Ever wondered why engineers talk about "imaginary" numbers like they’re as real as the coffee in your mug?
💡 In Simple Words: A complex number is just a pair of ordinary numbers glued together – one tells you how far to move left‑right, the other tells you how far to move up‑down on a special graph. Together they let us solve equations that ordinary numbers can’t.
What is a Complex Number?
A complex number is a number that has two parts: a real part (the ordinary number you already know) and an imaginary part (a multiple of the symbol i, where i² = –1). Think of it like a two‑digit locker code – the first digit (real) tells you which row, the second digit (imaginary) tells you which column.
Standard Form: a + bi
We write a complex number as a + bi. Here a is the real part, b is the coefficient of the imaginary part, and i is the magic unit that squares to –1.
Seeing It on the Argand Plane
Imagine a city map where the east‑west street is the real axis and the north‑south street is the imaginary axis. A complex number a + bi is a point you reach by walking a blocks east (or west if negative) and then b blocks north (or south if negative). This picture is called the Argand plane.
Basic Operations with Complex Numbers
Addition & Subtraction
Just like adding two directions, you add the real parts together and the imaginary parts together.
Example: (3 + 2i) + (‑1 + 5i) = (3‑1) + (2+5)i = 2 + 7i.
Subtraction works the same way: (3 + 2i) – (‑1 + 5i) = (3+1) + (2‑5)i = 4 – 3i.
Multiplication
Multiply like ordinary binomials, but remember that i² = –1.
Example: (2 + 3i)(1 – 4i) = 2·1 + 2·(‑4i) + 3i·1 + 3i·(‑4i) = 2 – 8i + 3i –12i² = 2 –5i –12(‑1) = 14 –5i.
Division
To divide, we get rid of the imaginary part in the denominator by multiplying numerator and denominator by the conjugate (same real part, opposite sign on the imaginary part).
Example: \(\frac{3+2i}{1‑i}\). Conjugate of denominator is (1+ i).
\(\frac{3+2i}{1‑i}·\frac{1+i}{1+i}=\frac{(3+2i)(1+i)}{1²‑i²}=\frac{3+3i+2i+2i²}{1‑(‑1)}=\frac{3+5i‑2}{2}=\frac{1+5i}{2}=\frac12+\frac{5}{2}i\).
Powers and Roots
Because i² = –1, higher powers of i repeat every four steps:
- i⁰ = 1
- i¹ = i
- i² = –1
- i³ = –i
- i⁴ = 1 again
So i⁷ = i³ = –i, i¹⁰ = i² = –1, and so on.
Finding the n‑th root of a complex number uses the polar form (r(cosθ + i sinθ)). The roots are spaced evenly around a circle – just like slicing a pizza into equal slices.
Quick Reference Table
| Operation | Rule | Example |
|---|---|---|
| Addition | (a+bi)+(c+di) = (a+c)+(b+d)i | (3+2i)+(‑1+5i)=2+7i |
| Subtraction | (a+bi)‑(c+di) = (a‑c)+(b‑d)i | (3+2i)‑(‑1+5i)=4‑3i |
| Multiplication | (a+bi)(c+di)= (ac‑bd)+(ad+bc)i | (2+3i)(1‑4i)=14‑5i |
| Division | \(\frac{a+bi}{c+di}=\frac{(a+bi)(c‑di)}{c²+d²}\) | \(\frac{3+2i}{1‑i}=\frac12+\frac{5}{2}i\) |
| Power of i | i⁴k = 1, i⁴k+1 = i, i⁴k+2 = –1, i⁴k+3 = –i | i⁷ = –i |
📝 Likely Exam Questions
- Find the sum and product of (5‑2i) and (‑3+4i).
Answer: Sum = 2+2i, Product = (5·‑3‑(‑2·4)) + (5·4 + (‑2)·‑3)i = (‑15+8)+(20+6)i = ‑7+26i. - Express \(\frac{7‑i}{2+i}\) in a+bi form.
Answer: Multiply by conjugate (2‑i): \(\frac{(7‑i)(2‑i)}{2²+1}=\frac{14‑7i‑2i+ i²}{5}=\frac{13‑9i}{5}=\frac{13}{5}-\frac{9}{5}i\). - If z = 1+ i, compute z³.
Answer: First find z² = (1+i)² = 1+2i+i² = 1+2i‑1 = 2i. Then z³ = z²·z = 2i(1+i)=2i+2i² = 2i‑2 = ‑2+2i. - Write the polar form of the complex number –√3 + i and find its principal argument (angle).
Answer: r = √[(√3)²+1²]=2. θ = atan(1/‑√3)=‑π/6 (or 5π/6 in the principal range 0‑π). So polar form = 2(cos5π/6 + i sin5π/6). - Find all the cube roots of 8.
Answer: Write 8 as 8(cos0 + i sin0). Cube roots have magnitude 8^{1/3}=2 and angles (0+2kπ)/3 for k=0,1,2 → 0, 2π/3, 4π/3. Thus roots: 2, 2(cos2π/3 + i sin2π/3)=‑1+√3 i, and 2(cos4π/3 + i sin4π/3)=‑1‑√3 i.