks. $\sqrt{16}$ has a radical sign, but it equals $4$, so it is rational. $\sqrt{7}$ is not a perfect square, so it is irrational. $\sqrt{0.25}$ equals $0.5 = \frac{1}{2}$, so it is rational. Do not assume that every square root is irrational; check whether the radicand is a perfect square first.

The imaginary unit and its powers

The imaginary unit is defined by $i^2 = -1$. Powers of $i$ cycle in a pattern of four:

After $i^4 = 1$ the pattern repeats. To reduce a high power, divide the exponent by $4$ and keep only the remainder, because $i$ to any multiple of $4$ equals $1$.

Worked Example. Simplify $i^{29}$.

Divide the exponent by $4$ and use the remainder: $29 = 4(7) + 1$, so $i^{29} = (i^4)^7 \cdot i^1 = 1 \cdot i = i$.

When two powers of $i$ are multiplied, add the exponents first, then reduce: $i^7 \cdot i^{13} = i^{20} = 1$. A remainder of $0$ means the answer is $1$, not $0$.

Complex arithmetic

A complex number has the form $a + bi$, where $a$ and $b$ are real numbers.

• Add or subtract by combining the real parts together and the imaginary parts together, separately. When subtracting, distribute the minus sign to both parts of the second number: $(3 + 4i) - (1 - 2i) = 2 + 6i$.
• Multiply with ordinary distribution (FOIL), then replace every $i^2$ with $-1$. For example $(5 + 3i)(2 - 4i) = 10 - 20i + 6i - 12i^2 = 10 - 14i + 12 = 22 - 14i$.
• Divide by multiplying the numerator and denominator by the conjugate of the denominator. The conjugate of $a + bi$ is $a - bi$, and the product $(a + bi)(a - bi) = a^2 + b^2$ is always a real number. So $(12 - 6i) \div (3 + i)$ becomes $\frac{(12-6i)(3-i)}{(3+i)(3-i)} = \frac{30 - 30i}{10} = 3 - 3i$.

When the denominator is a pure imaginary number such as $2i$, multiply the top and bottom by $-i$ (the conjugate of $2i$ is $-2i$, which is the same idea): $\frac{6 + 2i}{2i} \cdot \frac{-i}{-i} = \frac{-6i - 2i^2}{-2i^2} = \frac{2 - 6i}{2} = 1 - 3i$.

A quick pattern worth memorizing: $(a + bi)(a - bi) = a^2 + b^2$. So $(1 + i)(1 - i) = 1 - i^2 = 2$ and $(2 - i)(2 + i) = 5$. The answer is always a real number.

Complex solutions and the discriminant

A quadratic equation has complex (non-real) solutions exactly when its discriminant $b^2 - 4ac$ is negative. This links complex numbers to the algebra in the next Part: a question can ask for the range of a parameter that forces complex solutions, which reduces to solving an inequality built from the discriminant.

Worked Example. For what values of $c$ does $x^2 + 6x + c = 0$ have complex solutions?

Complex solutions require a negative discriminant: $36 - 4c < 0$, so $c > 9$.

Watch the direction of the inequality when you divide by a negative number, and remember that when the discriminant is exactly zero the equation has one repeated real solution, so the boundary value is not included.

Where this appears as a model

A quadratic model can predict no real outcome. If a launched object's height is modeled by $-16t^2 + 20t + 5$, solving for when it reaches $50$ feet gives a negative discriminant of $-2480$. The complex result confirms the object never actually reaches that height.

Common traps

• Subtracting complex numbers: forgetting to flip the sign of the imaginary part of the second number, so $-(1 - 2i)$ becomes $-1 - 2i$ instead of $-1 + 2i$.
• Stopping at $i^2$: an answer such as $4 - i^2$ is not finished until it becomes $4 + 1 = 5$.
• Reducing $i^{n}$ with the wrong remainder: always divide by $4$, not by $2$, and remember that a remainder of $0$ gives $1$.
• Treating the discriminant condition as an equation: "complex solutions" means an inequality, so the answer is a range of values, not a single number.

Strategy summary

• Rational or irrational? Simplify first, then decide.
• Powers of $i$: divide the exponent by $4$ and use the remainder ($0 \to 1$, $1 \to i$, $2 \to -1$, $3 \to -i$).
• Always replace $i^2$ with $-1$ before writing a final answer, and write it in $a + bi$ form.
• Complex roots: set the discriminant less than zero.