There are lots of questions out there explaining how you can multiply two complex numbers with only three real number multiplications (instead of four) with adaptations of Karatsuba's algorithm.

eg. here and here.

But I'm trying to find a solution to squaring a complex in two multiplications and I feel like I'm missing something obvious.

This multiplies two complex numbers in three multiplications

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

So squaring seems obvious...

$$ (a+ib)^2=(a^2-b^2)+i[(a+b)^2-a^2-b^2]=(a^2-b^2)+i(2ab) $$

But that's just the identity for complex squares $(a+ib)^2=a^2+2iab-b^2$ and it uses three multiplications too (since it's based on the same algorithm for three real multiplications).

Has anyone got any suggestions for whatever connection I'm missing? I've googled for any leads but all I get is the Karatsuba algorithm explanation for two numbers in three multiplications.

The next step (I don't even know how it's going to work) is to combine both approaches and calculate $(a+ib)^2(c+id)^2$ in 5 real multiplications only.


1 Answer 1


You can compute $a^2-b^2 = (a+b)(a-b)$ in one multiplication and $2ab = (a+a)b$ in one multiplication.

Compute $(a+ib)^2(c+id)^2 = [(a+ib)(c+id)]^2$ with three multiplications for $(a+ib)(c+id)$ and then two multiplications to square the result.

  • $\begingroup$ Ah, nice. I hadn't even thought to use $(a+b)(a-b)=a^2-b^2$. Thanks $\endgroup$
    – chillman
    Commented Apr 8, 2017 at 3:28

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .