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Show that $e^{\operatorname{Log}(z)} = z$ and use this to evaluate the derivative of the function Log(z).

I have done the first part like this:

Letting $z = re^{i\theta}$,

$$ \begin{align} e^{\operatorname{Log}(z)} & = e^{\operatorname{Log}(re^i\theta)} \\ \\ & = e^{\log r + i(\theta + 2k\pi)} \\ \\ & = e^{\log(r)}e^{i(\theta + 2k\pi)} \\ \\ & = r[\cos(\theta + 2k\pi) + i\sin(\theta + 2k\pi)] \\ \\ & = r[\cos(\theta\pi) + i\sin(\theta)] \\ \\ & = re^{i\theta} = z \end{align} $$

But I can't see how I am supposed to make use of that to calculate the derivative of $\operatorname{Log}(z)$.

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Differentiate both sides of the first equation, maybe? – Javier Apr 25 '12 at 21:26
up vote 6 down vote accepted

We have $e^{\log z} = z$ from above. Differentiating both sides in respect to $z$ using chain rule, we have (knowing that $\frac{d}{dz}e^z=e^z$ and $\frac{d}{dz}z = 1$):

$$e^{\log z}\log' z = z \log'z=1 \Rightarrow \log' z = \frac{1}{z}$$

So thus we have

$$\frac{d}{dz}\log z = \frac{1}{z}$$

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Use the chain rule: $$ 1=\frac{d}{dz} z = \frac{d}{dz} e^{\operatorname{Log} z} = e^{\operatorname{Log} z} \; \operatorname{Log}' z = z\; \operatorname{Log}' z. $$ So $$ 1 = z\;\operatorname{Log}' z. $$ Divide both sides by $z$.

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