If a complex number z=x+iy represents a point | vedclass

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(C) Let $z = x + iy$. Then $\frac{z-3}{z+3i} = \frac{(x-3) + iy}{x + i(y+3)}$.To simplify,multiply the numerator and denominator by the conjugate of t

Vedclass · Accepted Answer

$x-y-3=0, (x, y) 
eq (0, -3)$

Vedclass · Answer

(C) Let $z = x + iy$. Then $\frac{z-3}{z+3i} = \frac{(x-3) + iy}{x + i(y+3)}$.To simplify,multiply the numerator and denominator by the conjugate of the denominator: $x - i(y+3)$.$\frac{((x-3) + iy)(x - i(y+3))}{x^2 + (y+3)^2} = \frac{x(x-3) - i(x-3)(y+3) + ixy + y(y+3)}{x^2 + (y+3)^2}$.The imaginary part is $\frac{xy - (x-3)(y+3)}{x^2 + (y+3)^2} = \frac{xy - (xy + 3x - 3y - 9)}{x^2 + (y+3)^2} = \frac{-3x + 3y + 9}{x^2 + (y+3)^2}$.Given the imaginary part is zero,we have $-3x + 3y + 9 = 0$,which simplifies to $x - y - 3 = 0$.Since the denominator cannot be zero,$z 
eq -3i$,so $(x, y) 
eq (0, -3)$.

If a complex number $z=x+iy$ represents a point $P(x, y)$ in the Argand plane and $z$ satisfies the condition that the imaginary part of $\frac{z-3}{z+3i}$ is zero,then the locus of the point $P$ is

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