Question

Utilizând eventual forma trigonometrica a numerelor complexe ,calculați :
a) (1-i/i)totul la 100
b) (sin t - icos t)totul la 25 (in functie de t ∈[0,2π)).
c)z totul la 11 , unde z=з+1/з² iar з este soluția complexa nereala a ecuatiei x³-1=0
DAU COROANA

Answer 1

a)
((1-i)/i)^100 = (-i-1)^100=(i+1)^100=((i+1)^2)^50=(2i)^50=2^50*i^50=2^50*(i^2)^25=2^50*(-1)^25=-2^50.

b)
(sint-i*cost)^25=(-i)^25*(cost+i*sint)^25=-i*(cost+i*sint)^25=-i*(cos(25t)+i*sin(25t))=sin(25t)-i*cos(25t).

c)
epsilon^3-1=0 <=> (epsilon-1)(epsilon^2+epsilon+1)=0 <=> epsilon^2+epsilon+1=0 <=> epsilon+1=-epsilon^2.

Deci z=(epsilon+1)/(epsilon^2)=(-epsilon^2)/(epsilon^2)=-1. => z^41=(-1)^41=-1.