Reflection of the line bar{a} z+a bar{z}=0 in | vedclass

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(C) Let $a = \alpha + i\beta$ and $z = x + iy$.The given equation is $\bar{a}z + a\bar{z} = 0$.Substituting the values, we get $(\alpha - i\beta)(x +

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

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(C) Let $a = \alpha + i\beta$ and $z = x + iy$.The given equation is $\bar{a}z + a\bar{z} = 0$.Substituting the values, we get $(\alpha - i\beta)(x + iy) + (\alpha + i\beta)(x - iy) = 0$.Expanding this, we have $(\alpha x + \beta y) + i(\alpha y - \beta x) + (\alpha x + \beta y) - i(\alpha y - \beta x) = 0$.This simplifies to $2(\alpha x + \beta y) = 0$, or $\alpha x + \beta y = 0$.This is a line passing through the origin with slope $m_1 = -\frac{\alpha}{\beta}$.Reflecting this line in the real axis ($x$-axis) changes the slope from $m_1$ to $m_2 = -m_1 = \frac{\alpha}{\beta}$.The equation of the reflected line is $y = \frac{\alpha}{\beta}x$, which can be written as $\alpha x - \beta y = 0$.Using $x = \frac{z+\bar{z}}{2}$ and $y = \frac{z-\bar{z}}{2i}$, we get $\alpha(\frac{z+\bar{z}}{2}) - \beta(\frac{z-\bar{z}}{2i}) = 0$.Multiplying by $2$, we have $\alpha(z+\bar{z}) + i\beta(z-\bar{z}) = 0$.Rearranging, $(\alpha + i\beta)z + (\alpha - i\beta)\bar{z} = 0$, which is $az + \bar{a}\bar{z} = 0$.

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

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