Question

1 - i√3 supra 1+i √3

Answer 1

[tex] \frac{1-i \sqrt{3} }{1+i \sqrt{3} }= \frac{(1-i \sqrt{3})(1-i \sqrt{3}) }{(1+i \sqrt{3})(1-i \sqrt{3}) }= \frac{1-2i \sqrt{3}-(i)^2( \sqrt{3})^2 }{1-(i \sqrt{3})^2 }= \frac{1-2 \sqrt{3}i+3 }{1+3}= \frac{4-2 \sqrt{3}i}{4}= [/tex]$\frac{2(2- \sqrt{3}i) }{4}= \frac{2- \sqrt{3}i }{2}$

Answer 2

Notam
a=1
b=i√3 si avem
(a-b)/(a+b)
Amplificam cu a+b
(a-b)*(a+b)/(a+b)²=(a²-b²)/(a²+2*a*b+b²)
Inlocuiesti pe a si b si tii cont ca i²=-1
Obtii (1-i√3)/(1+i√3)=2/(i√3-1)
Verifici calculele
Spor!