Réponse :
Bonsoir,
Explications étape par étape
1)
[tex]2z+i\ \overline{z}=4\\z=a+i\ b\\2(a+i\ b)+i\ (a-i\ b)=4\\2a+b+i\(2b+a)=4\\\\\left \{\begin{array}{ccc|c|c}2a+b&=&4&1&-2 \\a+2b&=&0&-2&1\\\end{array}\right. \\\\\\\left \{\begin{array}{ccc}-3b&=&4 \\-3a&=&-8\\\end{array}\right. \\\left \{\begin{array}{ccc}-3b&=&4 \\-3a&=&-8\\\end{array}\right.[/tex]
[tex]\left \{\begin{array}{ccc}b&=&-\dfrac{4}{3} \\\\a&=&\dfrac{8}{3}\\\\\end{array}\right. \\[/tex]
2)
[tex]2i\ z-\overline{z}=4i\\z=a+i\ b\\\\2i(a+i\ b)-(a-i\ b)=4i\\-2b-a+i\ (2a+b)=4i\\\\\left \{\begin{array}{ccc}-a-2b&=&0 \\2a+b&=&4\\\end{array}\right.[/tex]
[tex]\left \{\begin{array}{ccc}a&=&8 \\b&=&-4\\\end{array}\right.[/tex]
