Explanation

(d)
$\because\ \arg{\left( \mathcal{z -}3i \right) = \arg{(x + iy - 3i) = \frac{3\pi}{4}}}$
$\Rightarrow \ \ x < 0,\ y - 3 > 0\ \ \ \left( \because\ \frac{3\pi}{4}\ is\ in\ II\ quadrant \right)$
$\frac{y - 3}{x} = \tan{\frac{3\pi}{4} = - 1}$
$\Rightarrow \ y = —x + 3\ldots(i)$
$\forall x < 0$ and $y > 3$
and $\arg{(2}\mathcal{z} + 1 - 2i) = \arg{\left\lbrack (2x + 1) + i(2y - 2) \right\rbrack = \frac{\pi}{4}}$
$\Rightarrow 2x + 1 > 0,\ 2y - 2 > 0\ \ \ \left( \because\frac{\pi}{4}\ is\ in\ I\ quadrant \right)$
$\therefore\ \ \frac{2y - 2}{2x + 1} = \tan{\frac{\pi}{4} = 1}$
$\Rightarrow \ \ 2y - 2 = 2x + 1$
$\Rightarrow \ \ y = x + \frac{3}{2},\ \forall x > \frac{- 1}{2},\ y > 1$ …(ii)
From Eqs.(i) and(ii)
[image:5_image4.emf]
It is clear from the graph that there is no point of intersection