If P(x, y) denotes z = x + iy in the Argand p | vedclass

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(D) Given,$z = x + iy$ and $\left|\frac{z-1}{z+2i}\right| = 1$. Taking the modulus on both sides,we have $|z-1| = |z+2i|$. Substituting $z = x + iy$,w

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straight line

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(D) Given,$z = x + iy$ and $\left|\frac{z-1}{z+2i}\right| = 1$. Taking the modulus on both sides,we have $|z-1| = |z+2i|$. Substituting $z = x + iy$,we get $|(x-1) + iy| = |x + i(y+2)|$. Squaring both sides,we get $(x-1)^2 + y^2 = x^2 + (y+2)^2$. Expanding the squares,$x^2 - 2x + 1 + y^2 = x^2 + y^2 + 4y + 4$. Simplifying the equation,$-2x + 1 = 4y + 4$,which leads to $2x + 4y + 3 = 0$. This is the equation of a straight line.

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

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