Question

AJUTORRRRR
dau coroanaa, urgent​

Answer 1

Răspuns:

Explicație pas cu pas:

b) paremai 'usor"

z^6= 2(1+i)/(-1-i)=-2

z^6=2( cosπ+isinπ)

z1,2...6=2^(1/6)* (cos∝k+isin∝k)

αk=(2kπ+π)/6 , k∈0;1;2...;5

∝1=π/6

α2=π/2

α3=5π/6

α4=7π/6

α5=3π/2

α6=11π/6

a) 2(-1/2+i√3/2) *z^4=-1-i=√2(-1/√2-i/√2)

2(cos2π/3+isin2π/3)*z^4=√2(cos5π/4+isin5π/4)

z^4=(1/√2)* (cos(5π/4-2π/3)+isin((5π/4-2π/3))

z^4=2^(-1/2)* (cos(7π/12)+isin (7π/12))

z1,2,3,4= 2^(-1/8)(cos αk+isinαk),

αk=(7π/12+2kπ)/4, k∈0;1;2;3

α1=7π/48

α2 si α3 le calculezi tu

α4= (7π/12+6π)/4=7π/48+3π/2=7π/48+72π/48=79π/48<96π/48=2π, deci e bine