The reflection of the complex number (3 + 2i) | vedclass

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Question

(A) Let $z = x + iy$. The given line is $z = -i \bar{z}$.Substituting $z = x + iy$ and $\bar{z} = x - iy$:$x + iy = -i(x - iy) = -ix - i^2y = -ix + y$

Vedclass · Accepted Answer

$(-2 - 3i)$

Vedclass · Answer

(A) Let $z = x + iy$. The given line is $z = -i \bar{z}$.Substituting $z = x + iy$ and $\bar{z} = x - iy$:$x + iy = -i(x - iy) = -ix - i^2y = -ix + y$.Equating real and imaginary parts,we get $x = y$ and $y = -x$,which implies $x + y = 0$.We need to find the reflection of the point $(3, 2)$ in the line $x + y = 0$.Let the reflected point be $(\alpha, \beta)$.The midpoint of the line segment joining $(3, 2)$ and $(\alpha, \beta)$ lies on the line $x + y = 0$:$\frac{\alpha + 3}{2} + \frac{\beta + 2}{2} = 0 \Rightarrow \alpha + \beta + 5 = 0$.The slope of the line joining $(3, 2)$ and $(\alpha, \beta)$ must be perpendicular to the slope of $x + y = 0$ (which is $-1$):$\frac{\beta - 2}{\alpha - 3} = 1$ $\Rightarrow \beta - 2 = \alpha - 3$ $\Rightarrow \beta = \alpha - 1$.Substituting $\beta = \alpha - 1$ into $\alpha + \beta + 5 = 0$:$\alpha + (\alpha - 1) + 5 = 0$ $\Rightarrow 2\alpha + 4 = 0$ $\Rightarrow \alpha = -2$.Then $\beta = -2 - 1 = -3$.The reflected point is $(-2, -3)$,which corresponds to the complex number $(-2 - 3i)$.

The reflection of the complex number $(3 + 2i)$ in the straight line $z = -i \bar{z}$ is-

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