Solving imaginary numbers comes down to one core idea: the imaginary unit i equals the square root of -1, which means i² = -1. Every operation you’ll perform with imaginary numbers builds on that single fact. Once you internalize it, simplifying expressions, doing arithmetic, and solving equations with complex solutions all follow a clear set of rules.
Simplifying Square Roots of Negative Numbers
Whenever you see a square root of a negative number, your first step is to separate out the -1 and rewrite it as i. The process looks like this: split √(-a) into √(a) · √(-1), then replace √(-1) with i.
For example, √(-25) becomes √(25) · √(-1), which simplifies to 5i. Similarly, √(-48) becomes √(48) · i, and since √(48) simplifies to 4√3, the final answer is 4√3 · i. Always simplify the radical first, then attach the i.
Adding and Subtracting Complex Numbers
A complex number has the form a + bi, where a is the real part and bi is the imaginary part. To add or subtract two complex numbers, you combine the real parts with each other and the imaginary parts with each other, just like combining like terms in algebra.
Addition: (a + bi) + (c + di) = (a + c) + (b + d)i
Subtraction: (a + bi) – (c + di) = (a – c) + (b – d)i
So (3 + 4i) + (2 + 7i) = 5 + 11i. And (6 + 2i) – (1 + 5i) = 5 – 3i. There’s no trick here. If you can combine like terms, you can add and subtract complex numbers.
Multiplying Complex Numbers
Multiplication works the same way as multiplying two binomials (the FOIL method). Multiply each term in the first complex number by each term in the second, then combine. The key step is replacing i² with -1 whenever it appears.
Take (2 + 3i) · (1 + 5i). Using FOIL:
- First: 2 · 1 = 2
- Outer: 2 · 5i = 10i
- Inner: 3i · 1 = 3i
- Last: 3i · 5i = 15i²
Combine everything: 2 + 10i + 3i + 15i². Replace i² with -1: 2 + 13i + 15(-1) = 2 + 13i – 15 = -13 + 13i.
The general formula is (a + bi) · (c + di) = (ac – bd) + (ad + bc)i, but you don’t need to memorize it. Just FOIL and replace i² with -1.
Dividing Complex Numbers
Division is the one operation that requires an extra step. You can’t leave an imaginary number in the denominator, so you need to eliminate it. The tool for this is the complex conjugate: the conjugate of a + bi is a – bi. You simply flip the sign of the imaginary part.
To divide, multiply both the numerator and denominator by the conjugate of the denominator. This works because when you multiply a complex number by its conjugate, the imaginary parts cancel out and you’re left with a real number on the bottom.
For example, to compute (2 + 10i) ÷ (3 + 10i):
Multiply top and bottom by (3 – 10i). In the denominator, (3 + 10i)(3 – 10i) = 9 – 100i² = 9 + 100 = 109. That’s now a plain real number. Expand the numerator using FOIL, simplify, and write the result in a + bi form by dividing both the real and imaginary parts by 109.
The Repeating Cycle of Powers of i
The powers of i follow a four-step cycle that repeats forever:
- i¹ = i
- i² = -1
- i³ = –i
- i⁴ = 1
Then it starts over: i⁵ = i, i⁶ = -1, and so on. To simplify any large power of i, divide the exponent by 4 and look at the remainder. The remainder tells you which value in the cycle to use: remainder 0 gives 1, remainder 1 gives i, remainder 2 gives -1, remainder 3 gives –i.
For i²⁵, divide 25 by 4 to get a remainder of 1. So i²⁵ = i¹ = i. For i⁷³, 73 ÷ 4 leaves a remainder of 1, so i⁷³ = i. For i⁵⁰, 50 ÷ 4 leaves a remainder of 2, so i⁵⁰ = i² = -1. This trick works for any exponent, no matter how large.
Solving Quadratic Equations With Complex Roots
Imaginary numbers most commonly show up when you use the quadratic formula and get a negative number under the square root. That square root portion, b² – 4ac, is called the discriminant. When it’s negative, the equation has no real solutions, but it does have two complex solutions.
Take x² + 2x + 5 = 0. Plugging into the quadratic formula: x = (-2 ± √(4 – 20)) / 2 = (-2 ± √(-16)) / 2. Rewrite √(-16) as 4i, giving x = (-2 ± 4i) / 2 = -1 ± 2i. The two solutions are -1 + 2i and -1 – 2i.
Notice the solutions always come in conjugate pairs. If a + bi is a solution, then a – bi is the other. This is true for every quadratic with real coefficients and a negative discriminant.
Plotting Complex Numbers on a Plane
Complex numbers can be graphed on the complex plane, which looks like a standard coordinate grid with one change: the horizontal axis represents real numbers and the vertical axis represents imaginary numbers. The complex number 3 + 4i is plotted at the point (3, 4), while 3 – 4i sits at (3, -4). The number 5i with no real part lands at (0, 5), right on the imaginary axis.
Visualizing complex numbers this way helps in more advanced topics. The distance from the origin to the point gives you the number’s magnitude, and the angle from the real axis gives you its direction. For solving problems in an algebra class, though, knowing the plotting convention is mainly useful for checking that your conjugate pairs make sense: they should always be mirror images across the real axis.
Where Imaginary Numbers Are Actually Used
Imaginary numbers aren’t just an abstract math exercise. They’re essential in electrical engineering, where circuits powered by alternating current are analyzed using complex numbers. Each component in a circuit (resistors, capacitors, inductors) is described as a complex impedance, and engineers solve systems of complex linear equations to predict how current flows through the circuit. The imaginary portions capture the way capacitors and inductors shift the timing of electrical signals, something real numbers alone can’t describe. The same math shows up in signal processing, quantum physics, and control systems. So while “imaginary” is an unfortunate name, the numbers themselves solve very real problems.