z^(i)=1

**wowsomeme** · February 28th 2010, 07:09 PM

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?

**tonio** · March 1st 2010, 03:35 AM

Originally Posted by wowsomeme

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

I don't get it: $z^i=e^{iLn\, z}=e^{i(\ln|z|+i\arg(z))}=e^{i\ln|z|-\arg(z)}$ , so

$1=z^i=e^{i\ln|z|-\arg(z)}\Longrightarrow i\ln|z|-\arg(z)=0\Longrightarrow \ln|z|=1\,\,\,and\,\,\,\arg(z)=0\Longrightarrow z=1$

Tonio

...and that's what happens when you forget the exponential function you're dealing with is the complex one, not the real one! The above is wrong, obviously, though at least the answer given there is one of the answers...

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

is that right?

.

**drumist** · March 1st 2010, 04:53 AM

$z^i = e^{i(\ln |z| +i\arg(z))} = 1$

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

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

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

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

with $n \in \mathbb{Z}$

**Laurent** · March 1st 2010, 05:02 AM

This question also appeared in this thread.

To complete the answer I gave there (since Tonio's answer is incorrect): $z^i=1$ is equivalent to $e^{i\log z}=1$ (up to this point, the values of the left-hand sides depend on the determination of the logarithm), which is equivalent to $i\log z=2ik\pi$ for some $k\in\mathbb{Z}$ , and thus to $\log z=2k\pi$ , $k\in\mathbb{Z}$ . Hence the solutions are the real numbers $e^{2k\pi}$ for every $k\in\mathbb{Z}$ ; 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

**wowsomeme** · March 1st 2010, 11:57 AM

ok i get that, but how do u knoe arg is 0?

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