Multiplying Complex Numbers

Expand as if i were a letter

To multiply 2 + 3i by 1 + 4i, multiply every term of the first bracket by every term of the second, exactly as for (2 + 3x)(1 + 4x). There are four products.

2 × 1 = 2, 2 × 4i = 8i, 3i × 1 = 3i and $3i\times 4i=12i^2$3i × 4i = 12i². So $\left(2+3i\right)\left(1+4i\right)=2+8i+3i+12i^2$(2 + 3i)(1 + 4i) = 2 + 8i + 3i + 12i².

One piece for each pair of terms: 1 × 2 = 2, 1 × 3i = 3i, 4i × 2 = 8i and $4i\times 3i=12i^2$4i × 3i = 12i². The gold piece, the product of the two imaginary parts, is the one that changes when $i^2$i² is replaced.

Where the sign flips

Up to here nothing is new. The one new rule is $i^2=-1$i² = −1, and it changes only the last product: $12i^2=12\times \left(-1\right)=-12$12i² = 12 × (−1) = −12.

So $2+8i+3i+12i^2=2+11i-12=-10+11i$2 + 8i + 3i + 12i² = 2 + 11i − 12 = −10 + 11i. The $12i^2$12i² began as a product of two imaginary parts and ends up in the real part, with its sign flipped. That is the difference from adding, where the real and imaginary parts never mix: in a product they do, because i × i is real.

The rule for any two

For (a + bi)(c + di), the four products are ac, adi, bci and bdi². The last one is bd times $i^2$i², and since $i^2=-1$i² = −1 it is −bd. Collect the real terms and the imaginary terms:

(a + bi)(c + di) = (ac − bd) + (ad + bc)i.

The real part is ac − bd, and its −bd is the product of the two imaginary parts after $i^2=-1$i² = −1. The imaginary part, ad + bc, comes from the two cross products. For (2 + 3i)(1 + 4i): ac − bd = 2 − 12 = −10 and ad + bc = 8 + 3 = 11, which gives −10 + 11i again.

The four products sorted into two columns. $3i\times 4i=12i^2=-12$3i × 4i = 12i² = −12 lands in the real column, so the real part is 2 − 12 = −10 and the imaginary part is 8 + 3 = 11.

Squares

$\left(1+i\right)\left(1+i\right)=1+i+i+i^2=1+2i-1=2i$(1 + i)(1 + i) = 1 + i + i + i² = 1 + 2i − 1 = 2i. The real parts 1 and −1 cancel, and the square of 1 + i has no real part at all.

Squaring again: ${\left(1+i\right)}^4={\left(2i\right)}^2=4i^2=-4$(1 + i)⁴ = (2i)² = 4i² = −4. And ${\left(2+i\right)}^2=4+2i+2i+i^2=4+4i-1=3+4i$(2 + i)² = 4 + 2i + 2i + i² = 4 + 4i − 1 = 3 + 4i.

A real factor, and a factor of i

A real number multiplies both parts: 3(2 − 5i) = 6 − 15i.

A factor of i swaps the parts and changes one sign: $i\left(3+2i\right)=3i+2i^2=-2+3i$i(3 + 2i) = 3i + 2i² = −2 + 3i. Multiplying by i again gives $i\left(-2+3i\right)=-2i+3i^2=-3-2i$i(−2 + 3i) = −2i + 3i² = −3 − 2i, which is −(3 + 2i). Two factors of i multiply by $i^2=-1$i² = −1, as they should.

Negative parts

$\left(3+2i\right)\left(1-5i\right)=3-15i+2i-10i^2=3-13i+10=13-13i$(3 + 2i)(1 − 5i) = 3 − 15i + 2i − 10i² = 3 − 13i + 10 = 13 − 13i. The last step is where mistakes happen: $-10i^2=-10\times \left(-1\right)=+10$−10i² = −10 × (−1) = +10.

$\left(4-i\right)\left(2+3i\right)=8+12i-2i-3i^2=8+10i+3=11+10i$(4 − i)(2 + 3i) = 8 + 12i − 2i − 3i² = 8 + 10i + 3 = 11 + 10i. Check with the rule: ac − bd = 8 − (−1)(3) = 11 and ad + bc = 12 − 2 = 10.

The usual mistakes

Taking i × i as 1. $3i\times 3i=9i^2$3i × 3i = 9i², and $i^2=-1$i² = −1, so the product is −9, not 9.

Keeping a single i. 3i × 3i is not 9i: two factors of i make $i^2$i², which is the real number −1.

Leaving $i^2$i² as it is, or as +1. (2 + 3i)(1 + 4i) is not 14 + 11i: $12i^2$12i² is −12.

Multiplying only the matching parts. (2 + 3i)(1 + 4i) is not 2 + 12i: every term of one bracket multiplies every term of the other, and the cross products 8i and 3i belong in the answer.