Question

The reflection of the complex number $\frac{4+3i}{1+2i}$ in the straight line $iz=$ is

• A. $2+i$
• B. $2-i$
• C. $1+2i$
• D. $1-2i$

Answer 1

Answer: (D)

We have,
$z_1=\frac{4+3i}{1+2i}=\frac{(4+3i)(1-2i)}{(1+2i)(1-2i)}$
$=\frac{10-5i}{5}=2-i$
which represents the point whose coordinates are $(2,-1)$
Also, we have,
$iz=\bar{z}$
$\Rightarrow i(x+iy)-(x-iy)=0$
$[$ Putting $z=x+iy]$
$\Rightarrow i(x+y)-(x+y)=0$
$\Rightarrow (i-1)(x+y)=0$
which represents the line $y=-x$
Hence, reflection of the point $(2,-1)$ in the line $y=-x$ gives the point $(1,-2)$ which is equivalent to $1-2i$ in the argand plane,