Question

Determinați numerele reale a si b, pentru care 10/3+i= a+ib ,unde i^2 =-1

Answer 1

Salut,

$\dfrac{10}{3+i}=\dfrac{10\cdot (3-i)}{(3+i)(3-i)}=\dfrac{30-10i}{9-i^2}=\dfrac{30-10i}{10}=3-i,\;deci\;a=3,\;b=-1.$

Green eyes.