Question

Exercitiul 17 va rog

Answer 1

Scri    numaratorul    si   numitorul  sub  forma    trigonometrica

z1=1-i

lz1l=√[1²+(-1)²]=√2

cosα=1/√2=√2/2

sinα=-1/√2=-√2/2

sinusul   negativ,   cosinusul pozitiv,   esti in   cadran 4

α=2π-π/4=7π/4

z1=√2(cos7π/4+isin7π/4)

z2=√3+i

lz2l=√(√3²+1²)=√4=2

cosα=√3/2

sinα=1/2

Esti    in    cadranul 1

α=π/4

z2=2*(cosπ/4+isinπ/4)

z1/z2=√2*(cos7π/4+isin 7π/4)/2*(cosπ/4+isinπ/4)=

√2/2[cos(7π/4-π/4)+isin(7π/4-π/4)]/(cosπ/4+isinπ/4)=

√2/2*[cos(7π/4-π/4)+isin(7π/4-π/4)]=

1/√2*[cos6π/4+i sin6π/4)]=1/√2(cos3π/2+isin3π/2)

z=z1/z2

radacina    de     ordinul    6 din   z=

(1/√2)^1/6*[cos(3π/2+2kπ)/6+isin(3π/2+2kπ)/6  k=1,2,...,5

k=1

⁶√z=1/(√2)^1/6[cos(3π/12+π/3)+isin(3π/12+π/3)]=

=1/2^1/12*[cosπ/4+π/3)+isin(π/4+π/3)]=

1/2^1/12*(cos 7π/12+isin 7π/12)

Continui    tu  pt    k=2 k=3, k=4(inlocuiesti   pe    k cu    numerele      de      mai    sus    si  faci    calculele

k=5

⁶√z=1/2^1/12*(cos( 3π/2+2*5π)/6+isin(3π/2+2*5π)/6=

1/√2^1/12*(cos 3π/12+10π/6)+isin(3π/12+10π/6)=

1/√2^1/12*cos(3π+20π)/12+isin(3π+20π)/12=

1/√2^1/12*cos23π/12+isiπn23π/12