Question

Rog ajutooorrrrrrrrrrr!!!!!

Answer 1

[tex]a) i^{3}=-3; i^{24}=1; i^{2002}=-1; \\ b)1+i+i^2+i^3+ i^{10}=1+i-1-i-1=-1 \\ 2)(3+2i)( -2+3i)^{-1}*(-i)+1=(3+2i) \frac{-2-3i}{13}*(-i)+1= \\ = \frac{(3+2i)(-1+2i)}{13} +1= \frac{-3+4i-4+13}{13}= \frac{6+4i}{13} \\ b)(2+i)^3+(2-i)^3=(2+i+2-i)(4+4i-1+5+4-4i-1)= \\ =4*11=44 \\ c)(1-i)(1+2i)-3(2-i)=1-i+2i+2-3+3i=4i [/tex]