The locus of the point z = x + iy satisfying | vedclass

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(A) Given the equation $\left|\frac{z-1}{z+1}\right| = 1$. Substitute $z = x + iy$: $\left|\frac{(x-1) + iy}{(x+1) + iy}\right| = 1$ Taking the modulu

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$x = 0$

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(A) Given the equation $\left|\frac{z-1}{z+1}\right| = 1$. Substitute $z = x + iy$: $\left|\frac{(x-1) + iy}{(x+1) + iy}\right| = 1$ Taking the modulus on both sides: $|(x-1) + iy| = |(x+1) + iy|$ Squaring both sides: $(x-1)^2 + y^2 = (x+1)^2 + y^2$ Expanding the squares: $x^2 - 2x + 1 + y^2 = x^2 + 2x + 1 + y^2$ Simplifying the equation: $-2x = 2x$ $4x = 0$ $x = 0$ Thus,the locus is the imaginary axis,represented by $x = 0$.

The locus of the point $z = x + iy$ satisfying the equation $\left|\frac{z-1}{z+1}\right| = 1$ is given by:

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