(D) Given $\left|\frac{z-1}{z+2i}\right| = 1$. This implies $|z-1| = |z+2i|$. Substituting $z = x + iy$: $|x + iy - 1| = |x + iy + 2i|$ $|(x-1) + iy|

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(D) Given $\left|\frac{z-1}{z+2i}\right| = 1$. This implies $|z-1| = |z+2i|$. Substituting $z = x + iy$: $|x + iy - 1| = |x + iy + 2i|$ $|(x-1) + iy| = |x + i(y+2)|$ Squaring both sides: $(x-1)^2 + y^2 = x^2 + (y+2)^2$ $x^2 - 2x + 1 + y^2 = x^2 + y^2 + 4y + 4$ $-2x + 1 = 4y + 4$ $2x + 4y + 3 = 0$. This is the equation of a straight line.

Class 11 Mathematics 4-1.Complex numbers Geometry of complex numbers

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If $P(x, y)$ denotes $z = x + iy$ where $x, y \in \mathbb{R}$ and $i = \sqrt{-1}$ in the Argand plane,and $\left|\frac{z-1}{z+2i}\right| = 1$,then the locus of $P$ is

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If $P(x, y)$ denotes $z = x + iy$ where $x, y \in \mathbb{R}$ and $i = \sqrt{-1}$ in the Argand plane,and $\left|\frac{z-1}{z+2i}\right| = 1$,then the locus of $P$ is

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