# Complex logarithm and derivatives

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... –  draks ... Apr 7 '12 at 11:22
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. –  Jim_CS Apr 7 '12 at 15:23
So you already have question (d) answered. –  draks ... Apr 7 '12 at 17:59
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). –  Jim_CS Apr 9 '12 at 7:51

(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