Question

If the complex number $z$ for which $arg\left(\frac{3z - 6 - 3i}{2z - 8 - 6i}\right) = \frac{\pi}{4}$ and $|z - 3 + i| = 3$ are $\left(a - \frac{4}{\sqrt{5}}\right) + i \left( 1 + \frac{2}{\sqrt{5}}\right)$ and $\left(a + \frac{4}{\sqrt{5}}\right) + i \left( 1 - \frac{2}{\sqrt{5}}\right)$ respectively, then ‘$a$’ must be equal to

• A. $5$
• B. $4$
• C. $3$
• D. $2$

Answer 1

Answer: (B)

As $arg\left(\frac{3z - 6 - 3i}{2z - 8 - 6i}\right) = \frac{\pi}{4}$
$\Rightarrow arg\left[\frac{3}{2}\left(\frac{z-2-i}{z-4-3i}\right)\right] = \frac{\pi}{4}$
$\Rightarrow arg\left(\frac{z-2-i}{z-4-3i}\right) = \frac{\pi}{4} $
$\left(\because arg \left(a + i0\right) = 0, \text{so} \,arg\left(32\right)= 0\right) $
$\Rightarrow arg\left[\frac{\left(x-2\right)+i\left(y-1\right)}{\left(x-4\right)+i\left(y-3\right)}\right] = \frac{\pi}{4} $
$\Rightarrow arg\left[\frac{\left[\left(x-2\right)+i\left(y-1\right)\right]\left[\left(x-4\right)-i\left(y-3\right)\right]}{\left(x-4\right)^{2} +\left(y-3\right)^{2}}\right] = \frac{\pi}{4}$
$\Rightarrow arg\left[\frac{\left[\left(x-2\right)\left(x-4\right)+\left(y-1\right)\left(y-3\right)\right]}{\left(x-4\right)^{2}+\left(y-3\right)^{2}} + \frac{i\left[\left(y-1\right)\left(x-4\right)-\left(x-2\right)\left(y-3\right)\right]}{\left(x-4\right)^{2} + \left(y-3\right)^{2}}\right] $
$= \frac{\pi}{4}$
$\Rightarrow tan^{-1} \left[\frac{\left(y-1\right)\left(x-4\right)-\left(x-2\right)\left(y-3\right)}{\left(x-2\right)\left(x-4\right)+\left(y-1\right)\left(y-3\right)}\right] = \frac{\pi}{4}$
$\Rightarrow (y - 1)(x - 4) - (x - 2)(y - 3) = (x - 2)( x - 4) + (y - 1)(y - 3)$
$\Rightarrow xy - x - 4y + 4 - xy + 2y + 3x - 6$
$ = x^2 + y^2 - 6x - 4y + 11$
$\Rightarrow x^2 + y^2 - 8x - 2y + 13 = 0\,\,...(i)$
Again $|z - 3 + i| = 3$
$\Rightarrow (x - 3)62 + (y + 1)^2 = 9$
$\Rightarrow x^2 + y^2 - 6x + 2y + 1 = 0\,\,....(ii)$
From (i) and (ii) we have $x = 6 - 2y$, putting this
value in (i) we get $5y^2 - 10y + 1 = 0$
$\Rightarrow y = 1 \pm \frac{2}{\sqrt{5}} $ and $x = 6 - 2y = 4 \mp \frac{4}{\sqrt{5}}$
Thus $\left( 4 \mp \frac{4}{\sqrt{5}}\right) + i\left(1 \pm \frac{2}{\sqrt{5}}\right) = \left(a \mp \frac{4}{5}\right) + i\left(1 \pm \frac{2}{\sqrt{5}}\right)$
$\Rightarrow a = 4$