Question:

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

Updated On: Apr 2, 2024
  • $ x-iy = \frac {1+i}{1-2i} $
  • $ x+iy = \frac {1-i}{1-2i} $
  • $x= \frac {1}{5}$
  • $x=- \frac {1}{5}$
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The Correct Option is B

Solution and Explanation

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 E (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.
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Concepts Used:

Complex Numbers and Quadratic Equations

Complex Number: Any number that is formed as a+ib is called a complex number. For example: 9+3i,7+8i are complex numbers. Here i = -1. With this we can say that i² = 1. So, for every equation which does not have a real solution we can use i = -1.

Quadratic equation: A polynomial that has two roots or is of the degree 2 is called a quadratic equation. The general form of a quadratic equation is y=ax²+bx+c. Here a≠0, b and c are the real numbers.