Gradient vector of u+iv=Logz

What's the gradient vector field for $u+iv= \mathrm{Log~} z$? I got:

$$\frac{\mathrm du}{\mathrm dr} = 1/r$$ $$\frac{\mathrm du}{\mathrm d\theta} = 0$$ and $$\frac{\mathrm dv}{\mathrm dr} = 0$$ $$\frac{\mathrm dv}{\mathrm d\theta} = 1$$ But the answer is $(1/r)ur$ and $(1/r)u\theta$

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Your problem is with the definition of gradient vector in orthogonal curvilinear coordinates: you need to include the scale factors. Specifically, $\nabla f = \sum_j \frac{1}{h_j} \frac{\partial f}{\partial x_j} {\bf u}_j$ where $x_j$ are the coordinates, ${\bf u}_j$ is the unit vector in the direction of increasing $x_j$, and $h_j = \left| \frac{\partial {\bf r}}{\partial x_j}\right|$. In particular for polar coordinates $(r,\theta)$, $$\nabla f = \frac{\partial f}{\partial r} {\bf u}_r + \frac{1}{r} \frac{\partial f}{\partial \theta} {\bf u}_\theta$$