Complex Log

[renegade05]

z is complex

Hmm... I am having problems setting this prove up.

Can someone give me the first step? and help me out.

Thanks!

 

[HallsofIvy]

Are we to presume that "Log", without a base shown, is base "e"?

Then $\displaystyle Log(-e^{-i\tau}z)= Log(-1)+ Log(e^{-i\tau})+ Log(z)= i\pi- i\tau+ Log(z)$
Adding $\displaystyle i(\tau+ \pi)= i\tau+ i\pi$ gives
$\displaystyle Log(-e^{-i\tau}z)= 2i\pi+ Log(z)$

But "Log(z)", like most complex functions is "multivalued". Specifically, if $\displaystyle z= re^{i\theta}$, $\displaystyle \theta$ is "modulo 2\pi" so we can write $\displaystyle z= re^{i(\theta+ 2n\pi)}$ for any integer, n. Then $\displaystyle log(z)= log(r)+ i(\theta+ 2n\pi)$ so that we can add any multiple of $\displaystyle 2i\pi$ to the logarithm.

 

[stapel]

\(\displaystyle \mbox{Show that }\, \log_{\tau}(z)\, =\, \mbox{Log}\left(-e^{-i\tau}z\right)\, +\, i(\tau\, +\, \pi)\)

z is complex

What is the difference between little-L "log" and big-L "Log"?

 

[renegade05]

What is the difference between little-L "log" and big-L "Log"?

Sorry, yes log is base e firstly.

Log is the principal branch - log is all possible solutions.

Im not sure if I follow your solution..... Is it complete?