If A = {z = x + iy : text{real part of } frac | vedclass

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

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a pair of lines passing through $(-1, 1)$

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(A) Given $z = x + iy$,then $\bar{z} = x - iy$. Substituting into the expression: $\frac{\bar{z}-1}{z-i} = \frac{(x-1) - iy}{x + i(y-1)}$. To find the real part,multiply the numerator and denominator by the conjugate of the denominator: $x - i(y-1)$. $\frac{((x-1) - iy)(x - i(y-1))}{x^2 + (y-1)^2} = \frac{x(x-1) - y(y-1) + i(\dots)}{x^2 + (y-1)^2}$. The real part is $\frac{x^2 - x - y^2 + y}{x^2 + (y-1)^2} = 2$. $x^2 - x - y^2 + y = 2(x^2 + y^2 - 2y + 1)$. $x^2 + 3y^2 + x - 5y + 2 = 0$. This equation represents a pair of straight lines passing through the point $(-1, 1)$.

If $A = \{z = x + iy : \text{real part of } \frac{\bar{z}-1}{z-i} = 2\}$,then the locus of the point $P(x, y)$ in the Cartesian plane is:

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