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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 ?

c1ron
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    Your mistake is assuming that $\frac{\partial r}{\partial z}$ is the reciprocal of $\frac{\partial z}{\partial r}$. That only works for functions from reals to reals. – Somos May 21 '20 at 11:17
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    $\frac{\partial a}{\partial b} = \frac{1}{\frac{\partial b}{\partial a}}$ doesn't always even hold for functions of two real variables, see, e.g., here. In the most commonly used cases, $\frac{\partial a}{\partial b} = \frac{1}{\frac{\partial b}{\partial a}}$ is going to require $a$ and $b$ to be the same dimension when viewed as real vectors (in your case $z$ is "two dimensional" while $r$ is "one dimensional") – Brevan Ellefsen May 21 '20 at 11:20
  • Thanks for the link and the explanation @BrevanEllefsen – c1ron May 21 '20 at 11:39

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