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

Question

If the conjugate of (x + iy)(1 - 2i) is (1 + i), then (A) x + iy = 1 - i (B) x + iy = (1 - i/1 - 2i)  (C) x - iy = (1 - i/1 + 2i)  (D) x + iy = (1 - i/1 + i)

Tardigrade Admin · Accepted Answer

We have, Conjugate of (x+i y)(1-2 i) is 1+i i.e. (x-i y)(1+2 i)=1+i x-i y=(1+i/1+2 i) Taking conjugate both sides, we get x+i y=(1-i/1-2 i)

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

x + iy = 1 - i

$x + i y = \frac{1 - i}{1 - 2 i}$

$x - i y = \frac{1 - i}{1 + 2 i}$

$x + i y = \frac{1 - i}{1 + i}$

We have,
Conjugate of $(x + i y) (1 - 2 i)$ is $1 + i$
i.e. $(x - i y) (1 + 2 i) = 1 + i$
$x - i y = \frac{1 + i}{1 + 2 i}$
Taking conjugate both sides, we get
$x + i y = \frac{1 - i}{1 - 2 i}$

Q. If the conjugate of (x+iy)(1−2i) is (1+i), then

x + iy = 1 - i

x+iy=1−2i1−i​

x−iy=1+2i1−i​

x+iy=1+i1−i​

We have, Conjugate of (x+iy)(1−2i) is 1+i i.e. (x−iy)(1+2i)=1+i x−iy=1+2i1+i​ Taking conjugate both sides, we get x+iy=1−2i1−i​

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

$x + i y = \frac{1 - i}{1 - 2 i}$

$x - i y = \frac{1 - i}{1 + 2 i}$

$x + i y = \frac{1 - i}{1 + i}$

We have,
Conjugate of $(x + i y) (1 - 2 i)$ is $1 + i$
i.e. $(x - i y) (1 + 2 i) = 1 + i$
$x - i y = \frac{1 + i}{1 + 2 i}$
Taking conjugate both sides, we get
$x + i y = \frac{1 - i}{1 - 2 i}$