For any complex number w = c + id,let arg ( w | vedclass

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(D) The condition $\arg \left(\frac{z+\alpha}{z+\beta}\right)=\frac{\pi}{4}$ represents an arc of a circle passing through the points $(-\alpha, 0)$ a

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$B, D$

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(D) The condition $\arg \left(\frac{z+\alpha}{z+\beta}ight)=\frac{\pi}{4}$ represents an arc of a circle passing through the points $(-\alpha, 0)$ and $(-\beta, 0)$,where the angle subtended by the chord joining these points at any point $z$ on the arc is $\frac{\pi}{4}$.Given the circle $x^2+y^2+5x-3y+4=0$,we find its intersection with the $x$-axis by setting $y=0$:$x^2+5x+4=0 \Rightarrow (x+1)(x+4)=0 \Rightarrow x=-1, x=-4$.Thus,the points $(-\alpha, 0)$ and $(-\beta, 0)$ are $(-1, 0)$ and $(-4, 0)$.This implies ${-\alpha, -\beta} = {-1, -4}$,so ${\alpha, \beta} = {1, 4}$.For $\arg \left(\frac{z+\alpha}{z+\beta}ight)=\frac{\pi}{4}$ to be positive,the order of points must be such that the rotation from the vector $(z+\beta)$ to $(z+\alpha)$ is counter-clockwise. Comparing with the standard form $\arg \left(\frac{z-z_1}{z-z_2}ight) = 	heta$,we identify $\alpha=1$ and $\beta=4$ (or vice versa depending on the arc orientation). Checking the options,$\beta=4$ and $\alpha\beta = 1 	imes 4 = 4$ are consistent with $(B)$ and $(D)$.

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