MCQ
Let point $P$ =$\alpha  + i\beta ,\alpha ,\beta  > 0\ $undergoes the following three transformations successively on argand plane

(I) Reflection about $amp\,(z)$ =$\frac{\pi }{4}$

(II) Transformation through a distance $'\beta'$ unit along the positive direction of real axis

(III) Rotation through an angle $\frac{\pi }{4}$ about origin in counter clockwise direction

If final position of the point is given by $Q= - \sqrt 2  + i\sqrt 6 $, then

  • A
    $\alpha  =  - \frac{1}{2} + \frac{{\sqrt 3 }}{2}$
  • B
    $\sqrt 3  - 1 = \beta$
  • C
    $\beta  = \frac{1}{2} + \frac{{\sqrt 3 }}{2}$
  • $\alpha  = \sqrt 3  + 1$

Answer

Correct option: D.
$\alpha  = \sqrt 3  + 1$
d
$\frac{Q}{|Q|}=\frac{(2 \beta+i \alpha) e^{i \pi / 4}}{\sqrt{(2 \beta)^{2}+\alpha^{2}}}$

$-\sqrt{2}+i \sqrt{6}=(2 \beta+i \alpha)\left(\frac{1}{\sqrt{2}}+\frac{i}{\sqrt{2}}\right)$

$\sqrt 2 \beta  - \frac{\alpha }{{\sqrt 2 }} =  - \sqrt 2 $

$\frac{\alpha }{{\sqrt 2 }} + \sqrt 2 \beta  = \sqrt 6 $

$\therefore \beta=\frac{\sqrt{3}-1}{2}, \alpha=2 \sqrt{3}-(\sqrt{3}-1)=\sqrt{3}+1$

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