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Solution

Let $z = x + iy$ Then $\left|\dfrac{z+i}{z-1}\right| = 1 \Rightarrow |z+i| = |z-1|$ $\Rightarrow (x)^2 + (y+1)^2 = (x-1)^2 + y^2$ $\Rightarrow 2x = 1 - 2y$ $\Rightarrow x + y = \dfrac{1}{2}$ Hence the locus is a straight line.