hey all, i got this problem

find all the solutions z^(i)=1

so i wrote it out as e^(i(ln(1)+i@)= 1

@= theta

and i end up getting e^(-@)=1

is that right?

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- Feb 28th 2010, 07:09 PMwowsomemez^(i)=1
hey all, i got this problem

find all the solutions z^(i)=1

so i wrote it out as e^(i(ln(1)+i@)= 1

@= theta

and i end up getting e^(-@)=1

is that right? - Mar 1st 2010, 03:35 AMtonio
- Mar 1st 2010, 04:53 AMdrumist
$\displaystyle z^i = e^{i(\ln |z| +i\arg(z))} = 1$

$\displaystyle \implies e^{i \ln |z| -\arg(z)} = 1$

$\displaystyle \implies i \ln |z| - \arg(z) = 2\pi n i$

$\displaystyle \implies \ln |z| = 2\pi n \mbox{ and } \arg(z)=0$

$\displaystyle \implies z=e^{2\pi n}$

with $\displaystyle n \in \mathbb{Z}$ - Mar 1st 2010, 05:02 AMLaurent
This question also appeared in this thread.

To complete the answer I gave there (since Tonio's answer is incorrect): $\displaystyle z^i=1$ is equivalent to $\displaystyle e^{i\log z}=1$ (up to this point, the values of the left-hand sides depend on the determination of the logarithm), which is equivalent to $\displaystyle i\log z=2ik\pi$ for some $\displaystyle k\in\mathbb{Z}$, and thus to $\displaystyle \log z=2k\pi$, $\displaystyle k\in\mathbb{Z}$. Hence the solutions are the real numbers $\displaystyle e^{2k\pi}$ for every $\displaystyle k\in\mathbb{Z}$; it is worth noting that since the answer is real, the choice of the determination of the logarithm is irrelevant (and you can forget about it if you don't know what I mean).

Edit: drumist was quicker than me, but I leave my post because of the remark on the definition of logarithm - Mar 1st 2010, 11:57 AMwowsomeme
ok i get that, but how do u knoe arg is 0?