Derivative of natural logarithm of a complex variable, using polar coordinates and the chain rule. Why do I fail?

Question

Let $z=r e^{i\theta}$ be a complex variable where $r,\theta$ are polar coordinates. The natural logarithm for $\theta\in(-\pi, \pi)$ is written $$\log(z)=\ln(r)+i\theta=u(r,\theta) + iv(r,\theta)$$ so $u(r,\theta)=\ln(r)$ and $v(r,\theta)=\theta$. We know the derivative should result in $1/z$, but if we do the following: $$\frac{d}{dz}\log(z)=\frac{\partial u}{\partial r}\frac{\partial r}{\partial z}+\frac{\partial u}{\partial \theta}\frac{\partial \theta}{\partial z}+i\left(\frac{\partial v}{\partial r}\frac{\partial r}{\partial z}+\frac{\partial v}{\partial \theta}\frac{\partial \theta}{\partial z}\right)$$ The partials involving $r,\theta$ and $z$ are $$\frac{\partial z}{\partial r}=e^{i\theta},\hspace{1cm}\frac{\partial z}{\partial \theta}=ire^{i\theta}=iz$$ or $$\frac{\partial r}{\partial z} = e^{-i\theta},\hspace{1cm}\frac{\partial \theta}{\partial z}=\frac{1}{iz}$$ Then, we get $$\frac{d}{dz}\log(z)=\frac{1}{r}e^{-i\theta} +\frac{1}{z}=\frac{2}{z}$$ Where am I wrong ?