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If log z is not defined at the origin and negative axis then does it mean that 1/z which is the the derivative of log z is not defined at negative axis and vice versa?

Heyo
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  • Yes. If the logarithm function is defined on the sheet for which $-\pi+2k\pi <\theta\le \pi+2k\pi$, then $\frac{d \log(z)}{dz}=\frac1z$, for $-\pi+2k\pi <\theta\le \pi+2k\pi$. – Mark Viola Nov 17 '16 at 22:36
  • What I mean is that logz is not continous at (-inf,0] but 1/z clearly is, why is that? – Heyo Nov 17 '16 at 23:50
  • $1/z$ is holomorphic in the punctured plane. And the integral $\int_1^z \frac{1}{w},dw=\log(z)$ when the path from $1$ to $z$ does not encircle the pole. See THIS ANSWER for details. – Mark Viola Nov 17 '16 at 23:53
  • I know the that 1/z is holomorphic on punctured plane but for example in real plane, when a function is differentiable at a point, it's continuous at at that point, like wise, 1/z is a derivative of log(z) and it's continuous on negative axis except at origin, but the function log z is not continous at (-inf,0]. Why is that? – Heyo Nov 18 '16 at 00:14
  • $1/z$ is NOT the derivative of $\log(z)$ along the branch cut.
    If a complex (or purely real function) is differentiable at a point, then it is continuous at that point. What is the contrapositive of that statement?     – Mark Viola Nov 18 '16 at 00:19
  • A real-valued function that is not differentiable might still be continuous. Take $f(x)=|x|$. It is not differentiable at $x=0$, but is everywhere continuous. Have you looked at the answer in the link I provided? – Mark Viola Nov 18 '16 at 00:30

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See THIS ANSWER for a detailed development of the evaluation of $\int_\gamma \frac1z'\,dz'$, where $\gamma$ is any rectifiable curve in $\mathbb{C} \setminus\{0\}$ from $1$ to $z$.

If the logarithm function is defined on the sheet for which $-\pi+2k\pi <\arg(z)\le \pi+2k\pi$, then

$$\log(z)=\log(|z|)+i(\arg(z))$$

and

$$\frac{d \log(z)}{dz}=\frac1z$$

for $-\pi+2k\pi <\arg(z)\le \pi+2k\pi$.

Inasmuch as $\log(z)$ is defined, but discontinuous on the branch cut, it is clearly not differentiable there.

Community
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Mark Viola
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No, it means that $1/z$ isn't the derivative of $\log z$ on the nonpositive axis. The derivative of $\log z$ isn't defined when $\log z$ isn't differentiable.

mrob
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This is usually why we have

$$\int\frac1xdx=\ln|x|+c$$

And by practical means, since it is impossible to integrate over $x=0$, this definition works fine.

Simply Beautiful Art
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