The locus of Z=x+iy such thatIm z-3iiz+4=0 is

z-3iiz+4=x+iy-3ii(x+iy)+4=x+i(y-3)(4-y)+ix ⇒z-3iiz+4=x+i(y-3) (4-y)-ix(4-y)+ix (4-y)-ix) =x(4-x)+x(y-3)(4-y)2+x2+iy-3 (4-y)-x2(4-y)2+x2 ⇒Im z-3iiz+4=0 ⇒ (y-3) (4-y)-x2=0 ⇒4y-y2-12+3y-x2=0 ⇒ x2+y2-7y+12=0

The locus of Z=x+iy such thatIm z-3iiz+4=0 is

z-3iiz+4=x+iy-3ii(x+iy)+4=x+i(y-3)(4-y)+ix ⇒z-3iiz+4=x+i(y-3) (4-y)-ix(4-y)+ix (4-y)-ix) =x(4-x)+x(y-3)(4-y)2+x2+iy-3 (4-y)-x2(4-y)2+x2 ⇒Im z-3iiz+4=0 ⇒ (y-3) (4-y)-x2=0 ⇒4y-y2-12+3y-x2=0 ⇒ x2+y2-7y+12=0

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The locus of $Z = x + iy$ such that
Im $(\frac{z - 3 i}{iz + 4}) = 0$ is

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$x^{2} - y^{2} + 7 y - 12 = 0$

$x^{2} + y^{2} - 7 y + 12 = 0$

$x^{2} + y^{2} - 7 y + 12 = 0 and (x, y) \neq (0, 4)$

$x^{2} - y^{2} + 7 y - 12 = 0 and (x, y) \neq (0, 4)$

answer is C.

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Detailed Solution

$\frac{z - 3 i}{iz + 4} = \frac{x + iy - 3 i}{i (x + iy) + 4} = \frac{x + i (y - 3)}{(4 - y) + ix} \Rightarrow \frac{z - 3 i}{iz + 4} = \frac{(x + i (y - 3)) ((4 - y) - ix)}{((4 - y) + ix) (4 - y) - ix)} = \frac{(x (4 - x) + x (y - 3))}{{(4 - y)}^{2} + x^{2}} + i \frac{((y - 3) (4 - y) - x^{2})}{{(4 - y)}^{2} + x^{2}} \Rightarrow Im (\frac{z - 3 i}{iz + 4}) = 0 \Rightarrow (y - 3) (4 - y) - x^{2} = 0 \Rightarrow 4 y - y^{2} - 12 + 3 y - x^{2} = 0 \Rightarrow x^{2} + y^{2} - 7 y + 12 = 0$