Let point P = alpha + ibeta,where alpha, beta | vedclass

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(D) Let the initial point be $P = \alpha + i\beta$.$(I)$ Reflection of $z = x + iy$ about $	ext{amp}(z) = \frac{\pi}{4}$ is $z' = i\bar{z} = i(x - iy

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$\alpha = \sqrt{3} + 1$

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(D) Let the initial point be $P = \alpha + i\beta$.$(I)$ Reflection of $z = x + iy$ about $	ext{amp}(z) = \frac{\pi}{4}$ is $z' = i\bar{z} = i(x - iy) = y + ix$. Thus,$P_1 = \beta + i\alpha$.$(II)$ Shifting $P_1$ by $\beta$ units along the positive real axis gives $P_2 = (\beta + \beta) + i\alpha = 2\beta + i\alpha$.$(III)$ Rotating $P_2$ by $\frac{\pi}{4}$ counter-clockwise about the origin gives $Q = P_2 \cdot e^{i\pi/4} = (2\beta + i\alpha) \left( \frac{1}{\sqrt{2}} + i\frac{1}{\sqrt{2}} ight)$.Given $Q = -\sqrt{2} + i\sqrt{6}$,we have:$-\sqrt{2} + i\sqrt{6} = \frac{1}{\sqrt{2}} (2\beta - \alpha) + i \frac{1}{\sqrt{2}} (2\beta + \alpha)$.Equating real and imaginary parts:$2\beta - \alpha = -2$  $(1)$$2\beta + \alpha = 2\sqrt{3}$ $(2)$Adding $(1)$ and $(2)$: $4\beta = 2\sqrt{3} - 2 \implies \beta = \frac{\sqrt{3} - 1}{2}$.Subtracting $(1)$ from $(2)$: $2\alpha = 2\sqrt{3} + 2 \implies \alpha = \sqrt{3} + 1$.Thus,the correct option is $D$.

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