Exercice 4 Ecrire sous forme algébrique les nombres complexes suivants : a=23+2i ;; b=1+i5−3ia=\dfrac{2}{3+2i} ~~~ ;; ~~~ b=\dfrac{1+i}{5-3i}a=3+2i2 ;; b=5−3i1+i Correction a=23+2i=2(3−2i)(3+2i)(3−2i)=6−4i32−(2i)2=6−4i9−(−4)=6−4i13=613−413i\begin{aligned} a&=\dfrac{2}{3+2i}=\dfrac{2(3-2i)}{(3+2i)(3-2i)}\\&=\dfrac{6-4i}{3^2-(2i)^2}=\dfrac{6-4i}{9-(-4)}\\&=\dfrac{6-4i}{13}=\dfrac{6}{13}-\dfrac{4}{13}i \end{aligned}a=3+2i2=(3+2i)(3−2i)2(3−2i)=32−(2i)26−4i=9−(−4)6−4i=136−4i=136−134i b=1+i5−3i=(1+i)(5+3i)(5−3i)(5+3i)=5+3i+5i+3i252−(3i)2=5+8i−325−(−9)=2+8i34=234+834i=117+417i\begin{aligned} b&=\dfrac{1+i}{5-3i}=\dfrac{(1+i)(5+3i)}{(5-3i)(5+3i)}\\ &=\dfrac{5+3i+5i+3i^2}{5^2-(3i)^2}\\ &=\dfrac{5+8i-3}{25-(-9)}=\dfrac{2+8i}{34}\\ &=\dfrac{2}{34}+\dfrac{8}{34}i=\dfrac{1}{17}+\dfrac{4}{17}i \end{aligned}b=5−3i1+i=(5−3i)(5+3i)(1+i)(5+3i)=52−(3i)25+3i+5i+3i2=25−(−9)5+8i−3=342+8i=342+348i=171+174i