The imaginary unit i, complex number operations, and why they exist
Reserve & Extensions • K-12
For centuries, mathematicians dismissed the square root of negative numbers as impossible. Then they asked: "What if we just invented a number whose square is −1?" That invention -- the imaginary unit i -- unlocked an entire new number system that now powers electrical engineering, quantum physics, and signal processing.
With this definition, we can now take the square root of any negative number:
Examples: √(−9) = 3i, √(−7) = i√7, √(−16) = 4i.
A complex number has the form:
where a is the real part and b is the imaginary part (both are real numbers).
| Complex Number | Real Part (a) | Imaginary Part (b) |
|---|---|---|
| 3 + 2i | 3 | 2 |
| −4i | 0 | −4 |
| 7 | 7 | 0 |
| −1 + i | −1 | 1 |
Notice that every real number is a complex number (with b = 0).
Adding/Subtracting: Combine real parts and imaginary parts separately.
(3 + 4i) + (2 − 7i) = (3 + 2) + (4 − 7)i = 5 − 3i
(6 − i) − (2 + 5i) = (6 − 2) + (−1 − 5)i = 4 − 6i
Multiplying: Use FOIL, then replace i2 with −1.
(3 + 2i)(1 − 4i)
= 3(1) + 3(−4i) + 2i(1) + 2i(−4i)
= 3 − 12i + 2i − 8i2
= 3 − 10i − 8(−1)
= 3 − 10i + 8
= 11 − 10i
The conjugate of a + bi is a − bi. Multiplying a complex number by its conjugate always produces a real number:
x2 + 1 = 0
x2 = −1
x = ±√(−1) = ±i
The solutions are x = i and x = −i. Notice these are conjugates of each other. Complex solutions to polynomial equations with real coefficients always come in conjugate pairs.
Never write √(−4) · √(−9) = √36) = 6. The rule √a · √b = √(ab) only works when a and b are non-negative. Instead: √(−4) · √(−9) = 2i · 3i = 6i2 = −6.
i1 = i, i2 = −1, i3 = −i, i4 = 1, i5 = i, ... The pattern repeats every 4. To find in, divide n by 4 and use the remainder: remainder 0 gives 1, remainder 1 gives i, remainder 2 gives −1, remainder 3 gives −i.
1. Simplify: √(−50)
√(−50) = i√50 = i · 5√2 = 5i√2
2. Compute: (5 + 3i) + (−2 + 7i)
(5 − 2) + (3 + 7)i = 3 + 10i
3. Multiply: (2 + i)(2 − i)
This is a conjugate pair: 22 + 12 = 4 + 1 = 5. (A real number, as expected.)
4. Simplify: i23
23 / 4 = 5 remainder 3. So i23 = i3 = −i.
5. Solve: x2 + 9 = 0
x2 = −9, so x = ±√(−9) = ±3i. Solutions: x = 3i and x = −3i.