3a
z=(1+2i)(2+i)/(1+i)=(2+i+4i-2)/(1+i)=5i(1-i)(1+i)(1-i)=(5i+5)/(1+1)=
=2,5+2,5i
4a
2√3(2+2√3i)+4(-√3+i)-(3-3i)=4√3+12i-4√3+4i-3+3i=-3+19i
4b
(3-3i)²=9-18i-9=-18i
4c
|z1|=√(4-12)=√-8=2√2i
4d
(2+2√3i)/(3-3i)=(2+2√3i)(3+3i)/(3-3i)(3+3i)=
=(6+6i+6√3i-6√3)/(9+9)=6(1+i+√3i-√3)/18=(1+i+√3i-√3)//3