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

d. Is it true that $\log(e^z)=z$ for complex numbers $z$? Justify your answer.

I don't know how to answer these questions, I get the concepts in my head but I don't know how to write it down on paper.

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    Which concepts would like to apply? So I assume that you know about branches and so on...2012-04-07
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    I know about branches yes. I know that e^z is a many to one mapping function. And I know that log z has infinite values, whereas Log (z) will have have definite answer within the range of it's principal argument. This is exam preparation.2012-04-07
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    So you already have question (d) answered.2012-04-07
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    I understand (d) put I dont know how to formally write it out as an answer in an exams. And I dont know how to answer (c).2012-04-09

1 Answers 1

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(c) By definition, $$Log\,z:=\log|z|+i\arg z$$ where $\,\arg z\,$ is defined only up to an integer multiple of $\,2\pi\,$, thus taking into account what for the corresponding real functions happens, we have: $$e^{Log\,z}=e^{\log|z|+i\arg z}=e^{\log|z|}e^{i\arg z}=|z|e^{i\arg z}=z$$ since the expression before the last to the right above is just the polar representation of the complex number $\,z\,$ .

From here, and knowing that $\,(e^z)'=e^z\,$, we get by the chain rule:

$$e^{Log\,z}=z\Longrightarrow \left(e^{Log\,z}\right)'=(z)'\Longrightarrow (Log\,z)'e^{Log\,z}=1\Longrightarrow (Log\,z)'=\frac{1}{e^{Log\,z}}=\frac{1}{z}$$

(d) Take $\,z=0\Longrightarrow \arg 0=\arg 1=2k\pi i\,\,,\,\,k\in\Bbb Z\,$ , so $$Log\,(e^0)=Log\,1=\log|1|+i\arg 1=2k\pi i\,$$ so the above value depends on the chosen branch for the logarithm and thus the equality is not necessarily true

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    Before using the chain rule, how do you know that lnz is differentiable?2016-09-17
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    Point (c) here asks to do something to calculate $\;(\log z)'\;$ , from which I deduce it already is given. Besides this, being the inverse of a differentiable function it is differentiable, too.2016-09-17
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    @DonAntonio What do you mean by "where $arg(z)$ is defined only up to an integer multiple of $2\pi$, thus taking into account what for the corresponding real functions happens, we have..." My understanding is that, by definition, $arg(z)$ takes on values in a set $\{Arg(z) + 2\pi k\ | k \in \mathbb{Z}\}$. That says something about how corresponding (which ones) real functions behave?2017-02-10
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    from c) can we conclude that derivative of any branch of logarithm function is $1/z$ ? @DonAntonio2017-05-11
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    @AaronMartinez Yes...but observe the value of the argument of $\;z\;$ , and thus of $\;\frac1z\;$ , will depend on the chosen branch.2017-05-11
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    @DonAntonio I have this exercise, and I thought I could answer it with your answer, tell me if I can do that. "Prove the derivative of every branch of logarithm function is 1/z, using Cauchy-Riemann equations in it's polar form"2017-05-11
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    @AaronMartinez Yo creo que sí puedes hacerlo: ¿cuál es el problema? ¿ Y qué tienen que ver las ecuaciones C-R con la *derivada* de alguna función? Las ecuaciones solo nos dan una condición necesaria para analicidad.2017-05-11
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    @DonAntonio pues quien sabe que tienen que ver esas ecuaciones, jajaja mi profe. me dejo ese ejercicio: Mostrar que la derivada de cualquier rama de la función logaritmo es 1/z, utilizando las ecs. de C-R en su forma polar2017-05-11
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    @DonAntonio por cierto que bien se siente hablar en español.. :D2017-05-11