write it in the form of a+ib
(1+i)^2 (1-i)^3
Hmmmm.
(1+i)(1-i)(1+i)(1-i)*(1-i)
taking them as pairs...
the first two give (1^-i^2) =2
so we have
2*2*(1-i)
can you take it from here?
Now if you want to do it in polar, the problem is
(1.414@45)^2(1.414@-45)^3=2@90*2@-90*2) * (1-i)=4(1-i) Amazing...same answer
or