Not so complex after all

Normally we would write

$(a+ib) \times (c+id)=(a \times c - b \times d )+ i(a \times d + b\times c)$

Count the $\times$ signs to see that this requires four multiplications.

But consider this alternative way to find the real and imaginary parts

$(a+b) \times c - b \times (c+d)=ac-bd$

$(a+b) \times c + a \times (d-c)=bc+ad$

At first sight this seems to require four multiplications as well (count the $\times$ signs!) - but notice that the first multiplication in the two lines is the same. You don't need to do it twice, so only three multiplications are required.

If you are working to high precision, adding and subtracting are easy compared to multiplication, and so this trick could save you about 25% of the work.