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

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Aicharya · Accepted Answer

$\begin{aligned} & \text {For } z=x+i y, x, y \in R,(x, y) \neq(0,-4) \ & \frac{2 z-3}{z+4 i}=\frac{(2 x-3)+2 i y}{x+i(y+4)} \times \frac{x-i(y+4)}{x-i(y+4)} \ & =\frac{\left(2 x^2-3 x+2 y^2+8 y\right)+i(2 x y-2 x y+3 y-8 x+12)}{x^2+(y+4)^2} \ & =\frac{\left(2 x^2-3 x+2 y^2+8 y\right)+i(12+3 y-8 x)}{x^2+(y+4)^2} \ & \text { So, } \arg \left(\frac{2 z-3}{z+4 i}\right)=\tan ^{-1}\left(\frac{12+3 y-8 x}{2 x^2-3 x+2 y^2+8 y}\right)=\frac{\pi}{4} \quad \text { (given) } \ & \Rightarrow \quad \frac{12+3 y-8 x}{2 x^2-3 x+2 y^2+8 y}=1 \ & \Rightarrow \quad 2 x^2+2 y^2+5 x+5 y-12=0 \end{aligned}$ Hence, option (1) is correct.

If \(z=x+i y, x, y \in R,(x, y) \neq(0,-4)\) and Arg \(\left(\frac{2 z-3}{z+4 i}\right)=\frac{\pi}{4}\),…

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