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Q.
If the conjugate of $(x + iy) (1 - 2i)$ is $1 + i,$ then
KCETKCET 2012Complex Numbers and Quadratic Equations
Solution:
Let $z = \left(x + iy\right)\left(1 - 2i\right)$
$= x + iy - 2xi - 2i^{2}y$
$= x + iy - 2xi + 2y$
$\Rightarrow z = x + 2y + i \left(y - 2x\right)$
$\therefore \bar{z}=x+ 2y-i \left(y-2x\right)$
According to the question,
$\bar{z} = 1+i$
$\Rightarrow x + 2y - i \left(y - 2x\right) = 1 + i$
On equating the real and imaginary parts from both sides, we get
$x+2y=1 \Leftrightarrow 2x+4y=2\quad ...\left(i\right)$
and $y - 2x = - 1\quad ... \left(ii\right)$
On adding Eqs. (i) and (ii), we get
$5y = 1 \Rightarrow y = \frac{1}{5}$
$\therefore $ From Eq. (i), we get
$2x+ 4\left(\frac{1}{5}\right) = 2$
$\Rightarrow 2x=2-\frac{4}{5}$
$\Rightarrow 2x = \frac{6}{5}$
$ \Rightarrow x = \frac{3}{5}$
Taking $z = x + iy = \frac{1-i}{1-2i}\times\frac{1+2i}{1+2i}$
$= \frac{1+2i-i-2i^{2}}{1-4i^{2}}$
$z = \frac{3+i}{5} = \frac{3}{5}+i \frac{1}{5}$
$\Rightarrow z = \frac{3}{5}+ i \frac{1}{5},$ which is true.