Number Systems
Understanding the Imaginary Unit i
The imaginary unit i is defined by i^2=-1. It extends real arithmetic so equations such as x^2+1=0 have solutions.
Reviewed
Powers of i repeat every four steps
Reduce a large exponent modulo 4 instead of multiplying i repeatedly.
| Exponent form | Remainder mod 4 | Value |
|---|---|---|
| i^(4k) | 0 | 1 |
| i^(4k+1) | 1 | i |
| i^(4k+2) | 2 | -1 |
| i^(4k+3) | 3 | -i |
Why the real numbers are not enough
Every real square is nonnegative, so no real number solves x^2=-1. Introducing i provides a consistent solution and creates the complex number system a+bi.
This extension preserves ordinary addition, multiplication, and distributive laws.
i^2=-1; sqrt(-a)=i sqrt(a) for a>0 under the principal-root conventionWhy introducing i is a consistent extension
No real number solves x^2 + 1 = 0 because every real square is nonnegative. The new symbol i is defined by i^2 = -1, then ordinary algebraic rules determine its consequences.
i^2 = -1This is the defining relation.(a + bi) + (c + di) = (a+c) + (b+d)iAddition remains component by component.(a + bi)(c + di) = (ac-bd) + (ad+bc)iExpand normally and replace i^2 by -1.(a+bi)(a-bi) = a^2 + b^2Conjugates produce a real, nonnegative product.
The extension preserves the familiar arithmetic of real numbers while making every quadratic polynomial solvable over a sufficiently broad number system.
Worked example 1
Simplify i^37
Divide the exponent by 4 and use the remainder.
37=4 x 9 + 1i^37=(i^4)^9 ii^4=1
Result: i^37=i
The four-step power cycle
Successive powers repeat every four exponents: i, -1, -i, 1. Reduce a large positive integer exponent modulo 4 to find its value quickly.
An exponent divisible by 4 gives 1. Remainders 1, 2, and 3 give i, -1, and -i respectively.
i^(n+4)=i^nDo not split every square root into factors
Incorrect chain: sqrt((-1)(-1)) = sqrt(-1)sqrt(-1) = i^2 = -1
The left side is sqrt(1) = 1.
The real-number product rule for square roots does not extend unchanged to negative or general complex factors.
Square roots and branch choice
The equation z^2=-9 has two solutions, 3i and -3i. The principal square-root symbol sqrt(-9) conventionally returns 3i.
As with positive square roots, distinguish a principal function value from all solutions of an equation.
Worked example 2
Solve x^2+16=0
Isolate the square.
x^2=-16x=+/-sqrt(-16)sqrt(-16)=4i
Result: x=+/-4i