How this instrument works
Multiplying (a + bi) by (c + di) starts exactly like multiplying two ordinary binomials: distribute each term across the other and collect four partial products, ac, adi, bci, and bdi². The only complex-specific step arrives at that last term, bdi² — because i² is defined to equal −1, bdi² does not stay a separate imaginary term at all, it flips sign and joins the real part instead. Collecting what is left gives ac − bd as the real part and ad + bc as the imaginary part, the two formulas this sheet computes directly rather than making you carry four cross-terms by hand.
There is a second, purely geometric way to read the same operation, described before the algebra above was standard notation: the Norwegian-Danish surveyor Caspar Wessel wrote in 1799 that multiplying two directed lengths in a plane multiplies their distances from the origin and adds their angles, years before Jean-Robert Argand's better-known diagram popularised the same picture. Multiplying by i alone is the cleanest example — since i sits one unit from the origin at a 90° angle, multiplying any complex number by i leaves its distance from the origin unchanged and rotates it a quarter turn counterclockwise, so 3 + 4i becomes −4 + 3i, not some new unrelated pair.
One edge case worth knowing: multiply any complex number by its own conjugate, so c = a and d = −b, and the imaginary part ad + bc collapses to a(−b) + ba = 0 every time, while the real part ac − bd becomes a² − b(−b) = a² + b², always a plain non-negative real number. That single fact — a number times its conjugate is real — is exactly the mechanism the companion division sheet on this site leans on to clear a complex denominator, reused here in reverse as a built-in check: if two of the four inputs form a matched conjugate pair, a Product: imaginary part reading anything other than 0 means one of the four numbers was mistyped.
- Enter the first factor's parts into First: real part, a and First: imaginary part, b, giving the number a + bi.
- Enter the second factor's parts into Second: real part, c and Second: imaginary part, d, giving the number c + di.
- Read Product: real part for ac − bd, the real component of the result.
- Read Product: imaginary part for ad + bc, the coefficient attached to i in the result.
- To sanity-check a conjugate pair, set c = a and d = −b: Product: imaginary part should land on exactly 0.
Worked example — (3 + 4i) times (2 − 5i)
Set First: real part, a to 3 and First: imaginary part, b to 4, describing 3 + 4i; set Second: real part, c to 2 and Second: imaginary part, d to −5, describing 2 − 5i. The four cross-terms are 3 × 2 = 6, 3 × (−5i) = −15i, 4i × 2 = 8i, and 4i × (−5i) = −20i², and since i² = −1 that last term becomes +20. Collecting real and imaginary pieces separately, 6 + 20 = 26 and −15 + 8 = −7, so Product: real part reads 26 and Product: imaginary part reads −7 — exactly what the shortcut formulas ac − bd = (3×2) − (4×−5) = 6 + 20 = 26 and ad + bc = (3×−5) + (4×2) = −15 + 8 = −7 compute in one step apiece.
An independent check runs through magnitudes rather than repeating the expansion: |3 + 4i| = √(3² + 4²) = 5 and |2 − 5i| = √(2² + 5²) = √29, so the product's own magnitude should equal 5 × √29 ≈ 26.926. Reading it straight from the result, √(26² + (−7)²) = √(676 + 49) = √725 ≈ 26.926 — the same figure, confirming the real and imaginary parts landed correctly without redoing a single cross-term.
Questions
What is the formula for multiplying two complex numbers?
For (a + bi)(c + di), the real part of the product is ac − bd and the imaginary part is ad + bc. Both come from distributing every term across the other and using i² = −1 to fold the bdi² cross-term back into the real part with its sign flipped — the two formulas this sheet applies directly rather than expanding four cross-terms by hand.
Why does the real part subtract bd instead of adding it?
Because one of the four cross-terms is bdi², and i² equals −1, not +1. That turns +bdi² into −bd, a real number, which then combines with ac to give ac − bd. Treating i like an ordinary variable and skipping that sign flip is the single most common mistake in a hand-worked complex multiplication — it produces ac + bd instead, one sign off from the correct answer.
What does multiplying by i do to a complex number?
It rotates the number a quarter turn counterclockwise around the origin without changing its distance from it. Because i itself sits exactly one unit from the origin at a 90° angle, multiplying 3 + 4i by i gives −4 + 3i — same distance from the origin, five units, just turned 90°. Multiplying by i twice, i.e. by i², rotates a further 90° and lands on a straight sign flip, matching i² = −1 exactly.
How is this different from FOIL-multiplying two ordinary binomials?
The distribution step is identical to FOIL — First, Outer, Inner, Last, four products from two two-term expressions. The difference shows up only in the Last term: for ordinary binomials bd just stays bd, but for complex numbers that term carries i², which equals −1, so it flips sign and moves into the real part instead of staying separate. Everything else about the four-term expansion carries over unchanged.
Is multiplying complex numbers the same as multiplying real numbers when b and d are both 0?
Yes — with b = 0 and d = 0, both formulas collapse to ordinary real multiplication: the real part becomes ac − 0 = ac, and the imaginary part becomes 0 + 0 = 0. A complex number with no imaginary part behaves exactly like the real number it actually is, and this instrument returns the familiar product with nothing extra attached.
Why does multiplying a number by its own conjugate always give a real result?
Set c = a and d = −b, describing the conjugate of a + bi. The imaginary part ad + bc works out to a(−b) + ba = 0 every time, since the two terms are equal in size and opposite in sign. What remains is the real part alone, ac − bd = a² + b², the number's squared distance from the origin — the same fact the companion division sheet relies on to clear a complex denominator.