If z=x+iy, x, y in R and the imaginary part o | vedclass

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(D) Given $z=x+iy$,then $\bar{z}=x-iy$. Substituting this into the expression: $\frac{\bar{z}-1}{\bar{z}-i} = \frac{(x-1)-iy}{x-i(y+1)}$. Multiplying

Vedclass · Accepted Answer

$x^2+y^2-x+3y+2=0, (x, y) 
eq (0, -1)$

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(D) Given $z=x+iy$,then $\bar{z}=x-iy$. Substituting this into the expression: $\frac{\bar{z}-1}{\bar{z}-i} = \frac{(x-1)-iy}{x-i(y+1)}$. Multiplying the numerator and denominator by the conjugate of the denominator $x+i(y+1)$: $\frac{[(x-1)-iy][x+i(y+1)]}{x^2+(y+1)^2} = \frac{x(x-1) + y(y+1) + i[(x-1)(y+1) - xy]}{x^2+(y+1)^2}$. The imaginary part is given as $1$: $\frac{(xy+x-y-1) - xy}{x^2+(y+1)^2} = 1$. $\frac{x-y-1}{x^2+(y+1)^2} = 1$. $x-y-1 = x^2+y^2+2y+1$. $x^2+y^2-x+3y+2=0$. Since the denominator cannot be zero,$x-i(y+1) 
eq 0$,which implies $(x, y) 
eq (0, -1)$. Thus,the locus is $x^2+y^2-x+3y+2=0, (x, y) 
eq (0, -1)$.

If $z=x+iy, x, y \in R$ and the imaginary part of $\frac{\bar{z}-1}{\bar{z}-i}$ is $1$,then the locus of $z$ is

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