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

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(D) Let $z = x+iy$. The given expression is $w = \frac{(x-2) + i(y+1)}{(x+1) + i(y+2)}$.For $w$ to be purely imaginary,the real part of $w$ must be ze

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a circle not containing the point $(-1,-2)$ and having its centre on the line $x+y+1=0$

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(D) Let $z = x+iy$. The given expression is $w = \frac{(x-2) + i(y+1)}{(x+1) + i(y+2)}$.For $w$ to be purely imaginary,the real part of $w$ must be zero.Multiply the numerator and denominator by the conjugate of the denominator: $(x+1) - i(y+2)$.The real part of the numerator is $(x-2)(x+1) + (y+1)(y+2) = 0$.Expanding this,we get $x^2 - x - 2 + y^2 + 3y + 2 = 0$,which simplifies to $x^2 + y^2 - x + 3y = 0$.This represents a circle with centre $(\frac{1}{2}, -\frac{3}{2})$.The point $z = -1-2i$ makes the denominator zero,so it is excluded from the locus.The centre $(\frac{1}{2}, -\frac{3}{2})$ satisfies the line $x+y+1 = \frac{1}{2} - \frac{3}{2} + 1 = 0$.

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

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