why $\frac{1}{z}$ is different from $e^{-Log(z)}$

Question

$\frac{1}{z}=z^{-1}=e^{-Logz}$,but they have different analytic region.Since $\frac{1}{z}$ is analytic on the whole complex plane except at $z=0$,but $ e^{-Logz} $is analytic on $\mathbb{C}- \{z\le 0\}$.what is problem with my thought?

Answer 1

Here's a simpler example: $$x+1 = \frac{x^2-1}{x-1} $$ Since no domain for $x$ has been specified, this is an incomplete statement: for some values of $x$ it is true, for other values it is not true.

If we try to fix this by quantifying $x$ carelessly, allowing all values for which the left hand side makes sense, we might end up with a false statement: $$\text{For all $x \in \mathbb R$}, \quad x+1 = \frac{x^2-1}{x-1} $$

We obtain a true statement only by carefully quantifying so that $x$ lies in the domain of both sides of the equation: $$\text{For all $x \in \mathbb R - \{1\}$}, \quad x+1 = \frac{x^2-1}{x-1} $$

In an ordinary precalculus or calculus course we are not generally very careful about specifying domains. So if we want our students that to prove that the derivative of $f(x)=x^2$ at $x=1$ is $f'(1)=2$ then we would not complain when they wrote the equation $\frac{x^2-1}{x-1}=x+1$.

But in a more advanced course which depends on more careful mathematical justification --- for example in a complex analysis course --- you have to be much more careful.

Answer 2

There's no problem: $e^{-\log z}$ is analytic on $\Bbb C - \{0\}$.

Just because $\log z$ is non-analytic on $\{z: z\le 0\}$, the composition with $\exp$ need not to be non-analytic.