Question

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Answer 1

Răspuns:

$z=x+yi,~\frac{z+1+i}{iz+2} =\frac{x+yi+1+i}{i(x+yi)+2}=\frac{(x+1)+(y+1)i}{(2-y)+ix}=\frac{[(x+1)+(y+1)i]*[(2-y)-ix]}{](2-y)+ix]*[(2-y)-ix]}=\frac{(x+1)(2-y)+x(y+1)+[(y+1)(2-y)-x(x+1)]i}{(2-y)^{2}+x^{2}}\\Partea~imaginara~trebuie~sa~fie~0\\(y+1)(2-y)-x(x+1)=0,~-y^{2}+y+2-x^{2}-x=0,~y^{2}-y-2+x^{2}+x=0,~y^{2}-2*y*\frac{1}{2}+(\frac{1}{2} )^{2}-(\frac{1}{2} )^{2}+x^{2}+2*x*\frac{1}{2}+(\frac{1}{2} )^{2}-(\frac{1}{2} )^{2}+2=0,~(y-\frac{1}{2})^{2}+(x+\frac{1}{2})^{2}=2+\frac{1}{4}+ \frac{1}{4}=$

=2,5

Explicație pas cu pas: