DPR
Took a quick look at the problem provided in class of z+2z3 . I think expanding it it is just basic algebra but I was wondering if there was an easier way maybe converting to polar with Euler's to avoid the cubic.
Here's what I tried quickly:
z+2z3(a+bi)+2(a+bi)3(a+bi)+2[(a3−3ab2)+i(3a2b−b3)](a+bi)+2a3−6ab2+6a2bi−2b3i(2a3−6ab2+a)+i(6a2b−2b3+b)