Question

If the conjugate of $(x+iy)(1-2i)$ is $1+i,$ then

• A. $x-iy=\frac{1+i}{1-2i}$
• B. $x+iy=\frac{1-i}{1-2i}$
• C. $x=\frac{1}{5}$
• D. $x=-\frac{1}{5}$

Answer 1

Answer: (B)

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
$x+2y=1\iff 2x+4y=2...\left(i\right)$
and $y-2x=-1...\left(ii\right)$
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.