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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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 +

Vedclass · Accepted Answer

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

Vedclass · Answer

(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

Similar Questions

If the four points $A, B, C, D$ in the Argand plane represented respectively by the complex numbers $2+i, 4+3i, 2+5i, 3i$ lie on a circle,then the centre of the circle is

The locus of the point $z=x+iy$ satisfying $\left|\frac{z-2i}{z+2i}\right|=1$ is

$A$ point $z$ moves on the Argand diagram in such a way that $|z - 3i| = 2$. Then its locus will be:

The maximum distance from the origin of coordinates to the point $z$ satisfying the equation $\left| z + \frac{1}{z} \right| = a$ is

Let $z = x + iy$ be a point in the Argand plane. If the amplitude of $\left(\frac{z - 3}{z + 2i}\right)$ is $\frac{\pi}{2}$,then the locus of $z$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module