If S = {z in mathbb{C} : frac{z-i}{z+2i} in m | vedclass

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(D) Let $z = x + iy$. The condition $\frac{z-i}{z+2i} \in \mathbb{R}$ implies that the argument of the complex number is $0$ or $\pi$ (or the number i

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$S$ is a straight line in the complex plane

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(D) Let $z = x + iy$. The condition $\frac{z-i}{z+2i} \in \mathbb{R}$ implies that the argument of the complex number is $0$ or $\pi$ (or the number is undefined). This represents the locus of points $z$ such that the vectors $(z-i)$ and $(z+2i)$ are collinear. Geometrically,this is the line passing through the points $i$ (which is $(0, 1)$) and $-2i$ (which is $(0, -2)$). Since these two points lie on the imaginary axis,the line is the imaginary axis itself (excluding the point $-2i$ where the expression is undefined). Thus,$S$ is a straight line in the complex plane.

If $S = \{z \in \mathbb{C} : \frac{z-i}{z+2i} \in \mathbb{R}\}$,then:

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