1. **Problem Statement:** Show that the function $w = \overline{z}$ (the complex conjugate of $z$) is not differentiable anywhere except possibly at the origin. 2. **Recall the definition of complex differentiability:** A function $f(z) = u(x,y) + iv(x,y)$ is complex differentiable at a point if it satisfies the Cauchy-Riemann equations: $$\frac{\partial u}{\partial x} = \frac{\partial v}{\partial y} \quad \text{and} \quad \frac{\partial u}{\partial y} = -\frac{\partial v}{\partial x}$$ 3. **Express $w = \overline{z}$ in terms of $x$ and $y$:** Let $z = x + iy$, then $$w = \overline{z} = x - iy$$ So, $$u(x,y) = x, \quad v(x,y) = -y$$ 4. **Compute partial derivatives:** $$\frac{\partial u}{\partial x} = 1, \quad \frac{\partial u}{\partial y} = 0$$ $$\frac{\partial v}{\partial x} = 0, \quad \frac{\partial v}{\partial y} = -1$$ 5. **Check Cauchy-Riemann equations:** $$\frac{\partial u}{\partial x} = 1 \neq -1 = \frac{\partial v}{\partial y}$$ $$\frac{\partial u}{\partial y} = 0 \neq 0 = -\frac{\partial v}{\partial x}$$ The first equation fails everywhere except possibly where both sides equal zero. 6. **At the origin $(0,0)$:** The derivatives are constants, so the equations do not hold even at the origin. 7. **Conclusion:** The function $w = \overline{z}$ does not satisfy the Cauchy-Riemann equations anywhere except possibly at isolated points, but since the derivatives are constant and the equations fail, it is not complex differentiable anywhere except possibly at the origin, where it is still not differentiable. **Final answer:** $w = \overline{z}$ is not differentiable anywhere except possibly at the origin, but even at the origin it fails the Cauchy-Riemann conditions, so it is nowhere complex differentiable except trivially at the origin.