# Breaking the Middle Term: Divide and Conquer

By Guassian

They have shown something like that:

(a + bi)(c + di) = ac − bd + (bc + ad)i-----(1)
=ac − bd + (bc + ad)i
=bc + ad = (a + b)(c + d) − ac – bd


In the above I can't understand how there are 3 multiplications because if we put the values back in (1), we would get something like:

 =ac -bd + [(a+b)(c+d) -ac -bd]i


Above looks like 5 multiplications:

I tried to go further:

= ac -bd -aci -bdi + (a+b)(c+d)i
=ac(1-i) -bd(1 + i) +(a+b)(c+d)i


I don't know if the above is correct to show that there are 3 multiplications.

Zulfi.

• You have got $ac$ and $bd$ twice ;-) Feb 11, 2020 at 15:53

The multiplies are

$$a\times c,$$

$$b\times d,$$

$$(a+b)\times(c+d).$$

Though you trade a single multiply for three adds, it is worth doing it because the computational complexity of naive multiplication ($$O(n^2)$$) is much higher than that of addition ($$O(n)$$).

Three multiplications: one in $$\color{blue}{\text{blue}}$$ and one in $$\color{red}{\text{red}}$$ and one in $$\color{green}{\text{green}}$$. The ones in $$\color{red}{\text{red}}$$ and $$\color{green}{\text{green}}$$ are written twice.

$$\color{red}{ac} -\color{green}{bd} + [\color{blue}{(a+b)(c+d)} -\color{red}{ac} -\color{green}{bd}]i$$

• Thanks a lot you solved my problem. I was not able to locate how we can have only 3 multiplications. Feb 11, 2020 at 18:59