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Numere complexe
Calculati:
1) [tex]\frac{1+i}{3-i}[/tex] ;
2) [tex]\frac{3-2i}{2+3i}[/tex] ;

Răspuns :

Maverickarcher

a) amplificam fractia cu 3+i=> (1+i)(3+i) / (9-i²) = (3+i+3i+i²)/(9+1) = (3+i+3i-1)/10 = (2+4i)/10 = [2(1+2i)]/10 = (1+2i)/5 = 1/5 + 2i/5

b) amplificam fractia cu 2-3i

=> (3-2i)(2-3i) / (4-9i²) = (6-9i-4i+6i²) / (4+9) = (6-9i-4i-6)/13 = -13i/13 = -i

M-am folosit in calcul de faptul ca i² = -1.

Mocanualexandrp2ikb6

1.  1+i /3-i =(1+i)·(3+i) /9-i² =3+i+3i+i² /9-i² =3+4i-1 /4 =2+4i /10 =1+2i /5=1/5 +2i /5 ;

2.  3-2i /2+3i =(3-2i)·(2-3i) /4-9i² =6-9i-4i+6i² /4-9i² =6-13i-6 /13 =-13i /13 =-i .

Observatie: i⁴ⁿ=1 ,i⁴ⁿ⁺¹=i ,i⁴ⁿ⁺²=-1 ,i⁴ⁿ⁺³=-i ,in cazul nostru am folosit i²=-1 .

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