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 $arg\left(\frac{3z-6-3i}{2z-8-6i}\right)=\frac{\pi }{4}$
$\Rightarrow arg\left[\frac{3}{2}\left(\frac{z-2-i}{z-4-3i}\right)\right]=\frac{\pi }{4}$
$\Rightarrow arg\left(\frac{z-2-i}{z-4-3i}\right)=\frac{\pi }{4}$
$\left(\because arg\left(a+i0\right)=0,\text{so}arg\left(32\right)=0\right)$
$\Rightarrow arg\left[\frac{\left(x-2\right)+i\left(y-1\right)}{\left(x-4\right)+i\left(y-3\right)}\right]=\frac{\pi }{4}$
$\Rightarrow arg\left[\frac{\left[\left(x-2\right)+i\left(y-1\right)\right]\left[\left(x-4\right)-i\left(y-3\right)\right]}{{\left(x-4\right)}^2+{\left(y-3\right)}^2}\right]=\frac{\pi }{4}$
$\Rightarrow arg\left[\frac{\left[\left(x-2\right)\left(x-4\right)+\left(y-1\right)\left(y-3\right)\right]}{{\left(x-4\right)}^2+{\left(y-3\right)}^2}+\frac{i\left[\left(y-1\right)\left(x-4\right)-\left(x-2\right)\left(y-3\right)\right]}{{\left(x-4\right)}^2+{\left(y-3\right)}^2}\right]$
$=\frac{\pi }{4}$
$\Rightarrow ta{n}^{-1}\left[\frac{\left(y-1\right)\left(x-4\right)-\left(x-2\right)\left(y-3\right)}{\left(x-2\right)\left(x-4\right)+\left(y-1\right)\left(y-3\right)}\right]=\frac{\pi }{4}$
$\Rightarrow (y-1)(x-4)-(x-2)(y-3)=(x-2)(x-4)+(y-1)(y-3)$
$\Rightarrow xy-x-4y+4-xy+2y+3x-6$
$=x^2+y^2-6x-4y+11$
$\Rightarrow x^2+y^2-8x-2y+13=0...(i)$
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 $\left(4\mp \frac{4}{\sqrt{5}}\right)+i\left(1\pm \frac{2}{\sqrt{5}}\right)=\left(a\mp \frac{4}{5}\right)+i\left(1\pm \frac{2}{\sqrt{5}}\right)$
$\Rightarrow a=4$