If a point P denotes the complex number z=x+i | vedclass

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(A) Let $z = x+iy$. The given expression is $\frac{(x-2)+i(y-1)}{(x+1)+i(y-2)}$.To make this purely real,the imaginary part of the product of the nume

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the line $x+3y-5=0$ excluding the point $(-1,2)$

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(A) Let $z = x+iy$. The given expression is $\frac{(x-2)+i(y-1)}{(x+1)+i(y-2)}$.To make this purely real,the imaginary part of the product of the numerator and the conjugate of the denominator must be zero.Multiplying by the conjugate of the denominator: $\frac{[(x-2)+i(y-1)][(x+1)-i(y-2)]}{(x+1)^2+(y-2)^2}$.The imaginary part is $(x+1)(y-1) - (x-2)(y-2) = 0$.Expanding this: $(xy - x + y - 1) - (xy - 2x - 2y + 4) = 0$.$xy - x + y - 1 - xy + 2x + 2y - 4 = 0$.$x + 3y - 5 = 0$.Since the denominator cannot be zero,$z 
eq -(1-2i)$,which means $x 
eq -1$ and $y 
eq 2$.Thus,the locus is the line $x+3y-5=0$ excluding the point $(-1,2)$.

If a point $P$ denotes the complex number $z=x+iy$ in the Argand plane and if $\frac{z-(2+i)}{z+(1-2i)}$ is purely real,then the locus of $P$ is

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