1
(1-5i)/(1+i)(2-i)³=(1-5i)(1-i)/(1+i)(1-i)=(1-i-5i-5)/(1+1)+8-12i-6+i=
=(-6i-4)/2+2-11i=-3i-2+2-11i=-14i
2 -
1)z0=1+3i,2)z0=1-3i
f(1)=-(1+3i)³³+4(1+3i)-5(1+3i)-1=-1-9i+27+i+4+24i-36-5-15i-1=-12+i
f(2)=-12-i
f(1)+f(2)=-12+i-12-i=-24
3
∛(-8i)=∛(8i³)=2i