Question

Am doua banalitati la care am obtinut raspuns diferit de cel de la sf. cartii. Presupun ca e greseala la sf., dar oricum... Sa se calculeze in C:

1) $\frac{i+1}{i-1} =$

2) (1+i)(2+i)(3-i)=

Answer 1

1) (i+1)(i+1)/(i+1)(i-1) = (i²+2i+1)/(i²-1²) = (-1+2i+1) / -1-1=
2i/-2= -i

2) (1+i)(2+i)(3-i)= (2+i+2i+i²)(3-i) = (2+3i-1)(3-i) = (1+3i)(3-i)=
=(3-i+9i-3i²) = (3+8i+3)=6+8i