Show how to multiply the complex numbers \(a + bi\) and \(c + di\) using only three multiplications of real numbers. The algorithm should take \(a\), \(b\), \(c\), and \(d\) as input and produce the real component \(ac - bd\) and the imaginary component \(ad + bc\) separately.

#### One Solution

Calculate the following products: \(ac\), \(bd\), and \((a + b)(c + d)\).

Real component = \(ac - bd\)

Imaginary component = \((a + b)(c + d) - ac - bd = ad + bc\)

#### 2020 Hindsight

Earlier version of the solution (posted around 2015), started with the following disclaimer:

I think there is no “methodical” way to show this except for manipulating the variables. If there is, please leave a comment - I’ll add it here.

In 2020, a reader left another alternate solution in the comments section (without mentioning how it was derived), which was much more structured and less arbitrary than the one shared above.

It also made me think: there must be a logical chain of thought to come up with such a solution, rather than using algebraic manipulation based on either intuition or experience.

After reading up a bit on this, I was able to form a structured logical thought process to ** show** the work.

#### Here We Go

Note that, we have four initial products: \(ac\), \(-bd\), \(ad\), and \(bc\).

And we are trying to find three final products: \(P_1\), \(P_2\), and \(P_3\)

Let’s think of a reduced version of this: *how can we reduce 2 multiplications to just 1?*

If we have some common factor, then it can be done.

For example, \(\underline {x \cdot y} + \underline {y \cdot z} = \underline {y \cdot (x + z)}\).

Fortunately our initial products fits the bill.

Let’s take any two initial products that have one common factor and create our first final product. Say, \(ac\) and \(bc\).

\[ac + bc = \bm{a(b + c)} = \bm{P_1}\]Now, note that \(ac\) is also part of the real component we need to calculate. We can use \(P_1\) to find the real component. At this point we could also use the relation of \(bc\) and imaginary component to move forward as well.

\[\begin{aligned} \R &= ac - bd \\ &= ac - bd + bc - bc \\ &= P_1 - bd - bc \\ &= P_1 - \bm{b(c + d)} \\ &= P_1 - \bm{P_2} \end{aligned}\]At this point, we can use either \(P_1\) or \(P_2\) to construct imaginary component and find \(P_3\) similarly. Let’s use \(P_1\):

\[\begin{aligned} \Z &= ad + bc \\ &= ad + bc + ac - ac \\ &= P_1 + ad - ac \\ &= P_1 - \bm{a(c - d)} \\ &= P_1 - \bm{P_3} \end{aligned}\]So, there we go. We have multiplied two complex numbers using only three multiplications of real numbers. But note that the solution is not unique. Based on which pair of common terms we choose, we can end up with different solutions.