Представить в алгебраической форме 2i1-i2-1+3i-2-3i3+i5-4i21-i

1-i2=1-2i+i2=1-2i-1=-2i
1-i2-1+3i=-2i-1+3i=-1+i
2i1-i2-1+3i=2i-1+i=2i-1+i∙-1-i-1-i=-2i-2i21-i2=-2i+21+1=1-i
2-3i3=23-3∙22-3i+3∙2∙-3i2--3i3=
=8+36i+54-i2+27-i3=8+36i-54-27i=46+9i
5-4i2=52-2∙5∙4i+16-i2=25-40i-16=9-40i
i5-4i2=i9-40i=9i-40i2=40-9i
2-3i3+i5-4i2=46+9i+9-40i=55-31i
2-3i3+i5-4i21-i=55-31i1-i=55-31i1-i∙1+i1+i=55+55i-31i-31i21-i2
=55+24i+311+1=86+24i2=43+12i
2i1-i2-1+3i-2-3i3+i5-4i21-i=1-i-43-12i=-42-13i
Ответ:
2i1-i2-1+3i-2-3i3+i5-4i21-i=-42-13i