Question

Rezolvati in C: (1-i)z=1+i​

Answer 1

Bună! ⭐

_________

$z=a+bi$

_________

$(1-i)z=1+i <=>$

$<=> (1-i)(a+bi)=1+i <=>$

$<=> a+bi-ai-bi^{2} =1+i <=>$

$<=> a+bi-ai+b=1+i <=>$

$<=> a+b+i (b-a)=1+i =>$

$=> a+b=1$ $si$ $b-a=1$

$\left \{ {{a+b=1} \atop {-a+b=1}} \right.$

$====(+)$

  $/ 2b=2 / :2=> b=1$

$a+b=1 <=> a+1=1 / -1 => a=0$

__________

$z=i$