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Solution

Let $z = x + iy$ and $|z| = 1$. The given condition means $(z - 1)$ is perpendicular to $(z + i)$. \[ (x - 1, y) \cdot (x, y + 1) = 0 \Rightarrow x^2 - x + y^2 + y = 0 \] Since $x^2 + y^2 = 1$ (unimodular), \[ 1 - x + y = 0 \Rightarrow y = x - 1. \] Substitute in $x^2 + y^2 = 1$: \[ x^2 + (x - 1)^2 = 1 \Rightarrow 2x^2 - 2x = 0 \Rightarrow x(x - 1) = 0. \] Hence $x = 0$ or $x = 1$. For $x = 0$, $z = -i$ (denominator $z+i=0$). For $x = 1$, $z = 1$ (numerator $z-1=0$). Both make $\arg$ undefined. Therefore, no unimodular complex number satisfies the condition.