Question

(3-4i/1+2i)+(2+i^15)-(1+i/1-i)^100=

Answer 1

(3-4i/1+2i)+(2+i^15)-(1+i/1-i)^100

=（ 3- 4i*(1-2i) / ( 1+2i )( 1-2i ) ）+ ( 2 + ( i^2 ) ^7 * i ) - ( 1 / (1-i) )^100

= 3- 4i*(1-2i) / 5 + 2 + ( -1 ) ^7 * i - ( ( i+1)/2 )^100

= 3- 4i / 5 - 8/5 + 2 - i - ( i )^50

= 3- 4i / 5 - 8/5 + 2 - i + 1

= 22/5-9i/5

= ( 22-9i ) / 5