If z=x+iy,where x, y in mathbb{R},(x, y) neq | vedclass

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(A) Given $z=x+iy$,we have $\frac{2z-3}{z+4i} = \frac{(2x-3)+2iy}{x+i(y+4)}$.Multiplying the numerator and denominator by the conjugate of the denomin

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$2x^2+2y^2+5x+5y-12=0$

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(A) Given $z=x+iy$,we have $\frac{2z-3}{z+4i} = \frac{(2x-3)+2iy}{x+i(y+4)}$.Multiplying the numerator and denominator by the conjugate of the denominator $x-i(y+4)$:$\frac{2z-3}{z+4i} = \frac{((2x-3)+2iy)(x-i(y+4))}{x^2+(y+4)^2} = \frac{(2x^2-3x+2y^2+8y) + i(2xy-2xy+3y-8x+12)}{x^2+(y+4)^2}$.Since $	ext{Arg}\left(\frac{2z-3}{z+4i}ight) = \frac{\pi}{4}$,we have $	an\left(\frac{\pi}{4}ight) = \frac{	ext{Im}}{	ext{Re}} = 1$.Thus,$\frac{12+3y-8x}{2x^2-3x+2y^2+8y} = 1$.$12+3y-8x = 2x^2-3x+2y^2+8y$.Rearranging the terms gives $2x^2+2y^2+5x+5y-12=0$.

If $z=x+iy$,where $x, y \in \mathbb{R}$,$(x, y) \neq (0, -4)$ and $\text{Arg}\left(\frac{2z-3}{z+4i}\right)=\frac{\pi}{4}$,then the locus of $z$ is

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