For any complex number $w = c + id$, let $\arg(w) \in (-\pi, \pi]$, where $i = \sqrt{-1}$. Let $\alpha$ and $\beta$ be real numbers such that for all complex numbers $z = x + iy$ satisfying $\arg\left(\frac{z + \alpha}{z + \beta}\right) = \frac{\pi}{4}$, the ordered pair $(x, y)$ lies on the circle $x^{2} + y^{2} + 5x - 3y + 4 = 0$. Then which of the following statements is (are) TRUE?

$x^{2}+y^{2}+5x-3y+4=0$

$\arg\!\left(\dfrac{z+\alpha}{z+\beta}\right)=\dfrac{\pi}{4}, \qquad z=x+iy$

$\text{Find } \alpha, \beta \text{ so that this locus } = \text{ the given circle}$

$(1)\ \alpha = -1 \quad (2)\ \alpha\beta = 4 \quad (3)\ \alpha\beta = -4 \quad (4)\ \beta = 4$

$x^{2}+y^{2}+5x-3y+4=0$

$\left(x+\dfrac{5}{2}\right)^{2}+\left(y-\dfrac{3}{2}\right)^{2}=\dfrac{9}{2}$

$\text{Centre } C\!\left(-\dfrac{5}{2},\dfrac{3}{2}\right),\quad r=\dfrac{3}{\sqrt{2}}$

$\arg(z+\alpha)-\arg(z+\beta)=\dfrac{\pi}{4}$

$\angle \text{ at } z \text{ subtended by } (-\alpha,-\beta) = \dfrac{\pi}{4}$

$z=-\alpha \Rightarrow z+\alpha=0;\qquad z=-\beta \Rightarrow z+\beta=0$

$\text{So the circle passes through the real points } -\alpha,\ -\beta$

$y=0:\ x^{2}+5x+4=0 \Rightarrow (x+1)(x+4)=0$

$x=-1,\ -4$

$\{-\alpha,-\beta\}=\{-1,-4\} \Rightarrow \{\alpha,\beta\}=\{1,4\}$

$\alpha\beta = 4$

$\text{On the arc: } \arg(z+1) > \arg(z+4)$

$\Rightarrow \alpha = 1,\qquad \beta = 4$

$(1)\ \alpha=-1:\ \times;\qquad (2)\ \alpha\beta=4:\ \checkmark$

$(3)\ \alpha\beta=-4:\ \times;\qquad (4)\ \beta=4:\ \checkmark$

$\boxed{\text{Answer: (2) and (4)}}$