First the easy one: (2+0i) and (2+0i)
Looking at the imaginary parts of the sum equaling the product we have:
ad+bc=b+d
Clearly if b and d are both positive or both negative the only possibility is a=c=1
Looking at the real parts
ac  bd = a + c
becomes 1  bd = 2
bd=1 which yields the solution
(1+1i) and (11i)
If instead b and d are opposite signs
ac  bd = a+c
is only possible if ac < a+c (since bd becomes a sum)
which is only possible if a=1 or c=1
so letting c=1
ad + b = b + d
ad = d
a=1
which yields the already found solution.

Posted by Jer
on 20120829 13:57:00 