If operatorname{Re}left(frac{z-1}{2z+i}right) | vedclass

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(B) Given $\operatorname{Re}\left(\frac{z-1}{2z+i}\right)=1.$Substitute $z=x+iy$:$\frac{z-1}{2z+i} = \frac{(x-1)+iy}{2x+i(2y+1)} = \frac{((x-1)+iy)(2x

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circle whose diameter is $\frac{\sqrt{5}}{2}$

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(B) Given $\operatorname{Re}\left(\frac{z-1}{2z+i}ight)=1.$Substitute $z=x+iy$:$\frac{z-1}{2z+i} = \frac{(x-1)+iy}{2x+i(2y+1)} = \frac{((x-1)+iy)(2x-i(2y+1))}{(2x)^2+(2y+1)^2}$The real part is $\frac{2x(x-1)+y(2y+1)}{(2x)^2+(2y+1)^2} = 1.$$2x^2-2x+2y^2+y = 4x^2+4y^2+4y+1.$$2x^2+2y^2+2x+3y+1 = 0.$$x^2+y^2+x+\frac{3}{2}y+\frac{1}{2} = 0.$This is the equation of a circle with centre $\left(-\frac{1}{2}, -\frac{3}{4}ight)$ and radius $r = \sqrt{\left(-\frac{1}{2}ight)^2 + \left(-\frac{3}{4}ight)^2 - \frac{1}{2}} = \sqrt{\frac{1}{4} + \frac{9}{16} - \frac{1}{2}} = \sqrt{\frac{4+9-8}{16}} = \frac{\sqrt{5}}{4}.$The diameter is $2r = 2 	imes \frac{\sqrt{5}}{4} = \frac{\sqrt{5}}{2}.$

If $\operatorname{Re}\left(\frac{z-1}{2z+i}\right)=1,$ where $z=x+iy,$ then the point $(x, y)$ lies on a

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