The locus of z=x+iy, such that operatorname{I | vedclass

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(C) Given $z=x+iy$ and $\operatorname{Im}\left(\frac{z-3i}{iz+4}\right)=0$.Substituting $z=x+iy$, we get $\frac{x+i(y-3)}{i(x+iy)+4} = \frac{x+i(y-3)}

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$x^2+y^2-7y+12=0$ and $(x,y) 
eq (0,4)$

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(C) Given $z=x+iy$ and $\operatorname{Im}\left(\frac{z-3i}{iz+4}ight)=0$.Substituting $z=x+iy$, we get $\frac{x+i(y-3)}{i(x+iy)+4} = \frac{x+i(y-3)}{(4-y)+ix}$.To find the imaginary part, multiply the numerator and denominator by the conjugate of the denominator: $(4-y)-ix$.The denominator becomes $(4-y)^2+x^2$.The numerator becomes $[x+i(y-3)][(4-y)-ix] = x(4-y)-ix^2+i(y-3)(4-y)+x(y-3)$.Simplifying the numerator: $4x-xy-ix^2+i(-y^2+7y-12)+xy-3x = x - i(x^2+y^2-7y+12)$.For the imaginary part to be zero, the coefficient of $i$ must be zero: $-(x^2+y^2-7y+12) = 0$, which implies $x^2+y^2-7y+12=0$.Also, the denominator must not be zero, so $(4-y)^2+x^2 
eq 0$, which means $(x,y) 
eq (0,4)$.

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

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