Question

Reflection of the line $\bar{a}z$+$a\bar{z}$=0 in the real axis is given by

• A. $az+\bar{az}=0$
• B. $\bar{a}z-a\bar{z}=0$
• C. $az-\bar{az}=0$
• D. $\frac{a}{z}+\frac{\bar a}{z}=0$

Answer 1

Answer: (A)

$az+\bar{az}=0$

Answer 2

The correct answer is option (A): a $z$+a $\bar z$=0

let $a=\alpha+i\beta$

$z=x+iy$

Now, $\bar{a}z+a \bar{z}=0$

$\Rightarrow (\alpha-i\beta)(x+iy)+(\alpha+i\beta)(x-iy)=0$

slope=$-\frac{\alpha x}{\beta}$

So, reflection slope = $\frac{\alpha}{\beta }x$

Line is $\alpha x-\beta y=0\rightarrow$reflection also passes through origin

$\Rightarrow (\frac{a+\bar{a}}{2})(\frac{z+\bar z}{2})-(\frac{a-\bar a}{2i})(\frac{z-\bar z}{2i})=0$

$\Rightarrow az+\bar{az}=0$