division by rationalizing

Dividing one complex number by another looks awkward at first — what does it mean to put an i in the denominator? The trick is the same one you use to clear a square root from the bottom of a real fraction: multiply top and bottom by the conjugate of the denominator. Because a denominator times its own conjugate is always a real number (c + i d times c - i d = c^2 + d^2), this single move pushes all the i's into the numerator and leaves a plain real number underneath.

Step by step, to compute (a + i b)/(c + i d): multiply numerator and denominator by c - i d. The bottom becomes c^2 + d^2 (real and positive, as long as the denominator was nonzero). The top becomes (a + i b)(c - i d) = (ac + bd) + i (bc - ad). So the answer is (ac + bd)/(c^2 + d^2) + i (bc - ad)/(c^2 + d^2) — a clean x + i y form, with the real and imaginary parts read straight off. A small example: 1/(1 + i) = (1 - i)/((1 + i)(1 - i)) = (1 - i)/2 = 1/2 - (1/2)i.

This is exactly how you build the multiplicative inverse of any nonzero complex number: 1/z = z-bar / |z|^2. Geometrically, taking the reciprocal flips the modulus to 1/|z| and reverses the argument. Division is the last of the four basic operations, and once you have it the complex numbers are a full number system — you can compute any rational expression in z without ever leaving the plane.