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Rezolvati in C: (1-i)z=1+i​

Răspuns :

Voden

Bună!

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[tex]z=a+bi[/tex]

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[tex](1-i)z=1+i <=>[/tex]

[tex]<=> (1-i)(a+bi)=1+i <=>[/tex]

[tex]<=> a+bi-ai-bi^{2} =1+i <=>[/tex]

[tex]<=> a+bi-ai+b=1+i <=>[/tex]

[tex]<=> a+b+i (b-a)=1+i =>[/tex]

[tex]=> a+b=1[/tex] [tex]si[/tex] [tex]b-a=1[/tex]

[tex]\left \{ {{a+b=1} \atop {-a+b=1}} \right.[/tex]

[tex]====(+)[/tex]

  [tex]/ 2b=2 / :2=> b=1[/tex]

[tex]a+b=1 <=> a+1=1 / -1 => a=0[/tex]

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[tex]z=i[/tex]